NCERT Solutions Class 7 Maths Chapter 6

Q.1

$\mathrm{In}\mathrm{\Delta }\mathrm{PQR},\mathrm{Disthemid}–\mathrm{point}\mathrm{of}\overline{\mathrm{QR}.}$

\begin{array}{l}\overline{\text{PM}}\text{is}\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_.\\ \overline{\text{PD}}\text{is}\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_.\\ \text{Is QM}=\text{MR}?\end{array}

Ans

\begin{array}{l}\overline{\text{PM}}\text{is}\boxed{\underset{\_}{\text{Altitude}}}.\\ \overline{\text{PD}}\text{is}\boxed{\underset{\_}{\text{Median}}}.\\ \text{No}\text{.}\end{array}

Q.2

$\begin{array}{l}\mathrm{Draw}\mathrm{rough}\mathrm{sketches}\mathrm{for}\mathrm{the}\mathrm{following}:\\ \left(\mathrm{a}\right)\mathrm{In}\mathrm{\Delta ABC},\mathrm{BE}\mathrm{is}\mathrm{a}\mathrm{median}.\\ \left(\mathrm{b}\right)\mathrm{In}\mathrm{\Delta }\mathrm{PQR},\mathrm{PQ}\mathrm{and}\mathrm{PR}\mathrm{are}\mathrm{a}\mathrm{ltitudes}\mathrm{of}\mathrm{the}\mathrm{triangle}.\\ \left(\mathrm{c}\right)\mathrm{In}\mathrm{\Delta }\mathrm{XYZ},\mathrm{YL}\mathrm{is}\mathrm{an}\mathrm{altitude}\mathrm{in}\mathrm{the}\mathrm{exterior}\mathrm{of}\mathrm{the}\mathrm{triangle}.\end{array}$

Ans

(a)

(b)

(c)

For Δ XYZ, YL is an altitude drawn exterior to side XZ which is extended up to point L.

Q.3

$\begin{array}{l}\mathrm{Verify}\mathrm{by}\mathrm{drawing}\mathrm{a}\mathrm{diagram}\mathrm{if}\mathrm{the}\mathrm{median}\mathrm{and}\mathrm{altitude}\mathrm{of}\mathrm{an}\\ \mathrm{isosceles}\mathrm{triangle}\mathrm{can}\mathrm{be}\mathrm{same}.\end{array}$

Ans

Draw a line segment AD perpendicular to BC. It is an altitude for the triangle.
The length of BD and DC are same.
Therefore, AD is also the median of this triangle.

Q.4

$\begin{array}{l}\text{Find the value of the unknown exterior angle x in the}\\ \text{following diagrams:}\end{array}$

Ans

\begin{array}{c}\text{(i)}x=70°+50°\left(\text{exterior angle theorem}\right)\\ =\boxed{120°}\\ \text{(ii)}x=65°+45°\left(\text{exterior angle theorem}\right)\\ =\boxed{110°}\\ \text{(iii)}x=40°+30°\left(\text{exterior angle theorem}\right)\\ =\boxed{70°}\\ \text{(iv)}x=60°+60°\left(\text{exterior angle theorem}\right)\\ =\boxed{120°}\\ \text{(v)}x=50°+50°\left(\text{exterior angle theorem}\right)\\ =\boxed{100°}\\ \text{(vi)}x=60°+30°\left(\text{exterior angle theorem}\right)\\ =\boxed{90°}\end{array}

Q.5

$\mathrm{Find}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{the}\mathrm{unknown}\mathrm{x}\mathrm{in}\mathrm{the}\mathrm{following}\mathrm{diagrams}:$

Ans

\begin{array}{l}\text{Since, the sum of all interior angles of a triangle is 180}°\text{.}\\ \text{So,}\\ \text{(i)}x+50°+60°=180°\\ x+110°=180°\\ x=180°-110°=\boxed{70°}\\ \text{(ii)}x+90°+30°=180°\\ x+120°=180°\\ x=180°-120°=\boxed{60°}\\ (iii)x+110°+30°=180°\\ x+140°=180°\\ x=180°-140°=\boxed{40°}\\ (iv)x+x+50°=180°\\ 2x+50°=180°\\ 2x=180°-50°\\ =130°\\ x=\frac{130°}{2}=\boxed{65°}\\ (v)x+x+x=180°\\ 3x=180°\\ x=\frac{180°}{3}=\boxed{60°}\\ (vi)x+2x+90°=180°\\ 3x+90°=180°\\ x=180°-90°\\ 3x=90°\\ x=\frac{90°}{3}=\boxed{30°}\end{array}

Q.6

$\begin{array}{l}\mathrm{Find}\mathrm{the}\mathrm{values}\mathrm{of}\mathrm{the}\mathrm{unknowns}\mathrm{x}\mathrm{and}\mathrm{y}\mathrm{in}\mathrm{the}\mathrm{following}\\ \mathrm{diagrams}:\end{array}$

Ans

\begin{array}{l}\text{(i)}\text{y+120}°\text{=180}°\text{(Linear pair)}\\ y=\text{180}°-\text{120}°=60°\\ \text{Now,}\\ x+y+50°=180°\text{(angle sum property)}\\ x+60°+50°=180°\\ x=180°-110°=\boxed{70°}\\ \text{(ii)}y=80°\text{(vertically opposite angle)}\\ y+x+50°=180°\\ 80°+x+50°=180°\\ x+130°=180°\\ x=180°-130°=\boxed{50°}\\ \text{(iii)}y+50°+60°=180°\text{(angle sum property)}\\ \text{y+110}°\text{=180}°\\ y=180°–110°=\boxed{\text{70}°}\\ \text{(iv)}x=60°\text{(vertically opposite angle)}\\ 30°+x+y=180°\\ 30°+60°+y=180°\\ y=180°-90°=\boxed{90°}\\ \text{(v)}y=90°\text{(vertically opposite angle)}\\ \text{x+x+y=180}°\\ 2x+90°=180°\\ 2x=180°-90°=90°\\ 2x=90°\\ x=\frac{90°}{2}=\boxed{45°}\\ \text{(vi)}y=x\text{(vertically opposite angle)}\\ a=x\text{(vertically opposite angle)}\\ b=x\text{(vertically opposite angle)}\\ a+b+y=180°\\ x+x+x=180°\\ 3x=180°\\ x=\frac{180°}{3}=60°\\ \text{So, we get}\boxed{x=y=60°}\end{array}

Q.7

$\begin{array}{l}\mathrm{Is}\mathrm{it}\mathrm{possible}\mathrm{to}\mathrm{have}\mathrm{a}\mathrm{triangle}\mathrm{with}\mathrm{the}\mathrm{following}\mathrm{sides}?\\ \left(\mathrm{i}\right)2\mathrm{cm},3\mathrm{cm},5\mathrm{cm}\\ \left(\mathrm{ii}\right)3\mathrm{cm},6\mathrm{cm},7\mathrm{cm}\\ \left(\mathrm{iii}\right)6\mathrm{cm},3\mathrm{cm},2\mathrm{cm}\end{array}$

Ans

\begin{array}{l}\text{For a triangle, the sum of lengths of either two sides is always}\\ \text{greater than the third side}\text{.}\\ \text{(i) Given sides of the triangle are 2 cm, 3 cm and 5 cm}\text{.}\\ \text{2 cm + 3 cm = 5 cm}\\ \text{So, the sum of two side is not greater than the third side}\text{.}\\ \text{Therefore, this triangle is not possible}\text{.}\\ \text{(ii) Given sides of the triangle are 3 cm, 6 cm and 7 cm}\text{.}\\ \text{3 cm + 6 cm = 9 cm}\\ \text{So, the sum of two side is greater than the third side}\text{.}\\ \text{Therefore, this triangle is possible}\text{.}\\ \text{(iii) Given sides of the triangle are 6 cm, 3 cm and 2 cm}\text{.}\\ \text{2 cm + 3 cm = 5 cm}\\ \text{So, the sum of two side is not greater than the third side}\text{.}\\ \text{Therefore, this triangle is not possible}\text{.}\end{array}

Q.8 Take any point O in the interior of a triangle PQR. Is
(i) OP + OQ > PQ?
(II) OQ + OR > QR?
( iii ) OR + OP > RP?

Ans

\begin{array}{l}\text{If O is a point in the interior of a given triangle, then three}\\ \text{triangles}\Delta \text{OPQ,}\Delta \text{OQR,}\text{and}\Delta \text{ORP can be constructed}\text{.}\\ \text{In a triangle, the sum of the lengths of either two sides}\\ \text{is always greater than the third side}\text{.}\\ \text{(i) Yes, as}\Delta \text{OPQ is a triangle with side OP, OQ and PQ}\text{.}\\ \text{So, OP + OQ > PQ}\\ \text{(ii) Yes, as}\Delta \text{OQR is a triangle with side OP, OQ and PQ}\text{.}\\ \text{So, OQ + OR > QR}\\ \text{(iii)}\text{Yes, as}\Delta \text{ORP is a triangle with side OP, OQ and PQ}\text{.}\\ \text{So, OP + OR > PR}\end{array}

Q.9

$\begin{array}{l}\mathrm{AM}\mathrm{is}\mathrm{a}\mathrm{median}\mathrm{of}\mathrm{a}\mathrm{triangle}\mathrm{ABC}.\\ \mathrm{Is}\mathrm{AB}+\mathrm{BC}+\mathrm{CA}>\; 2\mathrm{AM}?\end{array}$

(Consider the sides of triangles ΔABM and ΔAMC.)

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{ABM}\\ \text{AB}+\text{BM}>\text{AM}\dots \left(\text{i}\right)\\ \text{Similary in}\Delta \text{ACM, we get}\\ \text{AC}+\text{CM}>\text{AM}\dots \left(\text{ii}\right)\\ \text{Adding (i) and (ii) to get}\\ \text{AB}+\text{BM}+\text{AC}+\text{CM}>\text{2AM}\\ \text{AB}+\text{CA}+\text{(BM}+\text{CM)}>\text{2AM}\\ \text{AB}+\text{AC}+\text{BC}>\text{2AM}\\ \boxed{\text{AB}+\text{BC}+\text{CA}>\text{2AM}}\\ \text{Therefore, the given expression is true}\text{.}\end{array}

Q.10 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.11

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

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.12

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

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.13

$\begin{array}{l}\text{PQR is a triangle right angled at P. If PQ =10 cm}\\ \text{and PR=24cm,find\hspace{0.33em}QR.}\end{array}$

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.14

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

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.15

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

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.16

$\begin{array}{l}\mathrm{Which}\mathrm{of}\mathrm{the}\mathrm{following}\mathrm{can}\mathrm{be}\mathrm{the}\mathrm{sides}\mathrm{of}\mathrm{a}\mathrm{right}\mathrm{triangle}?\\ \left(\mathrm{i}\right)\mathrm{\hspace{0.33em}}2.5\mathrm{cm},6.5\mathrm{cm},6\mathrm{cm}.\\ \left(\mathrm{ii}\right)\mathrm{\hspace{0.33em}}2\mathrm{cm},2\mathrm{cm},5\mathrm{cm}.\\ \left(\mathrm{iii}\right)\mathrm{\hspace{0.33em}}1.5\mathrm{cm},2\mathrm{cm},2.5\mathrm{cm}\end{array}$

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.17

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

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.18

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

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.19

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

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.20

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

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}