Rd Sharma 2020 2021 Solutions for Class 7 Maths Chapter 17 Constructions are provided here with simple step-by-step explanations. These solutions for Constructions are extremely popular among Class 7 students for Maths Constructions Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma 2020 2021 Book of Class 7 Maths Chapter 17 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma 2020 2021 Solutions. All Rd Sharma 2020 2021 Solutions for class Class 7 Maths are prepared by experts and are 100% accurate.

#### Page No 17.1:

#### Question 1:

Draw an ∠*BAC* of measure 50° such that *AB* = 5 cm and *AC* = 7 cm. Through *C* draw a line parallel to *AB* and through *B* draw a line parallel to *AC*, intersecting each other at *D*. Measure *BD* and *CD*.

#### Answer:

Steps of construction:

- Draw angle BAC = 50$\xb0$ such that AB = 5 cm and AC = 7 cm.
- Cut an arc through C at an angle of 50$\xb0$.
- Draw a straight line passing through C and the arc. This line will be parallel to AB since $\angle CAB=\angle RCA=50\xb0$.
- Alternate angles are equal; therefore the line is parallel to AB.
- Again through B, cut an arc at an angle of 50$\xb0$ and draw a line passing through B and this arc and say this intersects the line drawn parallel to AB at D.
- $\angle SBA\hspace{0.17em}=\angle BAC=50\xb0$, since they are alternate angles. Therefore BD$\parallel $AC.
- Also we can measure BD = 7 cm and CD = 5 cm.

#### Page No 17.1:

#### Question 2:

Draw a line *PQ*. Draw another line parallel to *PQ* at a distance of 3 cm from it.

#### Answer:

1. Draw a line PQ.

2. Take any two points A and B on the line.

3. Construct $\angle PBF=90\xb0and\angle QAE=90\xb0$.

4. With A as centre and radius 3 cm cut AE at C.

5. With B as centre and radius 3 cm cut BF at D.

6. Join CD and produce it on either side to get the required line parallel to AB and at a distance of 5 cm from it.

#### Page No 17.1:

#### Question 3:

Take any three non-collinear points *A*, *B*, *C* and draw ∆*ABC*. Through each vertex of the triangle, draw a line parallel to the opposite side.

#### Answer:

Steps of construction:

#### Page No 17.2:

#### Question 4:

Draw two parallel lines at a distance 5 cm apart.

#### Answer:

Steps of construction:

3. Construct $\angle PBF=90\xb0and\angle QAE=90\xb0$.

4. With A as centre and radius 5 cm cut AE at C.

5. With B as centre and radius 5 cm cut BF at D.

6. Join CD and produce it on either side to get the required line parallel to AB and at a distance of 5 cm from it.

#### Page No 17.3:

#### Question 1:

Draw ∆ *ABC* in which *AB* = 5.5 cm, *BC* = 6 cm and *CA* = 7 cm. Also, draw perpendicular bisector of side *BC*.

#### Answer:

Steps of construction:

- Draw a line segment AB of length 5.5 cm.
- From B, cut an arc of radius 6 cm.
- With centre A, draw an arc of radius 7 cm intersecting the previously drawn arc at say, C.
- Join AC and BC to obtain the desired triangle.
- With centre B and radius more than $\frac{1}{2}BC$, draw two arcs on both sides of BC.
- With centre C and the same radius as in the previous step, draw two arcs intersecting the arcs drawn in the previous step at X and Y.
- Join XY to get the perpendicular bisector of BC.

#### Page No 17.3:

#### Question 2:

Draw ∆ *PQR* in which* PQ* = 3 cm, *QR* = 4 cm and *RP* = 5 cm. Also, draw the bisector of ∠*Q*.

#### Answer:

Steps of construction:

- Draw a line segment PQ of length 3 cm.
- With Q as centre and radius 4 cm, draw an arc.
- With P as centre and radius 5 cm, draw an arc intersecting the previously drawn arc at R.
- Join PR and QR to obtain the required triangle.
- From Q, cut arcs of equal radius intersecting PQ and QR at M and N, respectively.
- From M and N, cut arcs of equal radius intersecting at point S.

#### Page No 17.3:

#### Question 3:

Draw an equilateral triangle one of whose sides is of length 7 cm.

#### Answer:

Steps of construction:

- Draw a line segment AB of length 7 cm.
- With centre A, draw an arc of radius 7 cm.
- With centre B, draw an arc of radius 7 cm intersecting the previously drawn arc at C.
- Join AC and BC to get the required triangle.

#### Page No 17.3:

#### Question 4:

Draw a triangle whose sides are of lengths 4 cm, 5 cm and 7 cm. Draw the perpendicular bisector of the largest side.

#### Answer:

Steps of construction:

- Draw a line segment PR of length 7 cm.
- With centre P, draw an arc of radius 5 cm.
- With centre R, draw an arc of radius 4 cm intersecting the previously drawn arc at Q.
- Join PQ and QR to obtain the required triangle.
- From P, draw arcs with radius more than half of PR on either sides.
- With the same radius as in the previous step, draw arcs from R on either sides of PR intersecting the arcs drawn in the previous step at M and N.
- MN is the required perpendicular bisector of the largest side.

#### Page No 17.3:

#### Question 5:

Draw a triangle *ABC* with *AB* = 6 cm, *BC* = 7 cm and *CA* = 8 cm. Using ruler and compass alone, draw (i) the bisector *AD* of ∠*A* and (ii) perpendicular *AL* from *A* on *BC*. Measure *LAD*.

#### Answer:

Steps of construction:

- Draw a line segment BC of length 7 cm.
- With centre B, draw an arc of radius 6 cm.
- With centre C, draw an arc of radius 8 cm intersecting the previously drawn arc at A.
- Join AC and BC to get the required triangle.

^{o}.

#### Page No 17.3:

#### Question 6:

Draw ∆ *DEF* such that *DE* = *DF* = 4 cm and EF = 6 cm. Measure ∠*E* and ∠*F*.

#### Answer:

Steps of construction:

- Draw a line segment EF of length 6 cm.
- With E as centre, draw an arc of radius 4 cm.
- With F as centre, draw an arc of radius 4 cm intersecting the previous arc at D.
- Join DE and DF to get the desired triangle.DF, .

#### Page No 17.3:

#### Question 7:

Draw any triangle *ABC*. Bisect side *AB* at *D*. Through *D*, draw a line parallel to *BC*, meeting *AC* in *E*. Measure *AE* and *EC*.

#### Answer:

We first draw a triangle ABC with each side = 6 cm.

5. DE is the required parallel line.

#### Page No 17.5:

#### Question 1:

Draw ∆ *ABC* in which *AB* = 3 cm, *BC* = 5 cm and ∠*B* = 70°.

#### Answer:

Steps of construction:

- Draw a line segment AB of length 3 cm.
- Draw $\angle XBA=70\xb0$.
- Cut an arc on BX at a distance of 5 cm at C.
- Join AC to get the required triangle.

#### Page No 17.5:

#### Question 2:

Draw ∆ *ABC* in which ∠*A* = 70°, *AB* = 4 cm and *AC* = 6 cm. Measure *BC*.

#### Answer:

Steps of construction:

- Draw a line segment AC of length 6 cm.
- Draw $\angle $XAC = 70$\xb0$.
- Cut an arc on AX at a distance of 4 cm at B.
- Join BC to get the desired triangle.
- We see that BC = 6 cm.

#### Page No 17.5:

#### Question 3:

Draw an isosceles triangle in which each of the equal sides is of length 3 cm and the angle between them is 45°.

#### Answer:

Steps of construction:

- Draw a line segment PQ of length 3 cm.
- Draw $\angle QPX=45\xb0$.
- Cut an arc on PX at a distance of 3 cm at R.
- Join QR to get the required triangle.

#### Page No 17.5:

#### Question 4:

Draw ∆ *ABC* in which ∠*A* = 120°, *AB* = *AC* = 3 cm. Measure ∠*B* and ∠*C*.

#### Answer:

Steps of construction:

- Draw a line segment AC of length 3 cm.
- Draw $\angle XAC=120\xb0$.
- Cut an arc on AX at a distance of 3 cm at B.
- Join BC to get the required triangle.

By measuring,we get

$\angle B=\angle C=30\xb0$.

#### Page No 17.5:

#### Question 5:

Draw ∆ *ABC* in which ∠*C* = 90° and *AC* = *BC* = 4 cm.

#### Answer:

Steps of construction:

- Draw a line segment BC of length 4 cm.
- AT C, draw $\angle BCY=90\xb0$.
- Cut an arc on CY at a distance of 4 cm at A.
- Join AB.
- ABC is the required triangle.

#### Page No 17.5:

#### Question 6:

Draw a triangle *ABC* in which *BC* = 4 cm, *AB* = 3 cm and ∠*B* = 45°. Also, draw a perpendicular from *A* on *BC*.

#### Answer:

Steps of construction:

- Draw a line segment AB of length 3 cm.
- Draw an angle of 45$\xb0$ and cut an arc at this angle at a radius of 4 cm at C.
- Join AC to get the required triangle.
- With A as centre, draw intersecting arcs at M and N.
- With centre M and radius more that $\frac{1}{2}MN$, cut an arc on the opposite side of A.
- With N as centre and radius the same as in the previous step, cut an arc intersecting the previous arc at E.
- Join AE, it meets BC at D, then AE is the required perpendicular.

#### Page No 17.5:

#### Question 7:

Draw a triangle *ABC* with *AB* = 3 cm, *BC* = 4 cm and ∠*B* = 60°. Also, draw the bisector of angles *C* and *A* of the triangle, meeting in a point *O*. Measure ∠*COA*.

#### Answer:

Steps of construction:

- Draw a line segment BC = 4 cm.
- Draw $\angle CBX\hspace{0.17em}=60\xb0$.
- Draw an arc on BX at a radius of 3 cm cutting BX at A.
- Join AC to get the required triangle.

^{o}

#### Page No 17.6:

#### Question 1:

Construct ∆ *ABC* in which *BC* = 4 cm, ∠*B* = 50° and ∠*C* = 70°.

#### Answer:

Steps of construction:

- Draw a line segment BC of length 4 cm.
- Draw $\angle CBX\mathrm{such}\mathrm{that}\angle CBX\hspace{0.17em}=50\xb0$.
- Draw $\angle BCY$ with Y on the same side of BC as X such that $\angle $BCY = 70$\xb0$.
- Let CY and BX intersect at A.
- ABC is the required triangle.

#### Page No 17.6:

#### Question 2:

Draw ∆ *ABC* in which *BC* = 8 cm, ∠*B* = 50° and ∠*A* = 50°.

#### Answer:

$\angle ABC+\angle BCA+\angle CAB=180\xb0\phantom{\rule{0ex}{0ex}}\angle BCA=180\xb0-\angle ABC-\angle CAB\phantom{\rule{0ex}{0ex}}\angle BCA=180\xb0-100\xb0=80\xb0$

Steps of construction:

- Draw a line segment BC of length 8 cm.
- Draw $\angle CBX\mathrm{such}\mathrm{that}\angle CBX=50\xb0$.
- Draw $\angle $BCY with Y on the same side of BC as X such that $\angle $BCY = 80$\xb0$.
- Let CY and BX intersect at A.

#### Page No 17.6:

#### Question 3:

Draw ∆ *PQR* in which ∠*Q* = 80°, ∠*R* = 55° and *QR* = 4.5 cm. Draw the perpendicular bisector of side *QR*.

#### Answer:

Steps of construction:

- Draw a line segment QR = 4.5 cm.
- Draw $\angle RQX=80\xb0and\angle QRY=55\xb0$.
- Let QX and RY intersect at P so that PQR is the required triangle.
- With Q as centre and radius more that 2.25 cm, draw arcs on either sides of QR.
- With R as centre and radius more than 2.25 cm, draw arcs intersecting the previous arcs at M and N.
- Join MN; MN is the required perpendicular bisector of QR.

#### Page No 17.6:

#### Question 4:

Construct ∆ *ABC* in which *AB* = 6.4 cm, ∠*A* = 45° and ∠*B* = 60°.

#### Answer:

Steps of construction:

- Draw a line segment AB = 6.4 cm.
- Draw $\angle BAX=45\xb0$.
- Draw $\angle $ABY with Y on the same side of AB as X such that $\angle $ABY = 60$\xb0$.

#### Page No 17.6:

#### Question 5:

Draw ∆ ABC in which AC = 6 cm, ∠*A* = 90° and ∠*B* = 60°.

#### Answer:

We can see that ∠A+∠B+∠C = 180°. Therefore ∠C = 180$\xb0$ − 60$\xb0$ − 90$\xb0$ = 30°.

Steps of construction:

- Draw a line segment AC = 6 cm.
- Draw $\angle ACX=30\xb0$.
- Draw $\angle $CAY with Y on the same side of AC as X such that $\angle $CAY = 90$\xb0$.
- Join CX and AY. Let these intersect at B.
- ABC is the required triangle where angle $\angle $ABC = 60$\xb0$.

#### Page No 17.8:

#### Question 1:

Draw a right triangle with hypotenuse of length 5 cm and one side of length 4 cm.

#### Answer:

Steps of construction:

- Draw a line segment QR = 4 cm.
- Draw $\angle $QRX of measure 90$\xb0$.
- With centre Q and radius PQ = 5 cm, draw an arc of the circle to intersect ray RX at P.
- Join PQ to obtain the desired triangle PQR.
- PQR is the required triangle.

#### Page No 17.8:

#### Question 2:

Draw a right triangle whose hypotenuse is of length 4 cm and one side is of length 2.5 cm.

#### Answer:

Steps of construction:

- Draw a line segment QR = 2.5 cm.
- Draw $\angle $QRX of measure 90$\xb0$.
- With centre Q and radius PQ = 4 cm, draw an arc of the circle to intersect ray RX at P.
- Join PQ to obtain the desired triangle PQR.
- PQR is the required triangle.

#### Page No 17.8:

#### Question 3:

Draw a right triangle having hypotenuse of length 5.4 cm, and one of the acute angles of measure 30°.

#### Answer:

Let ABC be the right triangle at A such that hypotenuse BC = 5.4 cm. Let $\angle C=30\xb0$.

Therefore $\angle A+\angle B+\angle C=180\xb0\phantom{\rule{0ex}{0ex}}\angle B=180-30-90=60\xb0\phantom{\rule{0ex}{0ex}}$

Steps of construction:

- Draw a line segment BC = 5.4 cm.
- Draw angle CBY = 60
^{o}. - Draw angle BCX of measure 30
^{o}with X on the same side of BC as Y. - Let BY and CX intersect at A.

#### Page No 17.8:

#### Question 4:

Construct a right triangle *ABC* in which *AB* = 5.8 cm, *BC* = 4.5 cm and ∠C = 90°.

#### Answer:

Steps of construction:

- Draw a line segment BC = 4.5 cm.
- Draw $\angle $BCX of measure 90$\xb0$.
- With centre B and radius AB =5.8 cm, draw an arc of the circle to intersect ray BX at A.
- Join AB to obtain the desired triangle ABC.
- ABC is the required triangle.

#### Page No 17.8:

#### Question 5:

Construct a right triangle, right angled at C in which *AB* = 5.2 cm and *BC* = 4.6 cm.

#### Answer:

Steps of construction:

- Draw a line segment BC = 4.6 cm.
- Draw $\angle $BCX of measure 90$\xb0$.
- With centre B and radius AB = 5.2 cm, draw an arc of the circle to intersect ray CX at A.
- Join AB to obtain the desired triangle ABC.
- ABC is the required triangle.

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