NCERT Solutions Class 7 Maths

Q.1 ABCD is a quadrilateral. Is AB + BC + CD + DA>AC + BD?

Ans

\begin{array}{l}\text{Since, in a triangle, the sum of either two sides is always}\\ \text{greater than the third side}\text{.}\\ \text{So, in}\Delta \text{ABC}\\ \text{AB}+\text{BC}>\text{CA}\dots \left(\text{i}\right)\\ \text{Similary in}\Delta \text{BCD, we get}\\ \text{BC}+\text{CD}>\text{DB}\dots \left(\text{ii}\right)\\ \text{In}\Delta \text{CDA, we get}\\ \text{CD}+\text{DA}>\text{AC}\dots \left(\text{iii}\right)\\ \text{In}\Delta \text{DAB, we get}\\ \text{DA}+\text{AB}>\text{DB}\dots \left(\text{iv}\right)\\ \text{Adding (i), (ii), (iii) and (iv) to get}\\ \text{AB}+\text{BC}+\text{BC}+\text{CD}+\text{CD}+\text{DA}+\text{DA}+\text{AB}>\text{CA}+\text{DB}+\text{AC}+\text{DB}\\ \text{2}\left(\text{AB}+\text{BC}+\text{CD}+\text{DA}\right)+\text{2(AC}+\text{DB)}\\ \boxed{\text{AB}+\text{BC}+\text{CD}+\text{DA}>\text{AC+BD}}\\ \text{Therefore, the given expression is true}\text{.}\end{array}

Q.2

$\begin{array}{l}\mathrm{ABCD}\mathrm{is}\mathrm{a}\mathrm{quadrilateral}.\mathrm{Is}\\ \mathrm{AB}+\mathrm{BC}+\mathrm{CD}+\mathrm{DA}<2\left(\mathrm{AC}+\mathrm{BD}\right)?\end{array}$

Ans

Consider a quadrilateral ABCD.

Since, in a triangle, the sum of either two sides is always greater than the third side. So, in ΔOAB OA+OB>AB …( i )
Similary in ΔOBC, we get OC+OB>BC …( ii )
In ΔOCD, we get OD+OC>CD …( iii )
In ΔODA, we get OA+OD>DA …( iv )
Adding (i), (ii), (iii) and (iv) to get OA+OB+OC+OB+OD+OC+OA+OD>AB+BC+CD+DA
2( OA+OB+OC+OD )>2( AC+BD ) ( OA+OB+OC+OD )>( AC+BD )
Therefore, the given expression is true.

Q.3

$\begin{array}{l}\mathrm{The}\mathrm{lengths}\mathrm{of}\mathrm{two}\mathrm{sides}\mathrm{of}\mathrm{a}\mathrm{triangle}\mathrm{are}12\mathrm{cm}\mathrm{and}15\mathrm{cm}.\\ \mathrm{Between}\mathrm{what}\mathrm{two}\mathrm{measures}\mathrm{should}\mathrm{the}\mathrm{length}\mathrm{of}\mathrm{the}\mathrm{third}\\ \mathrm{side}\mathrm{fall}?\end{array}$

Ans

\begin{array}{l}\text{In a triangle, the sum of either two sides is always greater}\\ \text{than the third side and also, the difference of the lengths of}\\ \text{either two sides is always lesser than the third side}\text{.}\\ \text{Here, the third side will be lesser than the sum of these two}\\ \text{(i}\text{.e}\text{.,}12+15=27\text{) and also, it will be greater than the difference}\\ \text{of these two (i}\text{.e}\text{.,}15-12=3\text{)}\text{.}\\ \text{Therefore, those two measures are 27 cm and 3 cm}\text{.}\end{array}

Q.4

$\begin{array}{l}\text{PQR is a triangle right angled at P. If PQ =10 cm}\\ \text{and PR=24cm,findQR.}\end{array}$

Ans

By applying Pythagoras theorem in ΔPQR to get

(PQ)² + (PR)² = (QR)²
10² + 24² = QR²
100 + 576 = QR²
676 = QR²
QR =

$\sqrt{676}$

= 26 cm

Q.5

$\begin{array}{l}\mathrm{ABC}\mathrm{is}\mathrm{a}\mathrm{triangle}\mathrm{right}\mathrm{angled}\mathrm{at}\mathrm{C}.\mathrm{If}\mathrm{AB}=\; 25\mathrm{cm}\mathrm{and}\\ \mathrm{AC}=\; 7\mathrm{cm},\mathrm{find}\mathrm{BC}.\end{array}$

Ans

\begin{array}{l}\text{By Pythagoras theorem in}\Delta \text{ABC, we get}\\ {\left(\text{AC}\right)}^2+{\left(BC\right)}^2={\left(AB\right)}^2\\ {\left(BC\right)}^2={\left(AB\right)}^2-{\left(AC\right)}^2\\ ={24}^2-7^2\\ =625-49=576\\ \boxed{\text{BC}=\text{24}\text{cm}}\end{array}

Q.6

$\begin{array}{l}\mathrm{A}15\mathrm{m}\mathrm{long}\mathrm{ladder}\mathrm{reached}\mathrm{a}\mathrm{window}12\mathrm{m}\mathrm{high}\mathrm{from}\mathrm{the}\\ \mathrm{ground}\mathrm{on}\mathrm{placing}\mathrm{it}\mathrm{against}\mathrm{a}\mathrm{wall}\mathrm{at}\mathrm{a}\mathrm{distance}‘\mathrm{a}‘.\mathrm{Find}\mathrm{the}\\ \mathrm{distance}\mathrm{of}\mathrm{the}\mathrm{foot}\mathrm{of}\mathrm{the}\mathrm{ladder}\mathrm{from}\mathrm{the}\mathrm{wall}.\end{array}$

Ans

\begin{array}{l}\text{By Pythagoras theorem}\\ {\left(15\right)}^2={\left(12\right)}^2+a^2\\ 225-144=a^2\\ 81=a^2\\ \boxed{a=9\text{cm}}\\ \text{Therefore, the distance of the foot of the ladder from the wall}\\ \text{is 9 cm}\text{.}\end{array}

Q.7

$\begin{array}{l}\mathrm{Which}\mathrm{of}\mathrm{the}\mathrm{following}\mathrm{can}\mathrm{be}\mathrm{the}\mathrm{sides}\mathrm{of}\mathrm{a}\mathrm{right}\mathrm{triangle}?\\ \left(\mathrm{i}\right)\mathrm{}2.5\mathrm{cm},6.5\mathrm{cm},6\mathrm{cm}.\\ \left(\mathrm{ii}\right)\mathrm{}2\mathrm{cm},2\mathrm{cm},5\mathrm{cm}.\\ \left(\mathrm{iii}\right)\mathrm{}1.5\mathrm{cm},2\mathrm{cm},2.5\mathrm{cm}\end{array}$

Ans

\begin{array}{l}\left(\text{i}\right)\text{2}.\text{5 cm},\text{6}.\text{5 cm},\text{6 cm}.\\ {2.5}^2=6.25,{6.5}^2=42.25{\text{and 6}}^2=36\\ \text{Here, 2}{\text{.5}}^2+6^2={6.5}^2\\ \text{So, the square of the length of one side is the sum of the}\\ \text{squares of the lengths of remaining two sides}\text{.}\\ \text{Hence, these are the sides of a right}-\text{angled triangle}\text{.}\\ \left(\text{ii}\right)\text{2 cm},\text{2 cm},\text{5 cm}.\\ 2^2=4,2^2=4{\text{and 5}}^2=25\\ {\text{Here, 2}}^2+2^2\ne 5^2\\ \text{So, the square of the length of one side is not the sum of the}\\ \text{squares of the lengths of remaining two sides}\text{.}\\ \text{Hence, these are not the sides of a right}-\text{angled triangle}\text{.}\\ \left(\text{iii}\right)\text{1}.\text{5 cm},\text{2cm},\text{2}.\text{5 cm}\\ {1.5}^2=2.25,2^2=4\text{and 2}{\text{.5}}^2=6.25\\ \text{Here, 1}{\text{.5}}^2+2^2={2.5}^2\\ \text{So, the square of the length of one side is the sum of the}\\ \text{squares of the lengths of remaining two sides}\text{.}\\ \text{Hence, these are the sides of a right}-\text{angled triangle}\text{.}\end{array}

Q.8

$\begin{array}{l}\text{A tree is broken at a height of 5 m from the ground and its}\\ \text{top touches the ground at a distance of 12 m from the}\\ \text{Base of the tree.Find the original height of the tree.}\end{array}$

Ans

\begin{array}{l}\text{In above figure, BC represents the unbroken part of the tree}\text{.}\\ \text{Point C represents the point where the tree broke and CA}\\ \text{represents the broken part of the tree}\text{.}\\ \text{Triangle ABC thus formed is a right-angled triangle}\text{.}\\ \text{So, applying Pythagoras theorewm to get}\\ {\text{AC}}^2={\text{AB}}^2+{\text{BC}}^2\\ ={12}^2+5^2\\ =144+25\\ =169=13\text{cm}\\ \text{Thus,the original height of the tree is}=\text{AC}+\text{CB}\\ =\text{13 cm}+\text{5 cm}\\ =\boxed{\text{18 cm}}\end{array}

Q.9

$\begin{array}{l}\mathrm{Angles}\mathrm{Q}\mathrm{and}\mathrm{R}\mathrm{of}\mathrm{a}\mathrm{\Delta PQR}\mathrm{are}25º\mathrm{and}65º.\\ \mathrm{Write}\mathrm{which}\mathrm{of}\mathrm{the}\mathrm{following}\mathrm{is}\mathrm{true}:\\ \left(\mathrm{i}\right){\mathrm{PQ}}^2+{\mathrm{QR}}^2={\mathrm{RP}}^2\\ \left(\mathrm{ii}\right){\mathrm{PQ}}^2+{\mathrm{RP}}^2={\mathrm{QR}}^2\\ \left(\mathrm{iii}\right){\mathrm{RP}}^2+{\mathrm{QR}}^2={\mathrm{PQ}}^2\end{array}$

Ans

\begin{array}{l}\text{Since, the sum of all interior angle is 180}°\text{.}\\ \text{So, 25}°\text{+ 65}°+\angle \text{QPR}=180°\\ \angle \text{QPR}=180°-90°\\ =90°\\ \text{Therefore,}\Delta \text{PQR is a rigfht angled triangle at P}\text{.}\\ \text{Thus,}\boxed{{\text{PQ}}^{\text{2}}+{\text{RP}}^{\text{2}}={\text{QR}}^{\text{2}}}\\ \text{So, (ii) is true}\text{.}\end{array}

Q.10

$\begin{array}{l}\mathrm{Find}\mathrm{the}\mathrm{perimeter}\mathrm{of}\mathrm{the}\mathrm{rectangle}\mathrm{whose}\mathrm{length}\mathrm{is}40\mathrm{cm}\\ \mathrm{and}\mathrm{a}\mathrm{diagonal}\mathrm{is}41\mathrm{cm}.\end{array}$

Ans

\begin{array}{l}\text{In a rectangle},\text{all interior angles are of 9}0º\text{measure}.\\ \text{Therefore},\text{Pythagoras theorem can be applied here}.\\ {\left(\text{41}\right)}^{\text{2}}={\left(\text{4}0\right)}^{\text{2}}+x^{\text{2}}\\ \text{1681}=\text{16}00+x^{\text{2}}\\ x^{\text{2}}=\text{1681}-\text{16}00\\ =\text{81}\\ x=\text{9 cm}\\ \text{So, Perimeter}=\text{2}\left(\text{Length}+\text{Breadth}\right)\\ =\text{2}\left(x+\text{4}0\right)\\ =\text{2}\left(\text{9}+\text{4}0\right)\\ =\boxed{\text{98 cm}}\end{array}

Q.11

$\begin{array}{l}\mathrm{The}\mathrm{diagonals}\mathrm{of}\mathrm{a}\mathrm{rhombus}\mathrm{measure}16\mathrm{cm}\mathrm{and}30\mathrm{cm}.\\ \mathrm{Find}\mathrm{its}\mathrm{perimeter}.\end{array}$

Ans

\begin{array}{l}\text{Let ABCD be a rhombus}\left(\text{all sides are of equal length}\right)\text{and its}\\ \text{diagonals},\text{AC and BD},\text{are intersecting each other at point O}.\\ \text{Since, diagonals of a rhombus bisect each other at 9}0°.\\ \text{So,}\\ \text{By applying Pythagoras theorem in}\Delta \text{AOB},\\ {\text{OA}}^{\text{2}}+{\text{OB}}^{\text{2}}={\text{AB}}^{\text{2}}\\ {\text{8}}^{\text{2}}+{\text{15}}^{\text{2}}={\text{AB}}^{\text{2}}\\ \text{64}+\text{225}={\text{AB}}^{\text{2}}\\ \text{289}={\text{AB}}^{\text{2}}\\ \text{AB}=\text{17 cm}\\ \text{Therefore},\text{the length of the side of rhombus is 17 cm}.\\ \text{Perimeter of rhombus}=\text{4}\times \text{Side of the rhombus}\\ =\text{4}\times \text{17}\\ =\boxed{\text{68 cm}}\end{array}

Q.12

$\begin{array}{l}\mathrm{Complete}\mathrm{the}\mathrm{following}\mathrm{statements}:\\ \mathrm{a})\mathrm{Two}\mathrm{line}\mathrm{segments}\mathrm{are}\mathrm{congruent}\mathrm{if}\_\_\_\_\_\_\_\_.\\ \mathrm{b})\mathrm{Among}\mathrm{two}\mathrm{congruent}\mathrm{angles},\mathrm{one}\mathrm{has}\mathrm{a}\mathrm{measure}\mathrm{of}70°;\mathrm{the}\mathrm{measure}\mathrm{of}\mathrm{the}\mathrm{other}\mathrm{angle}\mathrm{is}\_\_\_\_\_\_\_\_\_.\\ \mathrm{c})\mathrm{When}\mathrm{we}\mathrm{write}\angle \mathrm{A}=\angle \mathrm{B},\mathrm{we}\mathrm{actually}\mathrm{mean}\_\_\_\_\_\_\_\_\_.\end{array}$

Ans

$\begin{array}{l}\text{Complete the following statements}:\\ \left(\text{a}\right)\text{Two line segments are congruent if}\\ \underset{\_}{\text{they have the same length}}.\\ \left(\text{b}\right)\text{Among two congruent angles},\text{one has a measure of 7}0°;\\ \text{the measure of the other angle is}\underset{\_}{\text{7}0°}.\\ \left(\text{c}\right)\text{When we write}\angle \text{A}=\angle \text{B},\text{we actually mean}\underset{\_}{\text{m}\angle \text{A}=\text{m}\angle \text{B}}.\end{array}$

Q.13

$\mathrm{Give}\mathrm{any}\mathrm{two}\mathrm{real}–\mathrm{life}\mathrm{examples}\mathrm{for}\mathrm{congruent}\mathrm{shapes}.$

Ans

$\begin{array}{l}\left(\text{i}\right)\text{Sheets of same letter pad}\\ \left(\text{ii}\right)\text{Biscuits in the same packet}\end{array}$

Q.14

$\begin{array}{l}\mathrm{If}\mathrm{\Delta ABC}\cong \mathrm{\Delta FED}\mathrm{under}\mathrm{the}\mathrm{correspondence}\mathrm{ABC}\leftrightarrow \mathrm{FED},\\ \mathrm{write}\mathrm{all}\mathrm{the}\mathrm{corresponding}\mathrm{congruent}\mathrm{parts}\mathrm{of}\mathrm{the}\mathrm{triangles}.\end{array}$

Ans

$\begin{array}{l}\text{If these triangles are congruent},\text{then the corresponding angles}\\ \text{and sides will be equal to each other}.\\ \text{So,}\\ \angle \text{A}\leftrightarrow \angle \text{F}\\ \angle \text{B}\leftrightarrow \angle \text{E}\\ \angle \text{C}\leftrightarrow \angle \text{D}\\ \overline{\text{AB}}\leftrightarrow \overline{\text{FE}}\\ \overline{\mathrm{BC}}\leftrightarrow \overline{\text{ED}}\\ \overline{\text{CA}}\leftrightarrow \overline{\text{DF}}\end{array}$

Q.15

$\begin{array}{l}\mathrm{If}\mathrm{\Delta DEF}\cong \mathrm{\Delta BCA},\mathrm{write}\mathrm{the}\mathrm{part}\left(\mathrm{s}\right)\mathrm{of}\mathrm{\Delta BCA}\mathrm{that}\\ \mathrm{correspond}\mathrm{to}\\ \left(\mathrm{i}\right)\angle \mathrm{E}\left(\mathrm{ii}\right)\overline{\mathrm{EF}}\left(\mathrm{iii}\right)\angle \mathrm{F}\left(\mathrm{iv}\right)\overline{\mathrm{DF}}\end{array}$

Ans

$\begin{array}{l}\text{(i})\angle \text{C}\\ \left(\text{ii}\right)\overline{\text{CA}}\\ \left(\text{iii}\right)\angle \text{A}\\ \left(\text{iv}\right)\overline{\text{BA}}\end{array}$

Q.16

$\begin{array}{l}\mathrm{Which}\mathrm{congruence}\mathrm{criterion}\mathrm{do}\mathrm{you}\mathrm{use}\mathrm{in}\mathrm{the}\mathrm{following}?\\ \left(\mathrm{a}\right)\mathrm{Given}:\mathrm{AC}=\mathrm{DF}\\ \mathrm{AB}=\mathrm{DE}\\ \mathrm{BC}=\mathrm{EF}\\ \end{array}$

So,ΔABC≅ΔDEF

$\begin{array}{l}\left(\mathrm{b}\right)\mathrm{Given}:\mathrm{ZX}=\mathrm{RP}\\ \mathrm{RQ}=\mathrm{ZY}\\ \angle \mathrm{PRQ}=\mathrm{XZY}\\ \mathrm{So},\mathrm{\Delta PQR}\cong \mathrm{\Delta XYZ}\end{array}$

$\begin{array}{l}\left(\mathrm{c}\right)\mathrm{Given}:\angle \mathrm{MLN}=\angle \mathrm{FGH}\\ \angle \mathrm{NML}=\angle \mathrm{GFH}\\ \mathrm{ML}=\mathrm{FG}\\ \mathrm{So},\mathrm{\Delta LMN}\cong \mathrm{\Delta GFH}\end{array}$

$\begin{array}{l}\left(\mathrm{d}\right)\mathrm{Given}:\mathrm{EB}=\mathrm{DB}\\ \mathrm{AE}=\mathrm{BC}\\ \angle \mathrm{A}=\angle \mathrm{C}=90°\\ \mathrm{So},\mathrm{\Delta ABE}\cong \mathrm{\Delta CDB}\end{array}$

Ans

$\begin{array}{l}\left(\text{a}\right)\boxed{\text{SSS}},\text{as the sides of}\mathrm{\Delta }\text{ABC are equal to the sides of}\mathrm{\Delta }\text{DEF}.\\ \left(\text{b}\right)\boxed{\text{SAS}},\text{as two sides and the angle included between these}\\ \text{sides of}\mathrm{\Delta }\text{PQR are equal to two sides and the angle}\\ \text{included between these sides of}\mathrm{\Delta }\text{XYZ}.\\ \left(\text{c}\right)\boxed{\text{ASA}},\text{as two angles and the side included between these}\\ \text{angles of}\mathrm{\Delta }\text{LMN are equal to two angles and the side}\\ \text{included between these angles of}\mathrm{\Delta }\text{GFH}.\\ \left(\text{d}\right)\boxed{\text{RHS}},\text{as in the given two right-angled triangles},\text{one side}\\ \text{and the hypotenuse are respectively equal.}\end{array}$

Q.17

$\begin{array}{l}\text{You want to show that ΔART@ΔPEN,}\\ \left(\text{a}\right)\text{If you have to use SSS criterion, then you need to show}\\ \left(\text{i}\right)\text{AR =}\left(\text{ii}\right)\text{RT =}\left(\text{iii}\right)\text{AT =}\\ \left(\text{b}\right)\text{If it is given that ∠T =}\angle \text{N and you are to use SAS criterion,}\\ \text{you need to have}\\ \left(\text{i}\right)\text{RT = and}\left(\text{ii}\right)\text{PN =}\\ \left(\text{c}\right)\text{If it is given that AT=PN and you are to use ASA criterion,}\\ \text{you need to have}\\ \left(\text{i}\right)\text{?}\left(\text{ii}\right)\text{?}\end{array}$

Ans

$\begin{array}{l}\left(\text{a}\right)\\ \left(\text{i}\right)\text{AR}=\text{PE}\left(\text{ii}\right)\text{RT}=\text{EN}\left(\text{iii}\right)\text{AT}=\text{PN}\\ \left(\text{b}\right)\\ \left(\text{i}\right)\text{RT}=\text{EN}\left(\text{ii}\right)\text{PN}=\text{AT}\\ \left(\text{c}\right)\\ \left(\text{i}\right)\angle \text{ATR}=\angle \text{PNE}\left(\text{ii}\right)\angle \text{RAT}=\angle \text{EPN}\end{array}$

Q.18

$\mathrm{You}\mathrm{have}\mathrm{to}\mathrm{show}\mathrm{that}\angle \mathrm{AMP}\cong \angle \mathrm{AMQ}.\mathrm{In}\mathrm{the}\mathrm{following}$

proof,supply the missing reasons.

Steps	Reasons
(i) PM = QM	(i)…
\left(ii\right)\angle PMA=\angle QMA	(ii)…
(iii) AM = AM	(iii)…
\left(iv\right)\Delta AMP\cong \Delta AMQ	(iv)…

Ans

Steps	Reasons
(i) PM = QM	(i)…Given
$\left(ii\right)\angle PMA=\angle QMA$	(ii)…Given
(iii) AM = AM	(iii)…Common
$\left(\mathrm{iv}\right)\mathrm{\Delta AMP}\cong \mathrm{\Delta AMQ}$	$\begin{array}{l}\left(\text{iv}\right)\text{SAS},\text{as the two sides and the angle}\\ \text{included between these sides of}\mathrm{\Delta }\text{AMP}\\ \text{are equal to two sides and the angle}\\ \text{included between these sides of}\mathrm{\Delta }\text{AMQ}\end{array}$

Q.19

$\begin{array}{l}\text{In}\mathrm{\Delta }\text{ABC},\angle \text{A}=\text{3}0°,\angle \text{B}=\text{4}0°\text{and}\angle \text{C}=\text{11}0°\\ \text{In}\mathrm{\Delta }\text{PQR},\angle \text{P}=\text{3}0°,\angle \text{Q}=\text{4}0°\text{and}\angle \text{R}=\text{11}0°\\ \text{A student says that}\mathrm{\Delta }\text{ABC}\cong \mathrm{\Delta }\text{PQR by AAA congruence}\\ \text{criterion}.\text{Is he justified}?\text{Why or why not?}\end{array}$

Ans

$\begin{array}{l}\text{No}.\text{This property represents that these triangles have their}\\ \text{respective angles of equal measure}.\text{However},\text{this gives no}\\ \text{information about their sides}.\text{The sides of these triangles have}\\ \text{a ratio somewhat different than 1}:\text{1}.\\ \text{Therefore},\text{AAA property does not prove the two}\\ \text{triangles are congruent.}\end{array}$

Q.20

$\begin{array}{l}\mathrm{In}\mathrm{the}\mathrm{figure},\mathrm{the}\mathrm{two}\mathrm{triangles}\mathrm{are}\mathrm{congruent}.\\ \mathrm{The}\mathrm{corresponding}\mathrm{parts}\mathrm{are}\mathrm{marked}.\\ \end{array}$

We can write ΔRAT≅?

Ans

$\begin{array}{l}\text{From the figure, we observe that:}\\ \angle \text{RAT}=\angle \text{WON}\\ \angle \text{ART}=\angle \text{OWN}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}AR}=\text{OW}\\ \text{Therefore},\mathrm{\Delta }\text{RAT}\cong \mathrm{\Delta }\text{WON},\text{by ASA criterion.}\end{array}$

Q.21

$\mathrm{Complete}\mathrm{the}\mathrm{congruence}\mathrm{statement}:$

Ans

$\begin{array}{l}\text{Given that},\\ \text{BC}=\text{BT}\\ \text{TA}=\text{CA}\\ \text{BA is common}.\\ \text{Therefore},\mathrm{\Delta }\text{BCA}\cong \mathrm{\Delta }\text{BTA}\\ \text{Similarly},\\ \text{PQ}=\text{RS}\\ \text{TQ}=\text{QS}\\ \text{PT}=\text{RQ}\\ \text{Therefore},\mathrm{\Delta }\text{QRS}\cong \mathrm{\Delta }\text{TPQ.}\end{array}$

Q.22

$\begin{array}{l}\mathrm{In}\mathrm{a}\mathrm{squared}\mathrm{sheet},\mathrm{draw}\mathrm{two}\mathrm{triangles}\mathrm{of}\mathrm{equalareas}\\ \mathrm{such}\mathrm{that}\\ \left(\mathrm{i}\right)\mathrm{the}\mathrm{triangles}\mathrm{are}\mathrm{congruent}.\\ \left(\mathrm{ii}\right)\mathrm{the}\mathrm{triangles}\mathrm{are}\mathrm{not}\mathrm{congruent}.\\ \left(\mathrm{iii}\right)\mathrm{What}\mathrm{can}\mathrm{you}\mathrm{say}\mathrm{about}\mathrm{the}\mathrm{irperimeters}?\end{array}$

Ans

(i)

Here,ΔABC and ΔPQR have the same area and are congruent to each other also. Also, the perimeter of both the triangles
will be the same.
(ii)

$\begin{array}{l}\text{Here},\text{the two triangles have the same height and base}.\\ \text{Thus},\text{their areas are equal}.\\ \text{However},\text{these triangles are not congruent to each other}.\\ \text{Also},\text{the perimeter of both the triangles will not be the same.}\end{array}$

Q.23

$\begin{array}{l}\mathrm{Draw}\mathrm{a}\mathrm{rough}\mathrm{sketch}\mathrm{of}\mathrm{two}\mathrm{triangles}\mathrm{such}\mathrm{that}\mathrm{they}\mathrm{have}\\ \mathrm{five}\mathrm{pairs}\mathrm{of}\mathrm{congruent}\mathrm{parts}\mathrm{but}\mathrm{still}\mathrm{the}\mathrm{triangles}\mathrm{are}\mathrm{not}\\ \mathrm{congruent}.\end{array}$

Ans

$\begin{array}{l}\text{A pair of triangles with threee equal angles, and two equal}\\ \text{sides are non congruent.}\end{array}$

Below is the example:

Q.24

$\begin{array}{l}\mathrm{If}\mathrm{\Delta ABC}\mathrm{and}\mathrm{\Delta PQR}\mathrm{are}\mathrm{to}\mathrm{be}\mathrm{congruent},\mathrm{name}\mathrm{one}\mathrm{additional}\\ \mathrm{pair}\mathrm{of}\mathrm{corresponding}\mathrm{parts}.\mathrm{What}\mathrm{criterion}\mathrm{did}\mathrm{you}\mathrm{use}?\end{array}$

Ans

$\begin{array}{l}\text{Here,}\\ \text{BC}=\text{QR}\\ \text{So,}\mathrm{\Delta }\text{ABC}\cong \mathrm{\Delta }\text{PQR}\left(\text{ASA criterion}\right)\end{array}$

Q.25

$\begin{array}{l}\mathrm{Explain},\mathrm{why}\\ \mathrm{\Delta ABC}\cong \mathrm{\Delta FED}\end{array}$

Ans

\begin{array}{l}\text{Given that},\\ \angle \text{ABC}=\angle \text{FED}\dots \dots \dots \dots \left(\text{1}\right)\\ \angle \text{BAC}=\angle \text{EFD}\dots \dots \dots \dots (\text{2})\\ \text{The two angles of}\Delta \text{ABC are equal to the two respective}\\ \text{angles of}\Delta \text{FED}.\text{Also},\text{the sum of all interior angles of a}\\ \text{triangle is 18}0º.\text{Therefore},\text{third angle of both triangles will}\\ \text{also be equal in measure}.\\ \angle \text{BCA}=\angle \text{EDF}\dots \dots \dots \dots \left(\text{3}\right)\\ \text{Also},\text{given that},\\ \text{BC}=\text{ED}\dots \dots \dots \dots \dots \dots \left(\text{4}\right)\\ \text{By using equation}\left(\text{1}\right),\left(\text{3}\right),\text{and}\left(\text{4}\right),\text{we obtain}\\ \boxed{\Delta \text{ABC}\cong \Delta \text{FED}}(\text{ASA criterion})\end{array}

Q.26

$\begin{array}{l}\mathrm{Find}\mathrm{the}\mathrm{ratio}\mathrm{of}:\\ \left(\mathrm{a}\right)\mathrm{Rs}5\mathrm{to}50\mathrm{paise}\left(\mathrm{b}\right)15\mathrm{kg}\mathrm{to}210\mathrm{g}\\ \left(\mathrm{c}\right)9\mathrm{m}\mathrm{to}27\mathrm{cm}\left(\mathrm{d}\right)30\mathrm{days}\mathrm{to}36\mathrm{hour}\end{array}$

Ans

\begin{array}{l}\left(\text{a}\right)\text{Rs 5 to 5}0\text{paise}\\ 1\text{ruppe}=\text{100 paise}\\ \text{So,}\text{5 rupee}=\text{500 paise}\\ \text{So,}\frac{\text{Rs}5}{50\text{paise}}=\frac{500}{50}=\frac{10}{1}\\ \text{Thus, the required ratio is}\boxed{\text{10:1}}.\\ \left(\text{b}\right)\text{15 kg to 21}0\text{g}\\ \text{1 kg}=\text{1000 g}\\ \text{So, 15 kg}=\text{15000 g}\\ \text{So,}\frac{15\text{kg}}{210\text{g}}=\frac{15000}{210}=\frac{500}{7}\\ \text{Thus, the required ratio is}\boxed{\text{500:7}}.\\ \left(\text{c}\right)\text{9 m to 27 cm}\\ \text{1 m}=\text{100 cm}\\ \text{So, 9 m}=\text{900 cm}\\ \text{So,}\frac{9\text{m}}{27\text{cm}}=\frac{900}{27}=\frac{100}{3}\\ \text{Thus, the required ratio is}\boxed{\text{100:3}}.\\ \left(\text{d}\right)\text{3}0\text{days to 36 hour}\\ \text{1 day}=\text{24 hours}\\ \text{So, 30 days}=\text{720 hours}\\ \text{So,}\frac{30\text{days}}{36\text{hour}}=\frac{720}{36}=\frac{20}{1}\\ \text{Thus, the required ratio is}\boxed{\text{20:1}}.\end{array}

Q.27

$\begin{array}{l}\mathrm{In}\mathrm{a}\mathrm{computer}\mathrm{lab},\mathrm{there}\mathrm{are}3\mathrm{computers}\mathrm{for}\mathrm{every}6\\ \mathrm{students}.\mathrm{How}\mathrm{many}\mathrm{computers}\mathrm{will}\mathrm{be}\mathrm{needed}\mathrm{for}24\\ \mathrm{students}?\end{array}$

Ans

\begin{array}{l}\text{For 6 students, number of computers required}=\text{3}\\ \text{So, for 1 student, number of computers required}=\frac{3}{6}=\frac{1}{2}\\ \text{Thus, for 24 students, number of computers required}=\frac{1}{2}\times 24\\ =\boxed{12}\\ \text{Therefore, 12 computers are needed for 24 students}\text{.}\end{array}

Q.28

$\begin{array}{l}\mathrm{Population}\mathrm{of}\mathrm{Rajasthan}=570\mathrm{lakhs}\mathrm{and}\\ \mathrm{population}\mathrm{of}\mathrm{UP}=1660\mathrm{lakh}\\ \mathrm{Area}\mathrm{of}\mathrm{Rajasthan}=\; 3\mathrm{lakh}{\mathrm{km}}^{\mathrm{2}}\mathrm{and}\\ \mathrm{area}\mathrm{of}\mathrm{UP}=\; 2\mathrm{lakh}{\mathrm{km}}^{\mathrm{2}}\mathrm{.}\\ \left(\mathrm{i}\right)\mathrm{How}\mathrm{many}\mathrm{people}\mathrm{are}\mathrm{there}\mathrm{per}{\mathrm{km}}^{\mathrm{2}}\mathrm{in}\mathrm{both}\mathrm{these}\mathrm{States}?\\ \left(\mathrm{ii}\right)\mathrm{Which}\mathrm{State}\mathrm{is}\mathrm{less}\mathrm{populated}?\end{array}$

Ans

\begin{array}{l}{\text{(i) Population of Rajasthan in 3 lakh km}}^{\text{2}}=\text{570 lakhs}\\ {\text{Population of Rajasthan in 1 lakh km}}^{\text{2}}=\frac{\text{570}}{3}\\ =\boxed{190}\\ {\text{Population of UP in 2 lakh km}}^{\text{2}}=\text{1660 lakhs}\\ {\text{Population of UP in 1 lakh km}}^{\text{2}}=\frac{1660}{2}\\ =\boxed{830}\\ \text{(ii)}\\ \text{From above data, clearly Rajasthan is less populated}\text{.}\end{array}

Q.29

$\begin{array}{l}\mathrm{Convert}\mathrm{the}\mathrm{given}\mathrm{fractional}\mathrm{numbers}\mathrm{to}\mathrm{per}\mathrm{cents}.\\ \left(\mathrm{a}\right)\frac{\mathrm{1}}{\mathrm{9}}\left(\mathrm{b}\right)\frac{\mathrm{5}}{\mathrm{4}}\left(\mathrm{c}\right)\frac{\mathrm{3}}{\mathrm{40}}\left(\mathrm{d}\right)\frac{\mathrm{2}}{\mathrm{7}}\end{array}$

Ans

\begin{array}{l}\text{(a})\frac{1}{8}\\ =\frac{1}{8}\times \frac{100}{100}=\frac{1}{8}\times 100\%=12.5\%\\ \left(\text{b}\right)\frac{5}{4}\\ =\frac{5}{4}\times \frac{100}{100}=\frac{5}{4}\times 100\%=125\%\\ \left(\text{c}\right)\frac{3}{40}\\ =\frac{3}{40}\times \frac{100}{100}=\frac{3}{40}\times 100\%=7.5\%\\ \left(\text{d}\right)\frac{2}{7}\\ =\frac{2}{7}\times \frac{100}{100}=\frac{2}{7}\times 100\%=28\frac{4}{7}\%\end{array}

Q.30 Convert the given decimal fractions to percents.
a. 0.65 b. 2.1 c. 0.02 d. 12.35

Ans

\begin{array}{l}\left(\text{a}\right)0.\text{65}\\ =0.65\times 100\%=\frac{65\times 100}{100}\%=65\%\\ \left(\text{b}\right)\text{2}.\text{1}\\ =2.1\times 100\%=\frac{21\times 100}{10}\%=210\%\\ \left(\text{c}\right)0.0\text{2}\\ =0.02\times 100\%=\frac{2\times 100}{100}\%=2\%\\ \left(\text{d}\right)\text{12}.\text{35}\\ =12.35\times 100\%=\frac{1235\times 100}{100}\%=1235\%\end{array}

Q.31 Estimate what part of the figures is coloured and hence find the percent which is coloured.

Ans

\begin{array}{l}\text{(i)}\\ \text{From the figure, we observe that 1 part out of 4 is shaded,}\\ \text{so its}\frac{1}{4}.\\ \frac{1}{4}\times \frac{100}{100}=\frac{1}{4}\times 100\%=25\%\\ \text{(ii)}\\ \text{From the figure, we observe that 3 part out of 5 is shaded,}\\ \text{so its}\frac{3}{5}.\\ \frac{3}{5}\times \frac{100}{100}=\frac{3}{5}\times 100\%=60\%\\ \text{(iii)}\\ \text{From the figure, we observe that 3 part out of 8 is shaded,}\\ \text{so its}\frac{3}{8}.\\ \frac{3}{8}\times \frac{100}{100}=\frac{3}{8}\times 100\%=37.5\%\end{array}

Q.32

$\begin{array}{l}\mathrm{Find}:\\ \left(\mathrm{a}\right)15\%\mathrm{of}250\left(\mathrm{b}\right)1\%\mathrm{of}1\mathrm{hour}\\ \left(\mathrm{c}\right)20\%\mathrm{of}\mathrm{Rs}2500\left(\mathrm{d}\right)75\%\mathrm{of}1\mathrm{k}\end{array}$

Ans

\begin{array}{l}\left(\text{a}\right)\text{15}\%\text{of 25}0\\ =\frac{15}{100}\times 250=\frac{75}{2}=37.5\\ \left(\text{b}\right)\text{1}\%\text{of 1 hour}\\ \text{1 hour}=\text{60 minutes}\text{.}\\ \text{So, 1}\%\text{of 1 hour}\\ =\frac{1}{100}\times 60=\frac{6}{10}=\frac{3}{5}\\ \left(\text{c}\right)\text{2}0\%\text{of Rs 25}00\\ =\frac{20}{100}\times 2500=20\times 25=500\\ \left(\text{d}\right)\text{75}\%\text{of 1 kg}\\ \text{1 kg}=\text{1000 g}\text{.}\\ \text{So, 75}\%\text{of 1 kg}\\ =\frac{75}{100}\times 1000=75\times 10=750\end{array}

Q.33

$\begin{array}{l}\mathrm{Find}\mathrm{the}\mathrm{whole}\mathrm{quantity}\mathrm{if}\\ \left(\mathrm{a}\right)5\%\mathrm{of}\mathrm{it}\mathrm{is}600.\left(\mathrm{b}\right)12\%\mathrm{of}\mathrm{it}\mathrm{is}1080.\\ \left(\mathrm{c}\right)40\%\mathrm{of}\mathrm{it}\mathrm{is}500\mathrm{km}.\left(\mathrm{d}\right)70\%\mathrm{of}\mathrm{it}\mathrm{is}14\mathrm{minutes}\\ \left(\mathrm{e}\right)8\%\mathrm{of}\mathrm{it}\mathrm{is}40\mathrm{litre}\end{array}$

Ans

\begin{array}{l}\left(\text{a}\right)\text{5}\%\text{of it is 6}00\text{is same as 5}\%\text{of x is 6}00.\\ \frac{5}{100}\times x=600\\ x=\frac{600\times 100}{5}=12000\\ \left(\text{b}\right)\text{12}\%\text{of it is}\text{1}0\text{8}0\text{is same as 12}\%\text{of x is}1080.\\ \frac{12}{100}\times x=1080\\ x=\frac{1080\times 100}{12}=9000\\ \left(\text{c}\right)\text{4}0\%\text{of it is 5}00\text{km is same as 40}\%\text{of x is}500\text{km}.\\ \frac{40}{100}\times x=500\\ x=\frac{500\times 100}{40}=1250\\ \left(\text{d}\right)\text{7}0\%\text{of it is 14 minutes is same as 70}\%\text{of x is}14\text{minutes}\\ \frac{70}{100}\times x=14\\ x=\frac{14\times 100}{70}=20\\ \left(\text{e}\right)\text{8}\%\text{of it is 4}0\text{litres is same as 8}\%\text{of x is}40\text{litres}\\ \frac{8}{100}\times x=40\\ x=\frac{40\times 100}{8}=500\end{array}

Q.34

$\begin{array}{l}\mathrm{Convert}\mathrm{given}\mathrm{per}\mathrm{cents}\mathrm{to}\mathrm{decimal}\mathrm{fractions}\mathrm{and}\mathrm{also}\mathrm{to}\\ \mathrm{fractions}\mathrm{in}\mathrm{simplest}\mathrm{forms}:\\ \left(\mathrm{a}\right)\mathrm{25\%}\left(\mathrm{b}\right)150\%\left(\mathrm{c}\right)\mathrm{20\%}\left(\mathrm{d}\right)\mathrm{5\%}\end{array}$

Ans

\begin{array}{l}\left(\text{a}\right)\text{25}\%\\ =\frac{25}{100}=\frac{1}{4}=0.25\\ \left(\text{b}\right)\text{15}0\%\\ =\frac{150}{100}=\frac{15}{10}=\frac{3}{2}=1.5\\ \left(\text{c}\right)\text{2}0\%\\ =\frac{20}{100}=\frac{2}{10}=\frac{1}{5}=0.2\\ \left(\text{d}\right)\text{5}\%\\ =\frac{5}{100}=\frac{1}{20}=0.05\end{array}

Q.35

$\begin{array}{l}\mathrm{In}\mathrm{a}\mathrm{city},\; 30\%\mathrm{are}\mathrm{females},\; 40\%\mathrm{are}\mathrm{males}\mathrm{and}\mathrm{remaining}\\ \mathrm{are}\mathrm{children}.\mathrm{What}\mathrm{percent}\mathrm{are}\mathrm{children}?\end{array}$

Ans

\begin{array}{l}\text{Children}=\left(100-\left(30+40\right)\right)\%\\ =\left(100-70\right)\%\\ =30\%\end{array}

Q.36

$\begin{array}{l}\mathrm{Out}\mathrm{of}15,000\mathrm{voters}\mathrm{in}\mathrm{a}\mathrm{constituency},\; 60\%\mathrm{voted}.\\ \mathrm{Find}\mathrm{the}\mathrm{percentage}\mathrm{of}\mathrm{voters}\mathrm{who}\mathrm{did}\mathrm{not}\mathrm{vote}.\mathrm{Can}\mathrm{you}\\ \mathrm{now}\mathrm{find}\mathrm{how}\mathrm{many}\mathrm{actually}\mathrm{did}\mathrm{not}\mathrm{vote}?\end{array}$

Ans

\begin{array}{l}\text{Percentage of voters who voted}=60\%\\ \text{Percentage of voters who did not vote}=\text{100\%-60\%}\\ =\text{40\%}\\ \text{Number of voters who did not vote}=\text{40\% of 15,000}\\ =\frac{40}{100}\times 15000\\ =6000\end{array}

Q.37

$\begin{array}{l}\mathrm{Meeta}\mathrm{saves}400\mathrm{from}\mathrm{her}\mathrm{salary}.\mathrm{If}\mathrm{this}\mathrm{is}10\%\mathrm{of}\mathrm{her}\\ \mathrm{salary}.\mathrm{What}\mathrm{is}\mathrm{her}\mathrm{salay}?\end{array}$

Ans

\begin{array}{l}\text{Let her salary be x}\text{.}\\ \text{Then according to the question, we have}\\ \text{10\% of x}=\text{400}\\ \frac{10}{100}\times x=400\\ x=\frac{400\times 100}{10}\\ =4000\\ \text{Therefore Meeta’s salary is 4000}\end{array}

Q.38 Draw a line, say AB, take a point C outside it. Through C, draw a line parallel to AB using ruler and compasses only.

Ans

\begin{array}{l}\text{The steps of construction are as follows:}\\ \text{(i) Draw a line AB}\text{. Take a point P on it}\text{. Take a point C outside}\\ \text{this line}\text{. Join C to P}\text{.}\end{array}

$$\begin{gathered} \text{(ii) Taking P as centre and with a convenient radius, draw arc intersecting line AB} \\ \text{at point D and PC at point E}\text{.} \end{gathered}$$

\text{(iii) Taking C as centre and with the same radius as before, draw an arc FG intersecting PC at H}\text{.}

\begin{array}{l}\text{(iv) Adjust the compasses up to the length of DE}\text{. Without}\\ \text{changing the opening of the compasses and taking H as the}\\ \text{cenre, draw an arc to intersect the previously draw arc}\\ \text{FG at po}\mathrm{int}\text{I}\text{.}\end{array}

\text{(v) Join the points C and I to draw a line}l.

\text{This is the required line which is parallel to AB}\text{.}

Q.39

$\begin{array}{l}\mathrm{A}\mathrm{local}\mathrm{cricket}\mathrm{team}\mathrm{played}20\mathrm{matches}\mathrm{in}\mathrm{one}\mathrm{season}.\\ \mathrm{It}\mathrm{won}25\%\mathrm{of}\mathrm{them}.\mathrm{How}\mathrm{many}\mathrm{matches}\mathrm{did}\mathrm{they}\mathrm{win}?\end{array}$

Ans

\begin{array}{l}\text{Since team played 20 matches,}\\ \text{So, number of matches won}=\text{25\% of 20}\\ =\frac{25}{100}\times 20\\ =\frac{500}{100}\\ =5\\ \text{So, number they won 5 matchces}\end{array}

Q.40

$\begin{array}{l}\mathrm{Tell}\mathrm{what}\mathrm{is}\mathrm{the}\mathrm{profit}\mathrm{or}\mathrm{loss}\mathrm{in}\mathrm{the}\mathrm{following}\mathrm{transactions}.\\ \mathrm{Also}\mathrm{find}\mathrm{profit}\mathrm{per}\mathrm{cent}\mathrm{or}\mathrm{loss}\mathrm{per}\mathrm{cent}\mathrm{in}\mathrm{each}\mathrm{case}.\\ \left(\mathrm{a}\right)\mathrm{Gardening}\mathrm{shears}\mathrm{bought}\mathrm{for}250\mathrm{and}\mathrm{sold}\mathrm{for}325.\\ \left(\mathrm{b}\right)\mathrm{A}\mathrm{refrigerater}\mathrm{bought}\mathrm{for}12,000\mathrm{and}\mathrm{sold}\mathrm{at}13,500.\\ \left(\mathrm{c}\right)\mathrm{A}\mathrm{cupboard}\mathrm{bought}\mathrm{for}2,500\mathrm{and}\mathrm{sold}\mathrm{at}3,000.\\ \left(\mathrm{d}\right)\mathrm{A}\mathrm{skirt}\mathrm{bought}\mathrm{for}250\mathrm{and}\mathrm{sold}\mathrm{at}150.\end{array}$

Ans

\begin{array}{l}\text{(a) Cost price}=\text{250}\\ \text{Selling price}=\text{325}\\ \text{Profit}=\text{325}-\text{250}\\ =\text{75}\\ \text{Profit\%}=\frac{\text{Profit}}{\text{Cost Price}}\times 100\\ =\frac{75}{250}\times 100\\ =30\%\\ \text{(b) Cost price}=\text{12000}\\ \text{Selling price}=\text{13,500}\\ \text{Profit}=\text{13500}-\text{1200}\\ =\text{1500}\\ \text{Profit\%}=\frac{\text{Profit}}{\text{Cost Price}}\times 100\\ =\frac{1500}{12000}\times 100\\ =12.5\%\\ \text{(c) Cost price}=\text{2500}\\ \text{Selling price}=\text{3,000}\\ \text{Profit}=\text{3000}-\text{2500}\\ =\text{500}\\ \text{Profit\%}=\frac{\text{Profit}}{\text{Cost Price}}\times 100\\ =\frac{500}{2500}\times 100\\ =20\%\\ \text{(d) Cost price}=\text{250}\\ \text{Selling price}=\text{150}\\ \text{Loss}=\text{250}-1\text{50}\\ =\text{1000}\\ \text{Loss\%}=\frac{\text{Loss}}{\text{Cost Price}}\times 100\\ =\frac{100}{250}\times 100\\ =40\%\end{array}

Q.41

$\begin{array}{l}\mathrm{Convert}\mathrm{each}\mathrm{part}\mathrm{of}\mathrm{the}\mathrm{ratio}\mathrm{to}\mathrm{percentage}:\\ \left(\mathrm{a}\right)3:1\left(\mathrm{b}\right)2:3:5\left(\mathrm{c}\right)1:4\left(\mathrm{d}\right)\mathrm{1:\; 2:5.}\end{array}$

Ans

\begin{array}{l}\left(\text{a}\right)\text{3}:\text{1}\\ \text{Here total parts}=3+1=4\\ 1^{\text{st}}\text{part}=\frac{3}{4}=\frac{3}{4}\times 100\%=75\%\\ 2^{\text{nd}}\text{part}=\frac{1}{4}=\frac{1}{4}\times 100\%=25\%\\ \left(\text{b}\right)\text{2}:\text{3}:\text{5}\\ \text{Here total parts}=2+3+5=10\\ 1^{\text{st}}\text{part}=\frac{2}{10}=\frac{2}{10}\times 100\%=20\%\\ 2^{\text{nd}}\text{part}=\frac{3}{10}=\frac{3}{10}\times 100\%=30\%\\ 3^{\text{rd}}\text{part}=\frac{5}{10}=\frac{5}{10}\times 100\%=50\%\\ \left(\text{c}\right)\text{1}:\text{4}\\ \text{Here total parts}=1+4=5\\ 1^{\text{st}}\text{part}=\frac{1}{5}=\frac{1}{5}\times 100\%=20\%\\ 2^{\text{nd}}\text{part}=\frac{4}{5}=\frac{4}{5}\times 100\%=80\%\\ \left(\text{d}\right)\text{1}:\text{2}:\text{5}\\ \text{Here total parts}=1+2+5=8\\ 1^{\text{st}}\text{part}=\frac{1}{8}=\frac{1}{8}\times 100\%=12.5\%\\ 2^{\text{nd}}\text{part}=\frac{2}{8}=\frac{2}{8}\times 100\%=25\%\\ 3^{\text{rd}}\text{part}=\frac{5}{8}=\frac{5}{8}\times 100\%=62.5\%\end{array}

Q.42

$\begin{array}{l}\mathrm{The}\mathrm{population}\mathrm{of}\mathrm{a}\mathrm{city}\mathrm{decreased}\mathrm{from}25,000\mathrm{to}24,500.\\ \mathrm{Find}\mathrm{the}\mathrm{percentage}\mathrm{decrease}.\end{array}$

Ans

\begin{array}{l}\text{Initial population}=2500\text{0}\\ \text{Final populaton}=24500\\ \text{Decrease}=25000-2\text{4500}\\ =500\\ \text{Decrease}\%=\frac{500}{25000}\times 100\\ =2\%\end{array}

Q.43

$\begin{array}{l}\mathrm{Arun}\mathrm{bought}\mathrm{a}\mathrm{car}\mathrm{for}3,50,000.\mathrm{The}\mathrm{next}\mathrm{year},\mathrm{the}\mathrm{price}\\ \mathrm{went}\mathrm{up}\mathrm{to}3,70,000.\mathrm{What}\mathrm{was}\mathrm{the}\mathrm{Percentage}\mathrm{of}\mathrm{price}\\ \mathrm{increase}?\end{array}$

Ans

\begin{array}{l}\text{Initial Price}=\text{3,50,000}\\ \text{Final Price}=\text{3,70,000}\\ \text{Increase}=\text{3,70,000}-\text{3,50,000}\\ =\text{20,000}\\ \text{Increase\%}=\frac{20,000}{3,50,000}\times 100\\ =5\frac{5}{7}\%\end{array}

Q.44

$\begin{array}{l}\mathrm{Draw}\mathrm{a}\mathrm{line}\mathrm{l}.\mathrm{Draw}\mathrm{a}\mathrm{perpendicular}\mathrm{to}\mathrm{l}\mathrm{at}\mathrm{any}\mathrm{point}\mathrm{on}\mathrm{l}.\\ \mathrm{On}\mathrm{this}\mathrm{perpendicular}\mathrm{choose}\mathrm{a}\mathrm{point}\mathrm{X},4\mathrm{cm}\mathrm{away}\mathrm{from}\mathrm{l}.\\ \mathrm{Through}\mathrm{X},\mathrm{draw}\mathrm{a}\mathrm{line}\mathrm{m}\mathrm{parallel}\mathrm{to}\mathrm{l}.\end{array}$

Ans

\begin{array}{l}\text{The steps of construction are as follows:}\\ \text{(i) Draw a line}l\text{and take a point P on the line}l\text{. Then draw a}\\ \text{perpendicular at point P}\text{.}\end{array}

\begin{array}{l}\text{(ii) Adjusting the compasses up to the length of 4 cm, draw an}\\ \text{arc to intersect this perpendicular at point X}\text{. Choose any point}\\ \text{Y on the line}l\text{. Join X to Y}\text{.}\end{array} \begin{array}{l}\end{array}

\begin{array}{l}\text{(iii) Taking Y as centre and with a convenient radius, draw}\\ \text{an arc intersecting}l\text{at A and XY at B}\text{.}\end{array} \begin{array}{l}\end{array}

\begin{array}{l}\text{(iv) Taking X as centre and with same radius as before, draw}\\ \text{an arc CD cutting XY at E}\text{.}\end{array} \begin{array}{l}\end{array}

\begin{array}{l}\text{(v) Adjust the compasses upto the length of AB}\text{. Wothout}\\ \text{changing the opening of the compasses and taking E as the}\\ \text{centre, draw an arc intersecting the previous arc CD t F}\text{.}\end{array}

\text{(vi) Join the point X and F to draw a line}m\text{.} \text{Line}m\text{is the required line parallel to}l\text{.}

Q.45

$\begin{array}{l}\mathrm{I}\mathrm{buy}\mathrm{a}\mathrm{T}.\mathrm{V}.\mathrm{for}\mathrm{10,000}\mathrm{and}\mathrm{sell}\mathrm{it}\mathrm{at}\mathrm{a}\mathrm{profit}\mathrm{of}20\%.\mathrm{How}\\ \mathrm{much}\mathrm{money}\mathrm{do}\mathrm{I}\mathrm{get}\mathrm{for}\mathrm{it}?\end{array}$

Ans

\begin{array}{l}\text{We know that}\\ \text{Profit\%}=\frac{\text{Profit}}{\text{Cost price}}\times 100\\ \text{So,}\\ 20=\frac{\text{Profit}}{10,000}\times 100\\ =\frac{\text{Profit}}{100\times \cancel{100}}\times \cancel{100}\\ \text{Profit}=\text{20}\times \text{100}\\ =\text{2000}\\ \text{Now,}\\ \text{Profit}=\text{Selling price}-\text{Cost price}\\ \text{2000}=\text{Selling price}-\text{10000}\\ \text{Selling price}=\text{10000}+\text{2000}\\ =\text{12000}\end{array}

Q.46

$\begin{array}{l}\mathrm{Let}\mathrm{l}\mathrm{be}\mathrm{a}\mathrm{line}\mathrm{and}\mathrm{P}\mathrm{be}\mathrm{a}\mathrm{point}\mathrm{not}\mathrm{on}\mathrm{l}.\mathrm{Through}\mathrm{P},\mathrm{draw}\\ \mathrm{a}\mathrm{line}\mathrm{m}\mathrm{parallel}\mathrm{to}\mathrm{l}.\mathrm{Now}\mathrm{join}\mathrm{P}\mathrm{to}\mathrm{any}\mathrm{point}\mathrm{Q}\mathrm{on}\mathrm{l}.\mathrm{Choose}\\ \mathrm{any}\mathrm{other}\mathrm{point}\mathrm{R}\mathrm{on}\mathrm{m}.\mathrm{Through}\mathrm{R},\mathrm{draw}\mathrm{a}\mathrm{line}\mathrm{parallel}\mathrm{to}\mathrm{PQ}.\\ \mathrm{Let}\mathrm{this}\mathrm{meet}\mathrm{l}\mathrm{at}\mathrm{S}.\mathrm{What}\mathrm{shape}\mathrm{do}\mathrm{the}\mathrm{two}\mathrm{sets}\mathrm{of}\mathrm{parallel}\\ \mathrm{lines}\mathrm{enclose}?\end{array}$

Ans

\begin{array}{l}\text{The steps of construction are as follows:}\\ \text{(i) Draw a line l and take a point A}\text{​}\text{on it}\text{. Take a point P not}\\ \text{on}l\text{and join A to P}\text{.}\end{array}

\begin{array}{l}\text{(ii) Taking A as centre and with a convenient radius, draw an}\\ \text{arc cutting}l\text{at B and AP at C}\text{.}\end{array} \begin{array}{l}\end{array}

\begin{array}{l}\text{(iii) Taking P as centre and with the same radius as before,}\\ \text{draw an arc DE to intersect AP at F}\text{.}\end{array} \begin{array}{l}\end{array}

\begin{array}{l}\text{(iv) Adust the compasses up to the length of BC}\text{. Without}\\ \text{changing the opening of compasses and taking F as the}\\ \text{centre, draw an arc to intersect the previously draw arc}\\ \text{DE at point G}\text{.}\end{array}

\text{(v) Join P to G to draw a line m}\text{. Line}m\text{will be parallel to line}l\text{.} \begin{array}{l}\end{array}

\begin{array}{l}\text{(vi) Join P to any point Q on line}\text{. Choose another point R on}\\ \text{line}m\text{. Similarly, a line can be drawn through point R and}\\ \text{parallel to PQ}\text{.}\end{array} \begin{array}{l}\end{array}

\begin{array}{l}\text{Let it meet line}l\text{at point S}\text{. In quadrilateral PQRS, opposite}\\ \text{lines are parallel to each other}\text{. PQ}âˆ\yen \text{RS and PR}âˆ\yen \text{QS}\text{.}\\ \text{Thus PQRS is a parallelogram}\text{.}\end{array}

Q.47

$\begin{array}{l}\mathrm{Juhi}\mathrm{sells}\mathrm{a}\mathrm{washing}\mathrm{machine}\mathrm{for}13,500.\mathrm{She}\mathrm{loses}20\%\mathrm{in}\\ \mathrm{the}\mathrm{bargain}.\mathrm{What}\mathrm{was}\mathrm{the}\mathrm{price}\mathrm{at}\mathrm{which}\mathrm{she}\mathrm{bought}\mathrm{it}?\end{array}$

Ans

\begin{array}{l}\text{Selling price}=\text{135}00\\ \text{Loss}\%=\text{2}0\%\\ \text{Let the cost price be}x.\\ \therefore \text{Loss}=\text{2}0\%\text{of}x\\ \text{Cost price}-\text{Loss}=\text{Selling price}\\ \text{x}-20\%\text{of x}=\text{13500}\\ \text{x}-\frac{20}{100}\times x=13500\\ \frac{100x-20x}{100}=13500\\ \frac{80x}{100}=13500\\ x=\frac{13500\times 100}{80}\\ =16875\end{array}

Q.48

$\begin{array}{l}\left(\mathrm{i}\right)\mathrm{Chalk}\mathrm{contains}\mathrm{calcium},\mathrm{carbon}\mathrm{and}\mathrm{oxygen}\mathrm{in}\mathrm{the}\mathrm{ratio}\\ 10:3:12.\mathrm{Find}\mathrm{the}\mathrm{percentage}\mathrm{of}\mathrm{carbon}\mathrm{in}\mathrm{chalk}.\\ \left(\mathrm{ii}\right)\mathrm{If}\mathrm{in}\mathrm{a}\mathrm{stick}\mathrm{of}\mathrm{chalk},\mathrm{carbon}\mathrm{is}3\mathrm{g},\mathrm{what}\mathrm{is}\mathrm{the}\mathrm{weight}\\ \mathrm{of}\mathrm{the}\mathrm{chalk}\mathrm{stick}.\end{array}$

Ans

\begin{array}{l}\text{(i) Given ratio}=\text{10:3:12}\\ \text{Total}=\text{10}+\text{3}+\text{12}\\ =\text{25}\\ \text{Percentage of Carbon}=\frac{3}{25}\times 100\\ =12\%\\ \text{(ii) Let the weight of the stick be x g}\\ \text{So, 12\% of x}=\text{3}\\ \frac{12}{100}\times x=3\\ x=\frac{300}{12}\\ =25\text{g}\end{array}

Q.49 Amina buys a book for 275 and sells it at a loss of 15%. How much does she sell it for?

Ans

\begin{array}{l}\text{Cost price of book}=\text{275}\\ \text{Loss\%}=\text{15}\\ \text{We have,}\\ \text{Loss\%}=\frac{\text{Loss}}{\text{Cost price}}\times 100\\ 15=\frac{\text{Loss}}{275}\times 100\\ \text{Loss}=\frac{15\times 275}{100}\\ =41.25\\ \text{Selling price}=\text{Cost price}-\text{Loss}\\ =\text{275}-\text{41}\text{.25}\\ =\text{233}\text{.75}\end{array}

Q.50 Find the amount to be paid at the end of 3 years in each case:
(a) Principal = 1,200 at 12% p.a.
(b) Principal = 7,500 at 5% p.a.

Ans

\begin{array}{l}\text{(a)}\\ \text{Principal (P)}=\text{1200}\\ \text{Rate (R)}=\text{12\%}\\ \text{Time (T)}=3\text{years}\\ \text{S}\text{.I}\text{.}=\frac{\text{P}\times \text{R}\times \text{T}}{100}\\ =\frac{1200\times 12\times 3}{100}\\ =432\\ \text{Amount}=\text{P}+\text{S}\text{.I}\text{.}\\ =\text{1200}+\text{432}=\text{1632}\\ \text{(b)}\\ \text{Principal (P)}=\text{7500}\\ \text{Rate (R)}=\text{5\%}\\ \text{Time (T)}=\text{3 years}\\ \text{S}\text{.I}\text{.}=\frac{\text{P}\times \text{R}\times \text{T}}{100}\\ =\frac{7500\times 5\times 3}{100}\\ =1125\\ \text{Amount}=\text{P}+\text{S}\text{.I}\text{.}\\ =\text{7500}+\text{1125}\\ =\text{8625}\end{array}

Q.51 What rate gives 280 as interest on a sum of 56000 in 2 years?

Ans

$\begin{array}{l}\mathrm{We}\mathrm{know}\mathrm{that}\\ \mathrm{S}.\mathrm{I}=\frac{\mathrm{P}\times \mathrm{R}\times \mathrm{T}}{100}\\ 280=\frac{\mathrm{56000}\times \mathrm{R}\times 2}{100}\\ \mathrm{R}=\frac{280\times 100}{56000\times 2}\\ =0.25\\ \mathrm{Therefore},\mathrm{the}\mathrm{rate}\mathrm{is}0.25\%.\end{array}$

Q.52

$\begin{array}{l}\mathrm{If}\mathrm{Meena}\mathrm{gives}\mathrm{an}\mathrm{interest}\mathrm{of}45\mathrm{for}\mathrm{one}\mathrm{year}\mathrm{at}9\%\\ \mathrm{rate}\mathrm{p}.\mathrm{a}.\mathrm{What}\mathrm{is}\mathrm{the}\mathrm{sum}\mathrm{she}\mathrm{has}\mathrm{borrowed}?\end{array}$

Ans

\begin{array}{l}\text{We know that}\\ \text{S}\text{.I}=\frac{\text{P}\times \text{R}\times \text{T}}{100}\\ \text{So, we get}\\ \text{45=}\frac{\text{P×9×1}}{\text{100}}\\ \text{P=}\frac{\text{45×100}}{\text{9}}\\ \text{=}\text{500}\\ \text{Therefore, she borrowed 500}\text{.}\end{array}

Q.53

$\mathrm{Construct}\mathrm{\Delta XYZ}\mathrm{in}\mathrm{which}\mathrm{XY}=4.5\mathrm{cm},\mathrm{YZ}=5\mathrm{cm}\mathrm{and}\mathrm{ZX}=6\mathrm{cm}.$

Ans

\begin{array}{l}\text{The steps of construction are as follows:}\\ \text{(i) Draw a line segment YZ of length 5 cm}\text{.}\end{array}

\begin{array}{l}\text{(ii) Point X is at a distance of 4}\text{.5 cm from point Y}\text{. Therefore}\\ \text{taking Y as centre, draw an arc of 4}\text{.5 cm radius}\text{.}\end{array} \begin{array}{l}\end{array}

\begin{array}{l}\text{(iii) Point X is at a distance of 6 cm from point Z}\text{.Therefore}\\ \text{taking Z as centre, draw an arc of 6 cm radius}\text{. Mark the}\\ \text{point of intersection of the arcs as X}\text{. Join XY and XZ}\text{.}\end{array}

\text{Therefore, XYZ is the required triangle}\text{.}

Q.54

$\mathrm{Construct}\mathrm{an}\mathrm{equilateral}\mathrm{triangle}\mathrm{of}\mathrm{side}5.5\mathrm{cm}.$

Ans

\begin{array}{l}\text{Since, we need to construct an equilateral triangle so all sides}\\ \text{should be equal}\text{.}\\ \text{So, AB}=\text{BC}=\text{CA}=\text{5}\text{.5 cm}\\ \text{The steps of construction are as follows:}\\ \text{(i) Draw a line segment BC of length 5}\text{.5 cm}\text{.}\end{array}

\text{(ii) Taking B as centre, draw an arc of 5}\text{.5 cm radius}\text{.}

\begin{array}{l}\text{(iii) Taking C as centre, draw an arc of 5}\text{.5 cm radius to}\\ \text{meet previous arc at point A}\text{.}\end{array}

\text{(iv) Join A}\text{​}\text{to B and C}\text{.}

\text{Therefore, ABC is the required triangle}\text{.}

Q.55

$\begin{array}{l}\mathrm{Draw}\mathrm{\Delta PQR}\mathrm{with}\mathrm{PQ}=4\mathrm{cm},\mathrm{QR}=3.5\mathrm{cm}\mathrm{and}\mathrm{PR}=4\mathrm{cm}.\\ \mathrm{What}\mathrm{type}\mathrm{of}\mathrm{triangle}\mathrm{is}\mathrm{this}?\end{array}$

Ans

\begin{array}{l}\text{The steps of construction are as follows:}\\ \text{(i) Draw a line segment QR of length 3}\text{.5 cm}\text{.}\end{array}

\text{(ii) Taking Q as centre, draw an arc of 4 cm radius}\text{.} \begin{array}{l}\end{array}

\begin{array}{l}\text{(iii) Taking R as centre, draw an arc of 4 cm radius to}\\ \text{meet previous arc at point P}\text{.}\end{array}

\text{(iv) Join P to Q and R}\text{.} \begin{array}{l}\end{array}

\begin{array}{l}\text{Therefore, PQR is the required triangle}\text{.}\\ \text{Since two sides of required triangle are equal,}\\ \text{so it is an isosceles triangle}\text{.}\end{array}

Q.56

$\begin{array}{l}\mathrm{Construct}\mathrm{\Delta ABC}\mathrm{such}\mathrm{that}\mathrm{AB}=2.5\mathrm{cm},\mathrm{BC}=6\mathrm{cm}\mathrm{and}\\ \mathrm{AC}=6.5\mathrm{cm}.\mathrm{Measure}\angle \mathrm{B}.\end{array}$

Ans

\begin{array}{l}\text{The steps of construction are as follows:}\\ \text{(i) Draw a line segment BC of length 6 cm}\text{.}\end{array}

\text{(ii) Taking C as centre, draw an arc of 6}\text{.5 cm radius}\text{.} \begin{array}{l}\end{array}

\begin{array}{l}\text{(iii) Taking B as centre, draw an arc of 2}\text{.5 cm radius to}\\ \text{meet previous arc at point A}\text{.}\end{array}

\text{(iv) Join A}\text{​}\text{to B and C}\text{.} \begin{array}{l}\end{array}

\begin{array}{l}\text{Therefore, ABC is the required triangle}\text{.}\\ \angle \text{B can be measured with the help of a protractor and}\\ \text{it comes out to be 90}°\text{.}\end{array}

Q.57

$\mathrm{Construct}\mathrm{\Delta DEF}\mathrm{such}\mathrm{that}\mathrm{DE}=5\mathrm{cm},\mathrm{DF}=3\mathrm{cm}\mathrm{and}\mathrm{m}\angle \mathrm{EDF}=90\mathrm{°}.$

Ans

\begin{array}{l}\text{The steps of construction are as follows:}\\ \text{(i) Draw a line segment DE of length 5 cm}\text{.}\end{array}

\text{(ii) At point D, draw a ray DX making an angle of 90}°\text{with DE}\text{.}

\begin{array}{l}\text{(iii) Taking D as centre, draw an arc of 3 cm radius}\text{. It will}\\ \text{intersect DX at point F}\end{array}

\text{(iv) Join F to E}

\text{Thus, DEF is the required triangle}\text{.}

Q.58

$\begin{array}{l}\mathrm{Construct}\mathrm{an}\mathrm{isosceles}\mathrm{triangle}\mathrm{in}\mathrm{which}\mathrm{the}\mathrm{lengths}\mathrm{of}\\ \mathrm{each}\mathrm{of}\mathrm{its}\mathrm{equal}\mathrm{sides}\mathrm{is}6.5\mathrm{cm}\mathrm{and}\mathrm{the}\mathrm{angle}\mathrm{between}\\ \mathrm{them}\mathrm{is}110\mathrm{°}.\end{array}$

Ans

\begin{array}{l}\text{An isosceles triangle PQR has to be constructed with}\\ \text{PQ}=\text{QR}=\text{6}\text{.5 cm}\text{.}\\ \text{The steps of construction are as follows:}\\ \text{(i) Draw a line segment QR of length 6}\text{.5 cm}\end{array}

\text{(ii) At point Q, draw a ray QX making an angle 110}°\text{with QR}\text{.} \begin{array}{l}\end{array}

\begin{array}{l}\text{(iii) Taking Q as centre, draw an arc of 6}\text{.5 cm radius}\text{. It}\\ \text{intersects QX at point P}\text{.}\end{array}

\text{(iv) Join P to R to obtain the required triangle PQR}\text{.}

Q.59

$\mathrm{Construct}\mathrm{\Delta ABC}\mathrm{with}\mathrm{BC}=7.5\mathrm{cm},\mathrm{AC}=5\mathrm{cm}\mathrm{and}\mathrm{m}\angle \mathrm{C}=60\mathrm{°}.$

Ans

\begin{array}{l}\text{The steps of constructions are as follows:}\\ \text{(i) Draw a line segment BC of length 7}\text{.5 cm}\text{.}\end{array}

\text{(ii) At point C, draw a ray CX making 60 with BC}\text{.} \begin{array}{l}\end{array}

\begin{array}{l}\text{(iii) Taking C as centre, draw an arc of 5 cm radius}\text{.}\\ \text{It intersect CX at point A}\text{.}\end{array}

\text{(iv) Join A to B to obtain the required triangle ABC}\text{.}

Q.60

$\mathrm{Construct}\mathrm{\Delta ABC},\mathrm{given}\mathrm{m}\angle \mathrm{A}=60\mathrm{°},\mathrm{m}\angle \mathrm{B}=30\mathrm{°}\mathrm{and}\mathrm{AB}=5.8\mathrm{cm}.$

Ans

\text{(i) Draw a line segment AB of length 5}\text{.8 cm}\text{.}

\text{(ii) At point A, draw a ray AX making 60}°\text{angle with AB}\text{.}

\text{(iii) At point B, draw a ray AX making 30}°\text{angle with AB}\text{.} \begin{array}{l}\end{array}

\begin{array}{l}\text{(iv) Point C has to lie on both the rays, AX and BY}\text{. Therefore,}\\ \text{C is the point of intersection of these two rays}\text{.}\end{array}

\text{Thus, ABC is the required triangle}\text{.}

Q.61

$\begin{array}{l}\mathrm{Construct}\mathrm{\Delta PQR}\mathrm{if}\mathrm{PQ}=5\mathrm{cm},\mathrm{m}\angle \mathrm{PQR}=105\mathrm{°}\mathrm{and}\mathrm{m}\angle \mathrm{QRP}=40\mathrm{°}.\\ (\mathrm{Hint}:\mathrm{Recallangle}–\mathrm{sum}\mathrm{property}\mathrm{of}\mathrm{a}\mathrm{triangle}).\end{array}$

Ans

\begin{array}{l}\text{To construct triangle PQR, we need to find}\angle \text{RPQ}\text{.}\\ \text{Using angle sum property of triangles to get}\\ \angle \text{PQR}+\angle \text{PRQ}+\angle \text{RPQ}=180°\\ 105°+40°+\angle \text{RPQ}=180°\\ \angle \text{RPQ}=180°-145°\\ =35°\\ \text{The steps of construction are as follows:}\\ \text{(i) Draw a line segment PQ of length 5 cm}\text{.}\end{array}

\text{(ii) At P, draw a ray PX making an angle of 35}°\text{with PQ}\text{.}

\text{(iii) At Q, draw a ray QY making an angle of 105}°\text{with PQ}\text{.} \begin{array}{l}\end{array}

\begin{array}{l}\text{(iv) Point R has to lie on both the ray, PX and QY}\text{. Therefore,}\\ \text{R is the point of intersectin of these two rays}\text{.}\end{array}

\text{Thus, PQR is the required triangle}\text{.}

Q.62

$\begin{array}{l}\mathrm{Examine}\mathrm{whether}\mathrm{you}\mathrm{can}\mathrm{construct}\mathrm{\Delta DEF}\mathrm{such}\mathrm{that}\\ \mathrm{EF}=7.2\mathrm{cm},\mathrm{m}\angle \mathrm{E}=110\mathrm{°}\mathrm{and}\mathrm{m}\angle \mathrm{F}=80\mathrm{°}.\mathrm{Justify}\mathrm{your}\mathrm{answer}.\end{array}$

Ans

\begin{array}{l}\text{We are given,}\\ \angle \text{E}=\text{110}°\text{and}\angle \text{F}=\text{80}°\\ \text{Using angle sum property to get}\\ \angle \text{D+}\angle \text{E+}\angle \text{F=180}°\\ \angle D+110°+80°=180°\\ \angle D+190°=180°\\ \text{Since}\angle \text{D}\text{can not be zero and it does not satisfy angle sum}\\ \text{property}\text{.}\\ \text{Thus, we can not construct}\Delta \text{DEF}\text{.}\end{array}

\begin{array}{l}\text{Also, it can be observed that point D should lie on both rays,}\\ \text{EX and FY, for the constructing the required triangle}\text{.}\\ \text{However, both rays are not intersecting each other}\text{.}\\ \text{Therefore, the required triangle can not be formed}\text{.}\end{array}

Q.63

$\mathrm{Construct}\mathrm{the}\mathrm{right}\mathrm{angled}\mathrm{\Delta PQR},\mathrm{where}\mathrm{m}\angle \mathrm{Q}=90\mathrm{°},\mathrm{QR}=8\mathrm{cm}\mathrm{and}\mathrm{PR}=10\mathrm{cm}.$

Ans

\begin{array}{l}\text{The steps of construction are as follows:}\\ \text{(i) Draw a line segment QR of length 8 cm}\text{.}\end{array}

\text{(ii) At point Q, draw a ray QX making 90}°\text{with QR}\text{.} \begin{array}{l}\end{array}

\begin{array}{l}\text{(iii) Taking R as centre, draw an arc of 10 cm radius to}\\ \text{intersect ray QX at point P}\text{.}\end{array}

\text{(iv) Join P to R}\text{.}

\text{Thus,}\Delta \text{PQR is the required right-angled triangle}\text{.}

Q.64 Construct a right-angled triangle whose hypotenuse is 6 cm long and one of the legs is 4 cm long.

Ans

\begin{array}{l}\text{A right-angled triangle ABC with hypotenuse 6 cm and one}\\ \text{of the legs as 4 cm has to be constructed}\text{.}\\ \text{The steps of construction are as follows:}\\ \text{(i) Draw a line segment BC of length 4 cm}\text{.}\end{array}

\text{(ii) At point B, draw a ray BX making 90}°\text{with BC}\text{.} \begin{array}{l}\end{array}

\begin{array}{l}\text{(iii) Taking C as centre, draw an arc of 6 cm radius to}\\ \text{intersect ray BX at point A}\text{.}\end{array}

\text{(iv) Join A to C}\text{.}

\text{Thus,}\Delta \text{ABC is the required right-angled triangle}\text{.}

Q.65

$$\begin{gathered} \mathrm{Construct}\mathrm{an}\mathrm{isosceles}\mathrm{right}–\mathrm{angled}\mathrm{triangle}\mathrm{ABC}, \\ \mathrm{where}\mathrm{m}\angle \mathrm{ACB}=90\mathrm{°}\mathrm{and}\mathrm{AC}=6\mathrm{cm}. \end{gathered}$$

Ans

\begin{array}{l}\text{Since, in a isosceles triangle, the length of any two side}\\ \text{are equal}\text{. So, AB}=\text{BC}=\text{6cm}\\ \text{The steps of construction are as follows:}\\ \text{(i) Draw a line segment AC of length 6 cm}\text{.}\end{array}

\text{(ii) At point C, draw a ray CX making 90}°\text{with AC}\text{.} \begin{array}{l}\end{array}

\begin{array}{l}\text{(iii) Taking C as centre, draw an arc of 6 cm radius to}\\ \text{intersect ray CX at point B}\text{.}\end{array}

\text{(iv) Join A to B}\text{.}

\text{Thus,}\Delta \text{ABC is the required triangle}\text{.}

Q.66

$\begin{array}{l}\text{Identify the nets which can be used to make cubes}\\ \text{(cut out copies of the nets and try it):}\end{array}$

Ans

$\left(i\right)\mathrm{The}\mathrm{given}\mathrm{net}\mathrm{can}\mathrm{be}\mathrm{folded}\mathrm{as}:$

\begin{array}{l}\text{When the faces that are in sky blue colur and in pink colour}\\ \text{are folded to make cube, they will be overlaping each other}\text{.}\end{array} \text{(ii) The given net can be folded as:}

\text{Thus, a cube can thus be formed in above way}\text{.} \text{(iii) The given net can be folded as:}

\text{Thus, a cube can thus be formed in above way}\text{.} \text{(iv) The given net can be folded as:}

\text{Thus, a cube can thus be formed in above way}\text{.} \text{(v) The given net can be folded as:} \begin{array}{l}\end{array}

\begin{array}{l}\text{When the faces are in blue colour and in red colour are folded}\\ \text{to make a cube, they will be overlapping each other}\text{.}\end{array} \begin{array}{l}\text{(vi) The given net can be folded as:}\end{array}

\text{Thus, a cube can thus be formed in above way}\text{.}

Q.67

$\begin{array}{l}\text{Dice are cubes with dots on each face. Opposite faces of}\\ \text{a die always have a total of seven dots on them.}\end{array}$

$\begin{array}{l}\text{Here are two nets to make dice}\left(\text{cubes}\right)\text{; the numbers}\\ \text{inserted in each square indicate the number of dots in}\\ \text{that box.}\end{array}$

$\begin{array}{l}\text{Insert suitable numbers in the blanks, remembering that}\\ \text{the number on the opposite faces should total to 7.}\end{array}$

Ans

\begin{array}{l}\text{(i) The numbers can be inserted as follows so as to make the}\\ \text{given net into a net of a dice}\text{.}\end{array}

\begin{array}{l}\text{It can be seen that the sum of opposite faces is 7}\text{.}\\ \\ \text{(ii) The numbers can be inserted as follows so as to make the}\\ \text{given net into a net of a dice}\text{.}\end{array}

\text{It can be seen that the sum of opposite faces is 7}\text{.}

Q.68

$\text{Can this be a net for a die? Explain your answer.}$

Ans

\text{The given net can be folded as:}

\begin{array}{l}\text{It can be observed that the opposite face of the dice so formed}\\ \text{have 2 and 5, 1 and 4, 3 and 6 on them}\text{.}\\ \text{The sum of numbers on the opposite faces comes to 7, 5 and 9}\\ \text{respectively}\text{.}\\ \text{However, in case of a dice, the sum of numbers on the opposite}\\ \text{faces should be7}\text{.}\\ \text{Therefore, this net is not of a dice}\text{.}\end{array}

Q.69

$\begin{array}{l}\text{Here is an incomplete net for making a cube. Complete}\\ \text{it in at least two different ways. Remember that a cube}\\ \text{has six faces. How many are there in the net here?}\\ \text{(Give two separate diagrams. If you like, you may use a}\\ \text{squared sheet for easy manipulation.)}\end{array}$

Ans

\text{There are 3 faces given in the net}\text{. The given net can be completed as:}

Q.70

$\mathrm{Match}\mathrm{the}\mathrm{nets}\mathrm{with}\mathrm{appropriate}\mathrm{solids}:$

Ans

\text{Consider net (i) It can be folded as:}

\text{Consider net (ii) It can be folded as:}

\text{It is a net of cube}\text{. Hence, (a) is the correct option}\text{.} \text{Consider net (iii) It can be folded as:}

\text{It is a net of cylinder}\text{. Hence, (b) is the correct option}\text{.} \text{Consider net (iv) It can be folded as:}

\text{It is a net of cone}\text{. Hence, (c) is the correct option}\text{.}

Q.71

$\mathrm{Use}\mathrm{isometric}\mathrm{dot}\mathrm{paper}\mathrm{and}\mathrm{make}\mathrm{an}\mathrm{isometric}\mathrm{sketch}\mathrm{for}\mathrm{each}\mathrm{one}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{shapes}:$

Ans

Q.72

$\begin{array}{l}\text{The dimensions of a cuboid are 5 cm, 3 cm and 2 cm.}\\ \text{Draw three different isometric sketches of this cuboid.}\end{array}$

Ans

Q.73

$\begin{array}{l}\text{Three cubes each with 2 cm edge are placed side by side}\\ \text{to form a cuboid. Sketch an oblique or isometric sketch}\\ \text{of this cuboid.}\end{array}$

Ans

Q.74

$\text{Make an oblique sketch for each one of the given isometric shapes:}$

Ans

Q.75

$\begin{array}{l}\text{Give}\left(\text{i}\right)\text{an oblique sketch and}\left(\text{ii}\right)\text{an isometric sketch for}\\ \text{each of the following:}\\ \left(\text{a}\right)\text{A cuboid of dimensions 5 cm, 3 cm and 2 cm.}\\ \left(\text{Is your sketch unique?}\right)\\ \left(\text{b}\right)\text{A cube with an edge 4 cm long.}\\ \text{An isometric sheet is attached at the end of the book.}\\ \text{You could try to make on it somecubes or cuboids of}\\ \text{dimensions specified by your friend.}\end{array}$

Ans

\text{(i) oblique sketch:}

\text{(ii) Isometric sketch:}

Q.76

$\begin{array}{l}\text{What cross-sections do you get when you give a}\\ \left(\text{i}\right)\text{vertical cut}\left(\text{ii}\right)\text{horizontal cut to the following solids?}\\ \left(\text{a}\right)\text{A brick}\\ \left(\text{b}\right)\text{Around apple}\\ \left(\text{c}\right)\text{A die}\\ \left(\text{d}\right)\text{A circular pipe}\\ \left(\text{e}\right)\text{A nice cream cone}\end{array}$

Ans

$\begin{array}{l}\text{(a) A brick}\\ \text{We can give a vertical cut to a brick in the following way:}\end{array}$

\begin{array}{l}\text{(b) A round apple}\\ \text{We can give a vertical cut to a roudn apple in the following way:}\end{array}

We can give a horizontal cut to a round apple in the following way:

(c) A die

We can give a vertical cut to a die in the following ways:

\text{We can give a horizontal cut to a die in the following way:} \begin{array}{l}\end{array}

\begin{array}{l}\text{(d) A circular pipe}\\ \text{We can give a vertical cut to a circular pipe in the following way:}\end{array}

\text{We can give a horizontal cut to a circular pipe in the following way:} \begin{array}{l}\end{array}

\begin{array}{l}\text{(e) An ice cream cone}\\ \text{We can give a vertical cut to an ice cream cone in the following way:}\end{array}

$\mathrm{We}\mathrm{can}\mathrm{give}\mathrm{a}\mathrm{horizontal}\mathrm{cut}\mathrm{to}\mathrm{an}\mathrm{ice}\mathrm{cream}\mathrm{cone}\mathrm{in}\mathrm{the}\mathrm{following}\mathrm{way}:$

Q.77

$\begin{array}{l}\text{A bulb is kept burning just right above the following solids.}\\ \text{Name the shape of the shadows obtained in each case.}\\ \text{Attempt to give a rough sketch of the shadow.(You may}\\ \text{try to experiment first and then answer these questions).}\end{array}$

Ans

\begin{array}{l}\text{The shapes of the shadows of these figures will be as follows:}\\ \text{(i) A ball}\end{array}

\text{The shape of the shadow of a ball will be a circle}\text{.} \text{(ii) A cylindrical pipe}

\text{The shape of the shadow of a cylindrical pipe will be a rectangle} \text{.} \text{(iii) A book}

\text{The shape of the shadow of a book will be a rectangle}\text{.}

Q.78

$\begin{array}{l}\text{Here are the shadows of some 3-D objects, when seen}\\ \text{under the lamp of an over head projector. Identify the}\\ \text{solid}\left(\text{s}\right)\text{that match each shadow. (There may be}\\ \text{multipleanswers for these!).}\end{array}$

Ans

\begin{array}{l}\text{The given shadows can be obtained in case of the following}\\ \text{objects}\text{.}\\ \text{(i) Compact disk (Sphere)}\\ \text{(ii) A}\text{​}\text{dice (Cube)}\\ \text{(iii) Triangular pyramid (Cone)}\\ \text{(iv) Note Book (Cuboid)}\end{array}

Q.79

$\begin{array}{l}\text{Examine if the following are true statements:}\\ \left(\text{i}\right)\text{The cube can cast a shadow in the shape of a rectangle.}\\ \left(\text{ii}\right)\text{The cube can cast a shadow in the shape of a hexagon.}\end{array}$

Ans

\begin{array}{l}\text{A cube can cast shadow only in the shape of a square}\text{.}\\ \text{Therefore, any other shapes are not possible}\text{.}\\ \text{So,}\\ \text{(i) False}\\ \text{(ii) False}\end{array}

Q.80 Find the area of a square park whose perimeter is 320 m.

Ans

\begin{array}{l}\text{Let the side of the square park be ‘}a\text{‘}\text{.}\\ \text{Since perimeter of a square}=\text{4}a\\ \text{So, we get}\\ 4a=3\text{20}\\ a=\frac{320}{4}\\ =80\text{m}\\ \text{Therefore, the area of the square park}=a^2\\ ={\left(80\text{m}\right)}^2\\ =6400{\text{m}}^2\end{array}

Q.81 Find the breadth of a rectangular plot of land, if its area is 440 m² and the length is 22 m. Also find its perimeter.

Ans

\begin{array}{l}\text{Let the breadth of the rectangular plot be}x\text{.}\\ \text{Area of the rectangular plot}=\text{length}\times \text{breadth}\\ \text{SO, we get}\\ {\text{440m}}^2=22\text{m}\times x\\ x=\frac{440{\text{m}}^2}{22\text{m}}\\ =20\text{m}\\ \text{Perimeter of the rectangular plot}=\text{2}\left(\text{length+breadth}\right)\\ =2\left(20\text{m}+22\text{m}\right)\\ =2\left(42\text{m}\right)\\ =84\text{m}\end{array}

Q.82 The perimeter of a rectangular sheet is 100 cm. If the length is 35 cm, find its breadth. Also find the area.

Ans

\begin{array}{l}\text{Let the breadth of the rectangular sheet be}x\text{.}\\ \text{Perimeter of the rectangular sheet}=\text{2}\left(\text{length+breadth}\right)\\ 100\text{cm}=2\left(35\text{cm}+x\right)\\ \frac{100}{2}\text{cm}=\left(35\text{cm+x}\right)\\ 50\text{cm}=35\text{cm+x}\\ x=50\text{cm}-35\text{cm}\\ =15\text{cm}\\ \text{Area of rectangular sheet}=\text{length}\times \text{breadth}\\ =\text{35 cm}\times \text{15 cm}\\ ={\text{525 cm}}^2\end{array}

Q.83

$\begin{array}{l}\mathrm{The}\mathrm{area}\mathrm{of}\mathrm{a}\mathrm{square}\mathrm{park}\mathrm{is}\mathrm{the}\mathrm{same}\mathrm{as}\mathrm{of}\mathrm{a}\mathrm{rectangular}\\ \mathrm{park}.\mathrm{If}\mathrm{the}\mathrm{side}\mathrm{of}\mathrm{the}\mathrm{square}\mathrm{park}\mathrm{is}60\mathrm{m}\mathrm{and}\mathrm{the}\mathrm{length}\\ \mathrm{of}\mathrm{the}\mathrm{rectangular}\mathrm{park}\mathrm{is}90\mathrm{m},\mathrm{find}\mathrm{the}\mathrm{breadth}\mathrm{of}\mathrm{the}\\ \mathrm{rectangular}\mathrm{park}.\end{array}$

Ans

\begin{array}{l}\text{Let the breadth of rectangular park be}x\text{.}\\ \text{Since, area of a square park is the same as of a rectangular}\\ \text{park}.\\ \text{Area of square}=\text{Area of rectangular park}\\ {\left(\text{60m}\right)}^2=90m\times x\\ 3600m^2=90m\times x\\ x=\frac{3600m^2}{90m}\\ =40m\end{array}

Q.84

$\begin{array}{l}\mathrm{A}\mathrm{wire}\mathrm{is}\mathrm{in}\mathrm{the}\mathrm{shape}\mathrm{of}\mathrm{a}\mathrm{rectangle}.\mathrm{Its}\mathrm{length}\mathrm{is}40\mathrm{cm}\\ \mathrm{and}\mathrm{breadth}\mathrm{is}22\mathrm{cm}.\mathrm{If}\mathrm{the}\mathrm{same}\mathrm{wire}\mathrm{is}\mathrm{rebent}\mathrm{in}\mathrm{the}\\ \mathrm{shape}\mathrm{of}\mathrm{a}\mathrm{square},\mathrm{what}\mathrm{will}\mathrm{be}\mathrm{the}\mathrm{measure}\mathrm{of}\mathrm{each}\mathrm{side}.\\ \mathrm{Also}\mathrm{find}\mathrm{which}\mathrm{shape}\mathrm{encloses}\mathrm{more}\mathrm{area}?\end{array}$

Ans

\begin{array}{l}\text{Perimeter of a rectangle}=\text{Perimeter of square}\\ \text{2}\left(\text{length+breadth}\right)=4\times \text{side}\\ \text{2}\left(40\text{m}+22\text{m}\right)=4\times \text{side}\\ \text{side}=\frac{124}{4}\\ =31\text{cm}\\ So,\\ \text{Area of rectangle}=\text{length}\times \text{breadth}\\ =\text{40cm}\times \text{22cm}\\ ={\text{880cm}}^2\\ \text{Area of square}={\left(\text{side}\right)}^2\\ ={\left(31\text{cm}\right)}^2\\ =961{\text{cm}}^2\\ \text{Therefore, the square-shaped wire encloses more area}\text{.}\end{array}

Q.85

$\begin{array}{l}\mathrm{The}\mathrm{perimeter}\mathrm{of}\mathrm{a}\mathrm{rectangle}\mathrm{is}130\mathrm{cm}.\mathrm{If}\mathrm{the}\mathrm{breadth}\\ \mathrm{of}\mathrm{the}\mathrm{rectangle}\mathrm{is}30\mathrm{cm},\mathrm{find}\mathrm{its}\mathrm{length}.\mathrm{Also}\mathrm{find}\mathrm{the}\\ \mathrm{area}\mathrm{of}\mathrm{the}\mathrm{rectangle}.\end{array}$

Ans

\begin{array}{l}\text{Let the length of the rectangle be}x\text{cm}\text{.}\\ \text{Perimeter of rectangle}=\text{2 (length}+\text{breadth)}\\ 130\text{cm}=\text{2(30 cm}+x\text{)}\\ \frac{130}{2}\text{cm}=\text{30 cm}+x\\ x=\text{65 cm}-\text{30 cm}\\ =\text{35 cm}\\ \text{Now,}\\ \text{area of rectangle}=\text{length}\times \text{breadth}\\ =\text{30 cm}\times \text{35 cm}\\ ={\text{1050 cm}}^2\end{array}

Q.86 Find the area of each of the following parallelogram :

Ans

\begin{array}{l}\text{(a)}\\ \text{Area of Parallelogram}=\text{Base}\times \text{Height}\\ \text{Height}=\text{4 cm}\\ \text{Base}=\text{7 cm}\\ \text{So,}\\ \text{area of parallelogram}=\text{7}\times \text{4}\\ ={\text{28cm}}^2\\ \text{(b)}\\ \text{Area of Parallelogram}=\text{Base}\times \text{Height}\\ \text{Height}=\text{3 cm}\\ \text{Base}=\text{5 cm}\\ \text{So,}\\ \text{area of parallelogram}=\text{5}\times \text{3}\\ ={\text{15 cm}}^2\\ \text{(c)}\\ \text{Area of Parallelogram}=\text{Base}\times \text{Height}\\ \text{Height}=\text{3}\text{.5 cm}\\ \text{Base}=\text{2}\text{.5 cm}\\ \text{So,}\\ \text{area of parallelogram}=\text{2}\text{.5}\times \text{3}\text{.5}\\ =\text{8}{\text{.75 cm}}^2\end{array} \begin{array}{l}\text{(d)}\\ \text{Area of Parallelogram}=\text{Base}\times \text{Height}\\ \text{Height}=\text{4}\text{.8 cm}\\ \text{Base}=\text{5 cm}\\ \text{So,}\\ \text{area of parallelogram}=\text{5}\times \text{4}\text{.8}\\ ={\text{24 cm}}^2\\ \text{(e)}\\ \text{Area of Parallelogram}=\text{Base}\times \text{Height}\\ \text{Height}=\text{4}\text{.4 cm}\\ \text{Base}=\text{2 cm}\\ \text{So,}\\ \text{area of parallelogram}=\text{2}\times \text{4}\text{.4}\\ =\text{8}{\text{.8 cm}}^2\end{array}

Q.87 Find the area of each of the following triangles :


Ans

\begin{array}{l}\text{(a) Area of a triangle}=\frac{1}{2}\times \text{Base}\times \text{Height}\\ =\frac{1}{2}\times 4\times 3\\ =\frac{1}{2}\times 12=6{\text{cm}}^2\\ \text{(b) Area of a triangle}=\frac{1}{2}\times \text{Base}\times \text{Height}\\ =\frac{1}{2}\times 5\times 3.2\\ =\frac{1}{2}\times 16=8{\text{cm}}^2\\ \text{(c) Area of a triangle}=\frac{1}{2}\times \text{Base}\times \text{Height}\\ =\frac{1}{2}\times 3\times 4\\ =\frac{1}{2}\times 12=6{\text{cm}}^2\\ \text{(d) Area of a triangle}=\frac{1}{2}\times \text{Base}\times \text{Height}\\ =\frac{1}{2}\times 3\times 2\\ =\frac{1}{2}\times 6=3{\text{cm}}^2\end{array}

Q.88 Find the missing values :

S. NO.	Base	Height	Area of the Parallelogram
a.	20 cm		246 cm2
b.		15 cm	154.5 cm2
c.		8.4 cm	78.72 cm2
d.	15.6 cm		16.38 cm2

Ans

\begin{array}{l}\text{Let the height be}h\text{and base be}b.\\ \text{(a) Area of parallelogram=Base}\times \text{Height}\\ \text{So,we get}\\ {\text{246 cm}}^{\text{2}}=\text{20 cm}\times \text{h}\\ \text{h=}\frac{246{\text{cm}}^2}{20\text{cm}}\\ =12.3\text{cm}\\ \text{(b) Area of parallelogram=Base}\times \text{Height}\\ \text{So,we get}\\ \text{154}{\text{.5 cm}}^{\text{2}}=\text{b}\times \text{15 cm}\\ \text{b=}\frac{154.5{\text{cm}}^2}{15\text{cm}}\\ =10.3\text{cm}\end{array} \begin{array}{l}\text{(c) Area of parallelogram=Base}\times \text{Height}\\ \text{So,we get}\\ \text{78}{\text{.72 cm}}^{\text{2}}=\text{b}\times \text{8}\text{.4 cm}\\ \text{b=}\frac{48.72{\text{cm}}^2}{8.4\text{cm}}\\ =5.8\text{cm}\\ \text{(d) Area of parallelogram=Base}\times \text{Height}\\ \text{So,we get}\\ \text{16}{\text{.38 cm}}^{\text{2}}=\text{15}\text{.6 cm}\times \text{h}\\ \text{h=}\frac{16.38{\text{cm}}^2}{15.6\text{cm}}\\ =1.05\text{cm}\\ \text{So, we get}\\ \begin{array}{cccc}\text{S}\text{.No}\text{.}& \text{Base}& \text{Height}& \text{Area of the Parallelogram}\\ \text{a}\text{.}& 20\text{cm}& 12.3\text{cm}& 246{\text{cm}}^2\\ \text{b}\text{.}& 10.3\text{cm}& 15\text{cm}& 154.5{\text{cm}}^2\\ \text{c}\text{.}& 5.8\text{cm}& 8.4\text{cm}& 78.72{\text{cm}}^2\\ \text{d}\text{.}& 15.6\text{cm}& 1.05\text{cm}& 16.38{\text{cm}}^2\end{array}\end{array}

Q.89 Find the missing values :

Base	Height	Area of Triangle
15 cm	……………..	84 cm2
……………..	31.4 mm	1256 mm2
22 cm	……………..	170.5 cm2

Ans

\begin{array}{l}\text{(a) Let the height be}h\text{and base be}b\text{.}\\ \text{Area of triangle}=\frac{1}{2}\times \text{base}\times \text{height}\\ {\text{87cm}}^2=\frac{1}{2}\times 15\text{cm}\times \text{h}\\ \text{h}=\frac{87{\text{cm}}^2\times 2}{15\text{cm}}\\ =11.6\text{cm}\\ \text{(b) Let the height be}h\text{and base be}b\text{.}\\ \text{Area of triangle}=\frac{1}{2}\times \text{base}\times \text{height}\\ {\text{1256 mm}}^2=\frac{1}{2}\times b\times \text{31}\text{.4 mm}\\ b=\frac{1256{\text{mm}}^2\times 2}{31.4\text{mm}}\\ =80\text{mm}\end{array} \begin{array}{l}\text{(c) Let the height be}h\text{and base be}b\text{.}\\ \text{Area of triangle}=\frac{1}{2}\times \text{base}\times \text{height}\\ \text{170}{\text{.5 cm}}^2=\frac{1}{2}\times 22\text{cm}\times \text{h}\\ \text{h}=\frac{170.5{\text{cm}}^2\times 2}{22\text{cm}}\\ =15.5\text{cm}\\ \text{So we get}\\ \begin{array}{ccc}\text{Base}& \text{Height}& \text{Area of Triangle}\\ 15\text{cm}& 11.6\text{cm}& 84{\text{cm}}^2\\ 80\text{mm}& 31.4\text{mm}& 1256{\text{mm}}^2\\ 22\text{cm}& 15.5\text{cm}& 170.5{\text{cm}}^2\end{array}\end{array}

Q.90

$\begin{array}{l}\mathrm{PQRS}\mathrm{is}\mathrm{a}\mathrm{parallelogram}.\mathrm{QM}\mathrm{is}\mathrm{the}\mathrm{height}\mathrm{from}\mathrm{Q}\mathrm{to}\mathrm{SR}\\ \mathrm{and}\mathrm{QN}\mathrm{is}\mathrm{the}\mathrm{height}\mathrm{from}\mathrm{Q}\mathrm{to}\mathrm{PS}.\mathrm{If}\mathrm{SR}=12\mathrm{cm}\mathrm{and}\\ \mathrm{QM}=\; 7.6\mathrm{cm}.\mathrm{Find}:\\ \left(\mathrm{a}\right)\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{parallegram}\mathrm{PQRS}\\ \left(\mathrm{b}\right)\mathrm{QN},\mathrm{if}\mathrm{PS}=\; 8\mathrm{cm}\end{array}$

Ans

\begin{array}{l}\text{(a)}\\ \text{Area of a parallelogram}=\text{Base}\times \text{Height}\\ =\text{SR}\times \text{QM}\\ =7.6\times 12\\ =91.2{\text{cm}}^2\\ \text{(b)}\\ \text{Area of a parallelogram}=\text{Base}\times \text{Height}\\ =\text{PS}\times \text{QN}\\ =91.2{\text{cm}}^2\\ \text{QN}\times 8=91.2\\ \text{QN}=\frac{91.2}{8}\\ =11.4\text{cm}\end{array}

Q.91

$\begin{array}{l}\mathrm{DL}\mathrm{and}\mathrm{BM}\mathrm{are}\mathrm{the}\mathrm{heights}\mathrm{on}\mathrm{sides}\mathrm{AB}\mathrm{and}\mathrm{AD}\mathrm{respectively}\\ \mathrm{of}\mathrm{parallelogram}\mathrm{ABCD}.\mathrm{If}\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{parallelogram}\mathrm{is}\\ 1470{\mathrm{cm}}^{\mathrm{2}},\mathrm{AB}=\; 35\mathrm{cm}\mathrm{and}\mathrm{AD}=\; 49\mathrm{cm},\mathrm{find}\mathrm{the}\mathrm{length}\mathrm{of}\\ \mathrm{BMand}\mathrm{DL}.\end{array}$

Ans

\begin{array}{l}\text{Area of parallelogram}=\text{Base}\times \text{Height}\\ =\text{AB}\times \text{DL}\\ \text{1470}=\text{35}\times \text{DL}\\ \text{DL}=\frac{1470}{35}\\ =42\text{cm}\\ \text{Also,}\\ 1470=\text{AD}\times \text{BM}\\ \text{1470}=\text{49}\times \text{BM}\\ \text{BM}=\frac{1470}{49}\\ =30\text{cm}\end{array}

Q.92

$\begin{array}{l}â–³\mathrm{ABC}\mathrm{is}\mathrm{right}\mathrm{angled}\mathrm{at}\mathrm{A}.\mathrm{AD}\mathrm{is}\mathrm{perpendicular}\mathrm{to}\mathrm{BC}.\\ \mathrm{If}\mathrm{AB}=5\mathrm{cm},\mathrm{BC}=13\mathrm{cm}\mathrm{and}\mathrm{AC}=12\mathrm{cm},\mathrm{Find}\mathrm{the}\mathrm{area}\mathrm{of}\\ â–³\mathrm{ABC}.\mathrm{Also}\mathrm{find}\mathrm{the}\mathrm{length}\mathrm{of}\mathrm{AD}.\end{array}$

Ans

\begin{array}{l}\text{Since, Area}=\frac{1}{2}\times \text{Base}\times \text{Height}\\ =\frac{1}{2}\times \text{5}\times \text{12}\\ ={\text{30 cm}}^2\\ \text{Also, area of triangle}=\frac{1}{2}\times \text{AD}\times \text{BC}\\ \text{30}=\frac{1}{2}\times \text{AD}\times \text{13}\\ \text{AD}=\frac{30\times 2}{12}\\ =4.6\text{cm}\end{array}

Q.93

$\begin{array}{l}â–³\mathrm{ABC}\mathrm{is}\mathrm{isosceles}\mathrm{with}\mathrm{AB}=\mathrm{AC}=7.5\mathrm{cm}\mathrm{and}\mathrm{BC}=9\mathrm{cm}.\\ \mathrm{The}\mathrm{height}\mathrm{AD}\mathrm{from}\mathrm{A}\mathrm{to}\mathrm{BC},\mathrm{is}6\mathrm{cm}.\mathrm{Find}\mathrm{the}\mathrm{area}\mathrm{of}â–³\mathrm{ABC}.\\ \mathrm{What}\mathrm{will}\mathrm{be}\mathrm{the}\mathrm{height}\mathrm{from}\mathrm{C}\mathrm{to}\mathrm{AB}\mathrm{i}.\mathrm{e}.,\mathrm{CE}?\end{array}$

Ans

\begin{array}{l}\text{Area of}\Delta \text{ABC}=\frac{1}{2}\times \text{Base}\times \text{Height}\\ =\frac{1}{2}\text{BC}\times \text{AD}=\frac{1}{2}\times \text{9}\times \text{6}={\text{27cm}}^{\text{2}}\\ \text{Also, area of}\Delta \text{ABC}=\frac{1}{2}\times \text{Base}\times \text{Height}\\ =\frac{1}{2}\text{AB}\times \text{CE}\\ 27=\frac{1}{2}\text{7}\text{.5}\times \text{CE}\\ \text{CE}=\frac{27\times 2}{7.5}=\text{7}\text{.2 cm}\end{array}

Q.94

$\begin{array}{l}\mathrm{Find}\mathrm{the}\mathrm{circumference}\mathrm{of}\mathrm{the}\mathrm{circles}\mathrm{with}\mathrm{the}\mathrm{following}\\ \mathrm{radius}:\left(\mathrm{Take}\mathrm{\pi }\mathrm{=}\frac{\mathrm{22}}{\mathrm{7}}\right)\\ \left(\mathrm{a}\right)14\mathrm{cm}\left(\mathrm{b}\right)28\mathrm{mm}\left(\mathrm{c}\right)21\mathrm{cm}\end{array}$

Ans

\begin{array}{l}\text{(a)}\\ \text{Circumference of a circle is}=\text{2}\pi \text{r}\\ =\text{2}\times \frac{22}{7}\times 14\\ =2\times 22\times 2\\ =88\text{cm}\\ \text{(b)}\\ \text{Circumference of a circle is}=\text{2}\pi \text{r}\\ =\text{2}\times \frac{22}{7}\times 28\\ =2\times 22\times 4\\ =176\text{mm}\end{array} \begin{array}{l}\text{(c)}\\ \text{Circumference of a circle is}=\text{2}\pi \text{r}\\ =\text{2}\times \frac{22}{7}\times 21\\ =2\times 22\times 3\\ =132\text{cm}\end{array}

Q.95

$\begin{array}{l}\mathrm{Find}\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{following}\mathrm{circles},\mathrm{given}\mathrm{that}:\\ \left(\mathrm{a}\right)\mathrm{radius}=\; 14\mathrm{mm}\left(\mathrm{Take}\mathrm{\pi }\mathrm{=}\frac{\mathrm{22}}{\mathrm{7}}\right)\\ \left(\mathrm{b}\right)\mathrm{diameter}=\; 49\mathrm{m}\left(\mathrm{c}\right)\mathrm{radius}=\; 5\mathrm{cm}\end{array}$

Ans

\begin{array}{l}\text{(a)}\\ \text{Area of a circle}=\pi {\text{r}}^2\\ =\frac{22}{7}{\left(14\right)}^2\\ =616{\text{mm}}^2\end{array} \begin{array}{l}\text{(b)}\\ \text{Radius}=\frac{\text{Diameter}}{2}=\frac{49}{2}=24.5\text{mm}\\ \text{Area of a circle}=\pi {\text{r}}^2\\ =\frac{22}{7}{\left(24.5\right)}^2\\ =1886.5{\text{m}}^2\\ \text{(c)}\\ \text{Area of a circle}=\pi {\text{r}}^2\\ =\frac{22}{7}{\left(5\right)}^2\\ =\frac{550}{7}{\text{cm}}^2\end{array}

Q.96 If the circumference of a circular sheet is 154 m, find its radius. Also find the area of the sheet. (Take π =

$\frac{22}{7}$

)

Ans

\begin{array}{l}\text{Since, circumference}=\text{2}\pi \text{r}\\ \text{So, we get}\end{array} \begin{array}{l}\text{154}=\text{2}\times \frac{22}{7}\times r\\ r=\frac{154\times 7}{2\times 22}\\ =\frac{49}{2}\text{m}=4.5\text{m}\\ \text{Now,}\\ \text{Area}=\pi {\text{r}}^2\\ =\frac{22}{7}\times \frac{49}{2}\times \frac{49}{2}\\ =1886.5{\text{m}}^2\end{array}

Q.97 From a circular sheet of radius 4 cm, a circle of radius 3 cm is removed. Find the area of the remaining sheet. (Take π = 3.14)

Ans

\begin{array}{l}\text{Outer radius of circular sheet}=\text{4 cm}\\ \text{Inner radius of circular sheet}=\text{3 cm}\end{array} \begin{array}{l}\text{Remaining area}=-\left(3.14\times 3\times 3\right)\\ =50.24-28.26\\ =21.98{\text{cm}}^2\end{array}

Q.98
< Find the perimeter of the adjoining figure, which is a semicircle including its diameter.

Ans

\begin{array}{l}\text{Radius}=\text{5 cm}\\ \text{Length of curved part}=\pi \text{r}\\ =\frac{22}{7}\times 5\\ =15.71\text{cm}\\ \text{Total permieter}=\text{Length of the curved part+Length of diameter}\\ =\text{15}\text{.71+10}\\ =\text{25}\text{.71 cm}\end{array}

Q.99

$\begin{array}{l}\mathrm{Shazli}\mathrm{took}\mathrm{a}\mathrm{wire}\mathrm{of}\mathrm{length}44\mathrm{cm}\mathrm{and}\mathrm{bent}\mathrm{it}\mathrm{into}\mathrm{the}\\ \mathrm{shape}\mathrm{of}\mathrm{a}\mathrm{circle}.\mathrm{Find}\mathrm{the}\mathrm{radius}\mathrm{of}\mathrm{that}\mathrm{circle}.\mathrm{Also}\mathrm{find}\\ \mathrm{its}\mathrm{area}.\mathrm{If}\mathrm{the}\mathrm{same}\mathrm{wire}\mathrm{is}\mathrm{bent}\mathrm{into}\mathrm{the}\mathrm{shape}\mathrm{of}\mathrm{a}\\ \mathrm{square},\mathrm{what}\mathrm{will}\mathrm{be}\mathrm{the}\mathrm{length}\mathrm{of}\mathrm{each}\mathrm{of}\mathrm{its}\mathrm{sides}?\\ \mathrm{Which}\mathrm{figure}\mathrm{encloses}\mathrm{more}.\end{array}$

Ans

\begin{array}{l}\text{Circumference}=\text{2}\pi \text{r}=\text{44 cm}\\ \text{2}\times \frac{22}{7}\times r=44\\ r=7\text{cm}\end{array} \begin{array}{l}\text{Area}=\pi {\text{r}}^2\\ =\frac{22}{7}\times 7\times 7\\ =154{\text{cm}}^2\\ \text{If the wire is bent into a square, then the length of each}\\ \text{side would be}=\frac{44}{4}=11\text{cm}\\ \text{Area of square}={\left(11\right)}^2\\ =121{\text{cm}}^2\\ \text{Therefore, circle encloses more area}\text{.}\end{array}

Q.100

$\begin{array}{l}\mathrm{From}\mathrm{a}\mathrm{circular}\mathrm{card}\mathrm{sheet}\mathrm{of}\mathrm{radius}14\mathrm{cm},\mathrm{two}\mathrm{circles}\mathrm{of}\\ \mathrm{radius}3.5\mathrm{cm}\mathrm{and}\mathrm{a}\mathrm{rectangle}\mathrm{of}\mathrm{length}3\mathrm{cm}\mathrm{and}\mathrm{breadth}\\ 1\mathrm{cm}\mathrm{are}\mathrm{removed}.\; (\mathrm{as}\mathrm{shown}\mathrm{in}\mathrm{the}\mathrm{adjoiningfigure}).\\ \mathrm{Find}\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{remaining}\mathrm{sheet}.\left(\mathrm{Take\pi }\mathrm{=}\frac{\mathrm{22}}{\mathrm{7}}\right)\end{array}$

Ans

\begin{array}{l}\text{Area of bigger cirle}=\frac{22}{7}\times 14\times 14\\ =616{\text{cm}}^2\\ \text{Area of 2 small circles}=\text{2}\times \pi {\text{r}}^2\\ =2\times \frac{22}{7}\times 3.5\times 3.5\\ =77{\text{cm}}^2\\ \text{Area of rectangle}=\text{Length}\times \text{Breadth}\\ =\text{3}\times \text{1}={\text{3 cm}}^2\\ \text{Remaining area of sheet}={\text{616 cm}}^2-{\text{77 cm}}^2-{\text{3 cm}}^2\\ ={\text{536 cm}}^2\end{array}

Q.101

$\begin{array}{l}\mathrm{A}\mathrm{circle}\mathrm{of}\mathrm{radius}2\mathrm{cm}\mathrm{is}\mathrm{cut}\mathrm{out}\mathrm{from}\mathrm{a}\mathrm{square}\mathrm{piece}\mathrm{of}\mathrm{an}\\ \mathrm{aluminium}\mathrm{sheet}\mathrm{of}\mathrm{side}6\mathrm{cm}.\mathrm{What}\mathrm{is}\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{left}\\ \mathrm{over}\mathrm{aluminium}\mathrm{sheet}?\left(\mathrm{Take}\mathrm{\pi }=3.14\right)\end{array}$

Ans

\begin{array}{l}\text{Area of square-shaped sheet}={\left(\text{side}\right)}^2\\ ={\left(6{\text{cm}}^2\right)}^2=36{\text{cm}}^2\\ \text{Area of circle}=\text{3}\text{.14}\times 2\times 2\\ =12.56{\text{cm}}^2\\ \text{Remaining area}={\text{36 cm}}^{\text{2}}-\text{12}{\text{.56 cm}}^{\text{2}}\\ =\text{23}{\text{.44 cm}}^{\text{2}}\end{array}

Q.102 The circumference of a circle is 31.4 cm. Find the radius and the area of the circle ? (Take π = 3.14)

Ans

\begin{array}{l}\text{Circumference}=\text{2}\pi r=31.4\text{cm}\\ \text{2}\times \text{3}\text{.14}\times \text{r}=\text{31}\text{.4}\\ \text{r}=\frac{31.4}{2\times 3.14}\\ =5\text{cm}\\ \text{So, Area}=\text{3}\text{.14}\times \text{5}\times \text{5}\\ =\text{78}{\text{.50 cm}}^2\end{array}

Q.103 A circular flower bed is surrounded by a path 4 m wide. The diameter of the flower bed is 66 m. What is the area of this path? (π = 3.14)

Ans

\begin{array}{l}\text{Radius of the flower bed}=\frac{66}{2}=33\text{m}\\ \text{So, radius of flower bed and path together}=\text{33+4}\\ =\text{37m}\\ \text{Area of flower baed and path together}=\text{3}\text{.14}\times 37\times 37\\ =4298.66{\text{m}}^2\\ \text{Area of flower bed}=\text{3}\text{.14}\times \text{33}\times \text{33}\\ =\text{3419}{\text{.46 m}}^2\\ \text{Area of path}=\text{4298}{\text{.66 m}}^{\text{2}}-\text{3419}{\text{.46m}}^{\text{2}}\\ =\text{879}{\text{.20m}}^2\end{array}

Q.104

$\begin{array}{l}\mathrm{A}\mathrm{circular}\mathrm{flower}\mathrm{garden}\mathrm{has}\mathrm{an}\mathrm{area}\mathrm{of}314{\mathrm{m}}^{\mathrm{2}}.\mathrm{A}\mathrm{sprinkler}\\ \mathrm{at}\mathrm{the}\mathrm{centre}\mathrm{of}\mathrm{the}\mathrm{garden}\mathrm{can}\mathrm{cover}\mathrm{an}\mathrm{area}\mathrm{that}\mathrm{has}\mathrm{a}\\ \mathrm{radius}\mathrm{of}12\mathrm{m}.\mathrm{Will}\mathrm{the}\mathrm{sprinkler}\mathrm{water}\mathrm{the}\mathrm{entire}\mathrm{garden}?\\ \left(\mathrm{Take\pi }=3.14\right)\end{array}$

Ans

\begin{array}{l}\text{Area}=\pi {\text{r}}^2=314{\text{m}}^2\\ 3.14\times r^2=314\\ r^2=100\\ r=10\text{m}\\ \text{Yes, the sprinkle will water the whole garden}\text{.}\end{array}

Q.105 Find the circumference of the inner and the outer circles, shown in the adjoining figure? (Take π = 3.14)

Ans

\begin{array}{l}\text{Radius of outer circle}=\text{19 m}\\ \text{Circumference}=\text{2}\pi \text{r}\\ =\text{2}\times \text{3}\text{.14}\times \text{19}\\ =\text{119}\text{.32 m}\\ \text{Radius of the inner circle}=\text{19}-\text{10}=\text{9 m}\\ \text{Circumference}=\text{2}\pi \text{r}\\ =\text{2}\times \text{3}\text{.14}\times \text{9}\\ =\text{56}\text{.52 m}\end{array}

Q.106 How many times a wheel of radius 28 cm must rotate to go 352 m? (take π =

$\frac{22}{7}$

)

Ans

\text{Radius, r}=\text{28 cm} \begin{array}{l}\text{Circumference}=\text{2}\pi \text{r}\\ =\text{2}\times \frac{22}{7}\times 28\\ =176\text{cm}\\ \text{Number of rotatinos}=\frac{\text{Total distance to be covered}}{\text{Circumference of the wheel}}\\ =\frac{352\text{m}}{176\text{cm}}\\ =\frac{35200}{176}\\ =200\\ \text{Therefore, it will rotate 200 times}\text{.}\end{array}

Q.107 The minute hand of a circular clock is 15 cm long. How far does the tip of the minute hand move in 1 hour. (Take π = 3.14)

Ans

\begin{array}{l}\text{Since, Distance travelled by the tip of minute hand}\\ =\text{Circumference of the clock}\\ =\text{2}\pi \text{r}\\ =\text{2}\times \text{3}\text{.14}\times 1\text{5}\\ =\text{94}\text{.2 cm}\\ \text{Therfore, minute hand move 94}\text{.2 cm in 1 hour}\text{.}\end{array}

Q.108 A garden is 90 m long and 75 m broad. A path 5 m wide is to be built outside and around it. Find the area of the path. Also find the area of the garden in hectare.

Ans

\begin{array}{l}\text{Length (L) of garden}=\text{90 m}\\ \text{Breadth (B) of garden}=\text{75 m}\\ \text{Area of garden}=\text{L}\times \text{B}\\ =\text{90 m}\times \text{75 m}\\ ={\text{6750 m}}^2\end{array} \begin{array}{l}\text{From the figure, it can be observed that the new length}\\ \text{and breadth of the garden,when path is also included are}\\ \text{100 m and 85 m respectively}\\ \text{Area of the garden including path}=\text{100 m}\times \text{80 m}\\ ={\text{8000 m}}^2\\ \text{Area of the path}\\ =\text{Area of the garden including path}-\text{Area of garden}\\ =8000{\text{m}}^2-{\text{6750 m}}^2\\ =1750{\text{m}}^2\\ 1\text{hectare}={\text{10000 m}}^2\\ \text{Therefore,}\\ \text{area of the garden in hectare}=\frac{6750}{10000}\\ =0.675\text{hectare}\end{array}

Q.109 A 3 m wide path runs outside and around a rectangular park of length 125 m and breadth 65m. Find the area of the path.

Ans

\begin{array}{l}\text{Length (L) of park}=\text{125 m}\\ \text{Breadth (B) of park}=\text{65 m}\\ \text{Area of park}=\text{L}\times \text{B}\\ =\text{125 m}\times \text{65 m}\\ ={\text{8125 m}}^2\end{array} \begin{array}{l}\text{From the figure, it can be observed that the new length}\\ \text{and breadth of the park,when path is also included are}\\ \text{131 m and 71 m respectively}\\ \text{Area of the park including path}=\text{131 m}\times \text{71 m}\\ ={\text{9301 m}}^2\\ \text{Area of the path}\\ =\text{Area of the park including path}-\text{Area of park}\\ =9301{\text{m}}^2-{\text{8125 m}}^2\\ =1176{\text{m}}^2\end{array}

Q.110 A picture is painted on a cardboard 8 cm long and 5 cm wide such that there is a margin of 1.5 cm along each of its sides. Find the total area of the margin.

Ans

\begin{array}{l}\text{Length (L) of cardboard}=\text{125 m}\\ \text{Breadth (B) of cardboard}=\text{65 m}\\ \text{Area of cardboard including margin}=\text{L}\times \text{B}\\ =\text{8 cm}\times \text{5 cm}\\ =40{\text{cm}}^2\end{array} \begin{array}{l}\text{From the figure, it can be observed that the new length}\\ \text{and breadth of the cardboard,when margin is not included}\\ \text{are 5 cm and 2 cm respectively}\\ \text{Area of the cardboard not including margin}=\text{5 cm}\times \text{2 m}\\ ={\text{10 cm}}^2\\ \text{Area of the margin}\\ =\text{Area of the cardboard including margin}\\ -\text{Area of cardboard not including margin}\\ =40{\text{cm}}^2-{\text{10 cm}}^2\\ =30{\text{cm}}^2\end{array}

Q.111

$\begin{array}{l}\mathrm{Two}\mathrm{cross}\mathrm{roads},\mathrm{each}\mathrm{of}\mathrm{width}10\mathrm{m},\mathrm{cut}\mathrm{at}\mathrm{right}\mathrm{angles}\\ \mathrm{through}\mathrm{the}\mathrm{centre}\mathrm{of}\mathrm{a}\mathrm{rectangular}\mathrm{park}\mathrm{of}\mathrm{length}700\mathrm{m}\\ \mathrm{and}\mathrm{breadth}300\mathrm{m}\mathrm{and}\mathrm{parallel}\mathrm{to}\mathrm{its}\mathrm{sides}.\mathrm{Find}\mathrm{the}\mathrm{area}\\ \mathrm{of}\mathrm{the}\mathrm{roads}.\mathrm{Also}\mathrm{find}\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{park}\mathrm{excluding}\mathrm{cross}\\ \mathrm{roads}.\mathrm{Give}\mathrm{the}\mathrm{answer}\mathrm{in}\mathrm{hectares}.\end{array}$

Ans

\begin{array}{l}\text{Length (L) of the park}=\text{700 m}\\ \text{Breadth (B) of the park}=\text{300 m}\\ \text{Area of park}=\text{700}\times \text{300}\\ ={\text{210000 m}}^2\\ \text{Length of road PQRS}=\text{700 m}\\ \text{Length of road ABCD}=\text{300 m}\\ \text{Width of each road}=\text{10 m}\\ \text{Area of the roads}=\text{area(PQRS)+area(ABCD)}-\text{area(KLMN)}\\ =\left(700\times 10\right)+\left(300\times 10\right)-\left(10\times 10\right)\\ =7000+3000-100\\ =9900{\text{m}}^2\\ =0.99\text{hectare}\end{array} \begin{array}{l}\\ \text{Area of park excluding roads}=\text{210000-9900}\\ ={\text{200100 m}}^2\\ =20.01\text{hectare}\end{array}

Q.112 Pragya wrapped a cord around a circular pipe of radius 4 cm (adjoining figure) and cut off the length required of the cord. Then she wrapped it around a square box of side 4 cm (also shown). Did she have any left? (π = 3.14)

Ans

\begin{array}{l}\text{Perimeter of the square}=\text{4}\times \text{side of the square}\\ =\text{4}\times \text{4}\\ =\text{16 cm}\\ \text{Perimeter of circular pip}=\text{2}\pi \text{r}\\ =\text{3}\times \text{3}\text{.14}\times \text{4}\\ =\text{25}\text{.12 cm}\\ \text{Therefore,}\\ \text{length of chord left with Pragya}=\text{25}\text{.12 cm}-\text{16 cm}\\ =\text{9}\text{.12 cm}\end{array}

Q.113 The adjoining figure represents a rectangular lawn with a circular flower bed in the middle. Find:
(i) the area of the whole land
(ii) the area of the flower bed
(iii) the area of the lawn excluding the area of the flower bed.
(iv) the circumference of the flower bed.

Ans

\begin{array}{l}\text{(i) Area of whole land}=\text{Length}\times \text{Breadth}\\ =\text{10}\times \text{5}\\ ={\text{50 m}}^2\\ \text{(ii) Area of flower bed}=\pi {\text{r}}^2\\ =3.14\times 2\times 2\\ =12.56{\text{m}}^2\\ \text{(iii) Area of the lawn excluding the flower bed}\\ =\text{Area of whole land}-\text{Area of flower bed}\\ =\text{50-12}\text{.56}\\ =\text{37}{\text{.44 m}}^2\end{array}

Q.114 In the following figures, find the area of the shaded portions:

Ans

\begin{array}{l}\text{(i)}\\ \text{Area of EFDC}=\text{area(ABCD)}-\text{area(BCE)}–\text{area(AFE)}\\ =\left(18\times 10\right)-\frac{1}{2}\left(10\times 8\right)-\frac{1}{2}\left(6\times 10\right)\\ =180-40-30\\ =110{\text{cm}}^2\\ \text{(ii)}\\ \text{area(QTU)}=\text{area(PQRS)}-\text{area(TSU)}-\text{area(RUQ)}-\text{area(PQT)}\\ =\left(20\times 20\right)-\frac{1}{2}\left(10\times 10\right)-\frac{1}{2}\left(20\times 10\right)-\frac{1}{2}\left(20\times 10\right)\\ =400-50-100-100\\ =150{\text{cm}}^2\end{array}

Q.115

$\begin{array}{l}\mathrm{Find}\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{quadrilateral}\mathrm{ABCD}.\\ \mathrm{Here},\mathrm{AC}=22\mathrm{cm},\mathrm{BM}=3\mathrm{cm},\mathrm{DN}=3\mathrm{cm},\\ \mathrm{and}\mathrm{BM}\perp \mathrm{AC},\mathrm{DN}\perp \mathrm{AC}.\end{array}$

Ans

\begin{array}{l}\text{area(ABCD)}=\text{area(ABC)+area(ADC)}\\ =\frac{1}{2}\left(3\times 22\right)\frac{1}{2}\left(3\times 22\right)\\ =33+33\\ =66{\text{cm}}^2\end{array}

Q.116

$\begin{array}{l}\mathrm{Get}\mathrm{the}\mathrm{algebraic}\mathrm{expressions}\mathrm{in}\mathrm{the}\mathrm{following}\mathrm{cases}\mathrm{using}\\ \mathrm{variables},\mathrm{constants}\mathrm{and}\mathrm{arithmetic}\mathrm{operations}.\\ \left(\mathrm{i}\right)\mathrm{Subtraction}\mathrm{of}\mathrm{z}\mathrm{from}\mathrm{y}.\\ \left(\mathrm{ii}\right)\mathrm{One}–\mathrm{half}\mathrm{of}\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{numbers}\mathrm{x}\mathrm{and}\mathrm{y}.\\ \left(\mathrm{iii}\right)\mathrm{The}\mathrm{number}\mathrm{z}\mathrm{multiplied}\mathrm{by}\mathrm{itself}.\\ \left(\mathrm{iv}\right)\mathrm{One}–\mathrm{fourth}\mathrm{of}\mathrm{the}\mathrm{product}\mathrm{of}\mathrm{numbers}\mathrm{p}\mathrm{and}\mathrm{q}.\\ \left(\mathrm{v}\right)\mathrm{Numbersxandyboth}\mathrm{squared}\mathrm{and}\mathrm{added}.\\ \left(\mathrm{vi}\right)\mathrm{Number}5\mathrm{added}\mathrm{to}\mathrm{three}\mathrm{times}\mathrm{the}\mathrm{product}\mathrm{of}\\ \mathrm{numbers}\mathrm{m}\mathrm{and}\mathrm{n}.\\ \left(\mathrm{vii}\right)\mathrm{Product}\mathrm{of}\mathrm{numbers}\mathrm{y}\mathrm{and}\mathrm{z}\mathrm{subtracted}\mathrm{from}10.\\ \left(\mathrm{viii}\right)\mathrm{Sum}\mathrm{of}\mathrm{numbers}\mathrm{a}\mathrm{and}\mathrm{b}\mathrm{subtracted}\mathrm{from}\mathrm{their}\\ \mathrm{product}.\end{array}$

Ans

\begin{array}{l}\text{(i) y}-\text{z}\\ \text{(ii)}\frac{1}{2}\left(x+y\right)\end{array} \begin{array}{l}{\text{(iii) z}}^2\\ (\text{iv)}\frac{1}{4}\left(pq\right)\\ {\text{(v) x}}^2+y^2\\ \text{(vi) 5+3}\left(mn\right)\\ \text{(vii) 10}-\text{yz}\\ \text{(viii) ab}-\left(a+b\right)\end{array}

Q.117

$\begin{array}{l}\left(\mathrm{i}\right)\mathrm{Identify}\mathrm{the}\mathrm{terms}\mathrm{and}\mathrm{their}\mathrm{factors}\mathrm{in}\mathrm{the}\mathrm{following}\\ \mathrm{expressions}.\\ \mathrm{Show}\mathrm{the}\mathrm{terms}\mathrm{and}\mathrm{factors}\mathrm{by}\mathrm{tree}\mathrm{diagrams}.\\ \left(\mathrm{a}\right)\mathrm{x}–3\left(\mathrm{b}\right)1+\mathrm{x}+{\mathrm{x}}^2\left(\mathrm{c}\right)\mathrm{y}–{\mathrm{y}}^3\\ \left(\mathrm{d}\right)5{\mathrm{xy}}^2+7{\mathrm{x}}^2\mathrm{y}\left(\mathrm{e}\right)–\mathrm{ab}+2{\mathrm{b}}^2–3{\mathrm{a}}^2\\ \left(\mathrm{ii}\right)\mathrm{Identify}\mathrm{terms}\mathrm{and}\mathrm{factors}\mathrm{in}\mathrm{the}\mathrm{expressions}\mathrm{given}\\ \mathrm{below}:\\ \left(\mathrm{a}\right)–4\mathrm{x}+5\left(\mathrm{b}\right)–4\mathrm{x}+5\mathrm{y}\left(\mathrm{c}\right)5\mathrm{y}+3{\mathrm{y}}^2\\ \left(\mathrm{d}\right)\mathrm{xy}+2{\mathrm{x}}^2{\mathrm{y}}^2\left(\mathrm{e}\right)\mathrm{pq}+\mathrm{q}\left(\mathrm{f}\right)1.2\mathrm{ab}–2.4\mathrm{b}+3.6\mathrm{a}\\ \left(\mathrm{g}\right)\frac{3}{4}\mathrm{x}+\frac{1}{4}\left(\mathrm{h}\right)0.1{\mathrm{p}}^2+0.2{\mathrm{q}}^2\end{array}$

Ans

(i)

a.