RD Sharma Solutions, Maths Chapter 17, for Class 7, help students who aspire to obtain a good academic score in the exam. The solutions are designed by experts to boost confidence among students in understanding the concepts covered in this chapter and methods to solve problems in a shorter period. These materials are prepared based on class 7 CBSE syllabus by our experts, keeping into consideration, the types of questions asked in the RD Sharma Solutions textbook.

Chapter 17 mainly explains constructions of triangles. It has four exercises. Students can easily access answers to the problems present in RD Sharma Solutions for Class 7. Let us have a look at some of the important concepts that are being discussed in this chapter.

- Drawing a line parallel to the given line
- SSS triangle constructions
- SAS triangle constructions
- ASA triangle constructions
- RHS triangle constructions

## Download the PDF of RD Sharma Solutions For Class 7 Maths Chapter 17 Constructions

### Access answers to Maths RD Sharma Solutions For Class 7 Chapter 17 – Constructions

Exercise 17.1 Page No: 17.1

**1. Draw a âˆ BAC of measure 50 ^{o} 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**

**Solution:**

Steps of construction:

1. Draw angle BAC = 50^{o} such that AB = 5 cm and AC = 7 cm.

Cut an arc through C at an angle of 50^{os}

2. Draw a straight line passing through C and the arc. This line will be parallel to AB since âˆ CAB =âˆ RCA=50^{o}

3. Alternate angles are equal; therefore the line is parallel to AB.

4. Again through B, cut an arc at an angle of 50^{o} and draw a line passing through B and this arc and say this intersects the line drawn parallel to AB at D.

5. âˆ SBA =âˆ BAC = 50^{o}, since they are alternate angles. Therefore BD parallel to AC

6. Also we can measure BD = 7 cm and CD = 5 cm.

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

**Solution:**

Steps of construction:

1. Draw a line PQ.

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

3. Construct âˆ PBF = 90^{o} and âˆ QAE = 90^{o}

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

5. With B as center 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.

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

**Solution:**

Steps of construction:

1. Mark three non collinear points A, B and C such that none of them lie on the same line.

2. Join AB, BC and CA to form triangle ABC.

Parallel line to AC

4. With A as center, draw an arc cutting AC and AB at T and U, respectively.

5. With center B and the same radius as in the previous step, draw an arc on the opposite side of AB to cut AB at X.

6. With center X and radius equal to TU, draw an arc cutting the arc drawn in the previous step at Y.

7. Join BY and produce in both directions to obtain the line parallel to AC.

Parallel line to AB:

8. With B as center, draw an arc cutting BC and BA at W and V, respectively.

9. With center C and the same radius as in the previous step, draw an arc on the opposite side of BC to cut BC at P.

10. With center P and radius equal to WV, draw an arc cutting the arc drawn in the previous step at Q.

11. Join CQ and produce in both directions to obtain the line parallel to AB.

Parallel line to BC:

12. With B as center, draw an arc cutting BC and BA at W and V, respectively (already drawn).

13. With center A and the same radius as in the previous step, draw an arc on the opposite side of AB to cut AB at R.

14. With center R and radius equal to WV, draw an arc cutting the arc drawn in the previous step at S.

15. Join AS and produce in both directions to obtain the line parallel to BC.

**4. Draw two parallel lines at a distance of 5kms apart.**

**Solution:**

Steps of construction:

1. Draw a line PQ.

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

3. Construct âˆ PBF = 90^{o} and âˆ QAE = 90^{o}

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

5. With B as center 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.

Exercise 17.2 Page No: 17.3

**1. Draw â–³ABC in which AB = 5.5 cm. BC = 6 cm and CA = 7 cm. Also, draw perpendicular bisector of side BC.**

**Solution:**

Steps of construction:

1. Draw a line segment AB of length 5.5 cm.

2. From B, cut an arc of radius 6 cm.

3. With center A, draw an arc of radius 7 cm intersecting the previously drawn arc at C.

4. Join AC and BC to obtain the desired triangle.

5. With center B and radius more than half of BC, draw two arcs on both sides of BC.

6. With center 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.

7. Join XY to get the perpendicular bisector of BC.

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

**Solution:**

Steps of construction:

1. Draw a line segment PQ of length 3 cm.

2. With Q as center and radius 4 cm, draw an arc.

3. With P as center and radius 5 cm, draw an arc intersecting the previously drawn arc at R.

4. Join PR and OR to obtain the required triangle.

5. From Q, cut arcs of equal radius intersecting PQ and QR at M and N, respectively.

6. From M and N, cut arcs of equal radius intersecting at point S.

7. Join QS and extend to produce the angle bisector of angle PQR.

8. Verify that angle PQS and angle SQR are equal to 45^{o} each.

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

**Solution:**

Steps of construction:

1. Draw a line segment AB of length 7 cm.

2. With center A, draw an arc of radius 7 cm.

3. With center B, draw an arc of radius 7 cm intersecting the previously drawn arc at C.

4. Join AC and BC to get the required triangle.

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

**Solution:**

Steps of construction:

1. Draw a line segment PR of length 7 cm.

2. With center P, draw an arc of radius 5 cm.

3. With center R, draw an arc of radius 4 cm intersecting the previously drawn arc at Q.

4. Join PQ and QR to obtain the required triangle.

5. From P, draw arcs with radius more than half of PR on either sides.

6. 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.

7. MN is the required perpendicular bisector of the largest side.

**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 âˆ A and (ii) perpendicular AL from A on BC. Measure LAD.**

**Solution:**

Steps of construction:

1. Draw a line segment BC of length 7 cm.

2. With center B, draw an arc of radius 6 cm.

3. With center C, draw an arc of radius 8 cm intersecting the previously drawn arc at A.

4. Join AC and BC to get the required triangle.

Angle bisector steps:

5. From A, cut arcs of equal radius intersecting AB and AC at E and F, respectively.

6. From E and F, cut arcs of equal radius intersecting at point H.

7. Join AH and extend to produce the angle bisector of angle A, meeting line BC at D.

8. Perpendicular from Point A to line BC steps:

9. From A, cut arcs of equal radius intersecting BC at P and Q, respectively (Extend BC to draw these arcs).

10. From P and Q, cut arcs of equal radius intersecting at M.

11. Join AM cutting BC at L.

12. AL is the perpendicular to the line BC.

13. Angle LAD is 15^{o}.

**6. Draw â–³DEF such that DE= DF= 4 cm and EF = 6 cm. Measure âˆ EÂ and âˆ F.**

**Solution:**

Steps of construction:

1. Draw a line segment EF of length 6 cm.

2. With E as center, draw an arc of radius 4 cm.

3. With F as center, draw an arc of radius 4 cm intersecting the previous arc at D.

4. Join DE and DF to get the desired triangle DEF.

5. By measuring we get, âˆ E= âˆ F= 40^{o}.

**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.**

**Solution:**

Steps of construction:

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

Steps to bisect line AB:

1. Draw an arc from A on either side of line AB.

2. With the same radius as in the previous step, draw an arc from B on either side of AB intersecting the arcs drawn in the previous step at P and Q.

3. Join PQ cutting AB at D. PQ is the perpendicular bisector of AB.

Parallel line to BC:

4. With B as center, draw an arc cutting BC and BA at M and N, respectively.

5. With center D and the same radius as in the previous step, draw an arc on the opposite side of AB to cut AB at Y.

6. With center Y and radius equal to MN, draw an arc cutting the arc drawn in the previous step at X.

7. Join XD and extend it to intersect AC at E.

8. DE is the required parallel line.

Exercise 17.3 Page No: 17.5

**1. Draw â–³ABCÂ in which AB = 3 cm, BC = 5 cm and âˆ QÂ = 70 ^{o}.**

**Solution:**

Steps of construction:

1. Draw a line segment AB of length 3 cm.

2. Draw âˆ XBA=70^{o}.

3. Cut an arc on BX at a distance of 5 cm at C.

4. Join AC to get the required triangle.

**2. Draw â–³ABC in which âˆ A=70^{o}. AB = 4 cm and AC= 6 cm. Measure BC.**

**Solution:**

Steps of construction:

1. Draw a line segment AC of length 6 cm.

2. Draw âˆ XAC=70^{o}.

3. Cut an arc on AX at a distance of 4 cm at B.

4. Join BC to get the desired triangle.

5. We see that BC = 6 cm.

**3. Draw an isosceles triangle in which each of the equal sides is of length 3 cm and the angle between them is 45 ^{o}.**

**Steps of construction:**

1. Draw a line segment PQ of length 3 cm.

2. Draw âˆ QPX=45^{o}.

3. Cut an arc on PX at a distance of 3 cm at R.

4. Join QR to get the required triangle.

**4. Draw â–³ABC in which âˆ A = 120 ^{o}, AB = AC = 3 cm. Measure âˆ B and âˆ C.**

**Solution:**

Steps of construction:

1. Draw a line segment AC of length 3 cm.

2. Draw âˆ XAC = 120^{o}.

3. Cut an arc on AX at a distance of 3 cm at B.

4. Join BC to get the required triangle.

5. By measuring, we get âˆ B = âˆ C = 30^{o}.

**5. Draw â–³ABCÂ in which âˆ C Â = 90^{o} and AC = BC = 4 cm.**

**Solution:**

Steps of construction:

1. Draw a line segment BC of length 4 cm.

2. At C, draw âˆ BCY=90Â°.

3. Cut an arc on CY at a distance of 4 cm at A.

4. Join AB. ABC is the required triangle.

**6. Draw a triangle ABC in which BC = 4 cm, AB = 3 cm and âˆ B Â = 45^{o}. Also, draw a perpendicular from A on BC.**

**Solution:**

Steps of construction:

1. Draw a line segment AB of length 3 cm.

2. Draw an angle of 45^{o} and cut an arc at this angle at a radius of 4 cm at C.

3. Join AC to get the required triangle.

4. With A as center, draw intersecting arcs at M and N.

5. With center M and radius more than half of MN, cut an arc on the opposite side of A.

6. With N as center and radius the same as in the previous step, cut an arc intersecting the previous arc at E.

7. Join AE, it meets BC at D, then AE is the required perpendicular.

**7. Draw a triangle ABC with AB = 3 cm, BC = 4 cm and âˆ B Â = 60^{o}. Also, draw the bisector of angles C and A of the triangle, meeting in a point O. Measure âˆ COA.**

**Solution:**

Steps of construction:

1. Draw a line segment BC = 4 cm.

2. Draw âˆ CBX = 60^{o}.

3. Draw an arc on BX at a radius of 3 cm cutting BX at A.

4. Join AC to get the required triangle.

Angle bisector for angle A:

5. With A as center, cut arcs of the same radius cutting AB and AC at P and Q, respectively.

6. From P and Q cut arcs of same radius intersecting at R.

7. Join AR to get the angle bisector of angle A.

Angle bisector for angle C:

8. With A as center, cut arcs of the same radius cutting CB and CA at M and N, respectively.

9. From M and N, cut arcs of the same radius intersecting at T

10. Join CT to get the angle bisector of angle C.

11. Mark the point of intersection of CT and AR as O.

12. Angle âˆ COA = 120^{o}.

Exercise 17.4 Page No: 16.3

**1. Construct âˆ†ABC in which BC = 4 cm, âˆ B = 50 ^{o} and âˆ C = 70^{o}.**

**Solution:**

Steps of construction:

1. Draw a line segment BC of length 4 cm.

2. Draw âˆ CBX such that âˆ CBX=50^{o}.

3. Draw âˆ BCY with Y on the same side of BC as X such that âˆ BCY=70^{o}.

4. Let CY and BX intersect at A.

5. ABC is the required triangle.

**2. Draw âˆ†ABC in which BC = 8 cm, âˆ B = 50 ^{o} and âˆ A = 50^{o}.**

**Solution:**

Steps of construction:

1. Draw a line segment BC of length 8 cm.

2. Draw âˆ CBX such that âˆ CBX = 50^{o}.

3. Draw âˆ BCY with Y on the same side of BC as X such that âˆ BCY = 80^{o}.

4. Let CY and BX intersect at A.

**3. Draw âˆ†ABC in which âˆ Q = 80 ^{o}, âˆ R = 55^{o} and QR = 4.5 cm. Draw the perpendicular bisector of side QR.**

**Solution:**

Steps of construction:

1. Draw a line segment QR = 4.5 cm.

2. Draw âˆ RQX = 80Â° and âˆ QRY = 55^{o}.

3. Let QX and RY intersect at P so that PQR is the required triangle.

4. With Q as center and radius more than 2.25 cm, draw arcs on either sides of QR.

5. With R as center and radius more than 2.25 cm, draw arcs intersecting the previous arcs at M and N.

6. Join MN

7. MN is the required perpendicular bisector of QR.

**4. Construct âˆ†ABC in which AB = 6.4 cm, âˆ A = 45 ^{o} and âˆ B = 60^{o}**

**Solution:**

Steps of construction:

1. Draw a line segment AB = 6.4 cm.

2. Draw âˆ BAX = 45^{o}.

3. Draw âˆ ABY with Y on the same side of AB as X such that âˆ ABY = 60^{o}.

4. Let AX and BY intersect at C.

5. ABC is the required triangle.

**5. Draw âˆ†ABC in which AC = 6 cm, âˆ A = 90 ^{o} and âˆ B = 60^{o}**

**Solution:**

Steps of construction:

1. Draw a line segment AC = 6 cm.

2. Draw âˆ ACX = 30^{o}.

3. Draw âˆ CAY with Y on the same side of AC as X such that âˆ CAY = 90^{o}.

4. Join CX and AY. Let these intersect at B.

5. ABC is the required triangle where angle âˆ ABC = 60^{o}.

Exercise 17.5 Page No: 16.3

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

**Solution:**

Steps of construction:

1. Draw a line segment QR = 4 cm.

2. Draw âˆ QRX of measure 90^{o}.

3. With center Q and radius PQ = 5 cm, draw an arc of the circle to intersect ray RX at P.

4. Join PQ to obtain the desired triangle PQR.

5. PQR is the required triangle.

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

**Solution:**

Steps of construction:

1. Draw a line segment QR = 2.5 cm.

2. Draw âˆ QRX of measure 90^{o}.

3. With center Q and radius PQ = 4 cm, draw an arc of the circle to intersect ray RX at P.

4. Join PQ to obtain the desired triangle PQR.

5. PQR is the required triangle.

**3. Draw a right triangle having hypotenuse of length 5.4 cm, and one of the acute angles of measure 30 ^{o}**

**Solution:**

Let ABC be the right triangle at A such that hypotenuse BC = 5.4 cm. Let C = 30^{o}.

Therefore âˆ A + âˆ B + âˆ C = 180^{o}

âˆ B = 180^{o }âˆ’ 30^{o}âˆ’ 90^{o} = 60^{o}

Steps of construction:

1. Draw a line segment BC = 5.4 cm.

2. Draw angle CBY = 60^{o}

3. Draw angle BCX of measure 30^{o} with X on the same side of BC as Y.

4. Let BY and CX intersect at A.

5. Then ABC is the required triangle.

**4. Construct a right triangle ABC in which AB = 5.8 cm, BC = 4.5 cm and âˆ C = 90 ^{o}.**

**Solution:**

Steps of construction:

1. Draw a line segment BC = 4.5 cm.

2. Draw âˆ BCX of measure 90^{o}

3. With center B and radius AB = 5.8 cm, draw an arc of the circle to intersect ray BX at A.

4. Join AB to obtain the desired triangle ABC.

5. ABC is the required triangle.

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

**Solution:**

Steps of construction:

1. Draw a line segment BC = 4.6 cm.

2. Draw âˆ BCX of measure 90^{o}

3. With center B and radius AB = 5.2 cm, draw an arc of the circle to intersect ray CX at A.

4. Join AB to obtain the desired triangle ABC.

5. ABC is the required triangle.