In this chapter, we provide RD Sharma Class 7 ex 17.2 Solutions Chapter 17 Constructions for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 7 ex 17.2 Solutions Chapter 17 Constructions Maths pdf, Now you will get step by step solution to each question.

Textbook | NCERT |

Class | Class 7 |

Subject | Maths |

Chapter | Chapter 17 |

Chapter Name | Constructions |

Exercise | Ex 17.2 |

Table of Contents

**RD Sharma Solutions for Class 7 Chapter 17 Constructions** **Ex 17.2 Download PDF**

**Constructions Exercise 17.2**

**Question: 1**

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

**Solution:**

**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 C.
- Join AC and BC to obtain the desired triangle.
- With centre B and radius more than half of 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.

**Question: 2**

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

**Solution:**

**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 OR 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.
- Join QS and extend to produce the angle bisector of angle PQR.
- Verify that angle PQS and angle SQR are equal to 45° each.

**Question: 3**

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

**Solution:**

**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.

**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.

**Solution:**

**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.

**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.

**Solution:**

**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.

**Angle bisector steps:**

- From A, cut arcs of equal radius intersecting AB and AC at E and F, respectively.
- From E and F, cut arcs of equal radius intersecting at point H.
- Join AH and extend to produce the angle bisector of angle A, meeting line BC at D.

**Perpendicular from Point A to line BC steps:**

- From A, cut arcs of equal radius intersecting BC at P and Q, respectively (Extend BC to draw these arcs).
- From P and Q, cut arcs of equal radius intersecting at M.
- Join AM cutting BC at L.
- AL is the perpendicular to the line BC.
- Angle LAD is 15°.

**Question: 6**

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

**Solution:**

**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 DEF.
- By measuring we get, ∠E= ∠F= 40°..

**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.

**Solution:**

**Steps of construction:**

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

**Steps to bisect line AB:**

- Draw an arc from A on either side of line AB.
- 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.
- Join PQ cutting AB at D. PQ is the perpendicular bisector of AB.

**Parallel line to BC:**

- With B as centre, draw an arc cutting BC and BA at M and N, respectively.
- With centre D and the same radius as in the previous step, draw an arc on the opposite side of AB to cut AB at Y.
- With centre Y and radius equal to MN, draw an arc cutting the arc drawn in the previous step at X.
- Join XD and extend it to intersect AC at E.
- DE is the required parallel line.

**All Chapter RD Sharma Solutions For Class 7 Maths**

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