NCERT Solutions Class 7 Maths Chapter 10

Q.1  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.2

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

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.3

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

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}\parallel \text{RS and PR}\parallel \text{QS}\text{.}\\ \text{Thus PQRS is a parallelogram}\text{.}\end{array}

Q.4

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

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.5

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

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.6

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

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.7

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

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.8

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

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.9

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

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.10

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

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.11

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

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.12

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

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.13

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

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.14

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

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.15 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.16

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

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{.}