RD Sharma Solutions for class 9 chapter 17 is given in detail for students to understand constructions in a more comprehensive manner. All the questions are solved with the help of figures for easy and better understanding. In geometry, constructions refer to the drawing of angles, lines, and various other shapes accurately using only a compass and a ruler. These constructions are considered as a pure form of geometric constructions since no other measuring devices like protractors are used. This method of construction was used by Euclid for various basic constructions including bisecting a line, drawing different angles, constructing shapes like triangles, etc. Students will get to learn here how to construct figures using various calculations in geometry. All questions of chapter 17 – Constructions are given according to the exercises in the RD Sharma class 9 Maths textbook.

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### Access Answers to Maths RD Sharma Solutions for Class 9 Chapter 17 Construction

### Exercise 17.1 Page No: 17.3

**Question 1: Draw a line segment of length 8.6 cm. Bisect it and measure the length of each part.**

**Solution: **

Step 1: Draw a line segment AB = 8.6 cm.

Step 2: Draw arcs on each side of AB using A as a center at any radius more than half of 8.6.

Step 3: Repeat Step 2 using B as a center and make sure these arcs cut the previous arcs.

Step 4: Join the points P and Q which intersects AB at M.

Therefore AM= MB = 4.3 cm

**Question 2: Draw a line segment AB of length 5.8cm. Draw the perpendicular bisector of this line segment.**

**Solution: **

Step 1: Draw a line segment AB = 5.8 cm.

Step 2: Draw arcs on each side of AB using A as a center at any radius more than half of 5.8.

Step 3: Repeat Step 2 using B as a center and make sure these arcs cut the previous arcs.

Step 4: Join the points P and Q.

Here, PQ is the perpendicular bisector of AB.

**Question 3: Draw a circle with center at point O and radius 5cm. Draw its chord AB, the perpendicular bisector of line segment AB. Does it pass through the center of the circle?**

**Solution: **

Step 1: Draw a circle choosing radius 5 cm and point O as center.

Step 2: Draw a chord AB using scale.

Step 3: Draw arcs one on each side of chord chossing A as center and radius more than half of 5 cm.

Step 4: Repeat step 3 using B as a centre and make sure these arcs cut the previous arcs.

Step 5: Join P and Q.

Therefore PQ is a perpendicular bisector of chord AB passes through the center of the circle.

### Exercise 17.2 Page No: 17.7

**Question 1: Draw an angle and label it as ∠BAC. Construct another angle, equal to ∠BAC.**

** Solution: **

Steps of construction:

Step 1: Draw any angle ABC.

Now will construct an angle equal to ∠BAC

Step 2: Draw a line segment QR.

Step 3: Draw an arc which intersects ∠BAC at E and D using A as center and choose any radius.

Step 4: With same measurements (set in step 2), Draw an arc from point Q.

Step 5: With S as center and radius equal to DE, draw an arc which intersects the previous arc at T.

Step 6: Join Q and T.

Therefore ∠PQR= ∠BAC

**Question 2: Draw an obtuse angle. Bisect it. Measure each of the angles so formed.**

** Solution:**

Steps of construction:

Step 1: Draw an obtuse angle. We choose ∠ABC = 120^{0}.

Step 2: Draw an arc which intersects AB at P and BC at Q, from center B and choose any radius.

Step 3: Draw an arc from point P by setting radius more than half of PQ.

Step 4: Repeat step 3 using Q as center and cut the previous arc at R.

Step 5: Join BR.

Therefore ∠ABR= ∠RBC = 60^{0}

**Question 3: Using your protractor, draw an angle of 108 ^{0}. With this given angle as given, draw an angle of 54^{0}.**

**Solution: **

Steps of construction:

Step 1: Draw ∠ABC = 108^{0}.

Step 2: Draw an arc which intersects AB at P and BC at Q from point B. (Choose any radius)

Step 3: Draw an arc from point P by setting radius more than half of PQ.

Step 4: Repeat Step 3 using Q as the centre and intersect the previous arc at R.

Step 5: Join BR.

Therefore ∠RBC = 54^{0}

**Question 4: Using the protractor, draw a right angle. Bisect it to get an angle of measure 45 ^{0}.**

** Solution:**

Steps of construction:

Step 1: Draw ∠ABC = 90^{0}.

Step 2: Draw an arc which intersects AB at P and BC at Q from point B. (Choose any radius)

Step 3: Draw an arc from point P by setting radius more than half of PQ.

Step 4: Repeat step 3 using Q as a centre and intersect the previous arc at R.

Step 5: Join RB.

Therefore ∠RBC= 45^{0}

### Exercise 17.3 Page No: 17.15

**Question 1: Construct a △ABC in which BC = 3.6 cm, AB + AC = 4.8 cm and ∠B = 60 ^{0}.**

**Solution:**

Steps of Construction:

Step 1: Draw a line segment BC = 3.6 cm.

Step 2: At the point B, draw ∠XBC = 60^{0}.

Step 3: Draw an arc which intersects XB at point D form point B and with radius 4.8 cm

Step 4: Join DC.

Step 5: Draw a perpendicular bisector of DC which intersects DB at A.

Step 6: Join AC.

Hence, △ABC is the required triangle.

**Question 2: Construct a △ABC in which AB + AC = 5.6 cm, BC = 4.5 cm and ∠B=45 ^{0}.**

**Solution:**

Steps of Construction:

Step 1: Draw a line segment BC = 4.5 cm.

Step 2: At the point B, draw ∠XBC = 45^{0}.

Step 3: Draw an arc which intersects XB at point D form point B and with radius 5.6 cm

Step 4: Join DC.

Step 5: Draw a perpendicular bisector of DC which intersects DB at A.

Step 6: Join AC.

Hence, △ABC is the required triangle.

**Question 3: Construct a △ABC in which BC = 3.4 cm, AB – AC = 1.5 cm and ∠B = 45 ^{0}.**

**Solution:**

Steps of Construction:

Step 1: Draw a line segment BC = 3.4 cm.

Step 2: Draw ∠XBC = 45^{0}.

Step 3: Draw an arc which intersects XB at point D form point B and with radius 1.5 cm. So, BD = 1.5 cm.

Step 4: Join line segment DC.

Step 5: Draw a perpendicular bisector of DC which intersects BX at A.

Step 6: Join line segment AC.

Hence, △ABC is the required triangle.

**Question 4: Using rulers and compasses only, construct a △ABC, given base BC = 7 cm, ∠ABC = 60 ^{0} and AB + AC = 12 cm.**

**Solution:**

Step 1: Draw a line segment BC = 7 cm.

Step 2: Draw an arc from point B cutting BC at N. (Choose any radius.)

Step 3: Keep compass at point N with same radius selected in step 2, cut the previous arc at M.

Step 4: Join line segment BM.

Step 5: Produce BM to any point P

Step 6: Cut BR = 12 cm, from BP.

Step 7: Join CR.

Step 8: Draw a perpendicular bisector of RC which intersects BR at A.

Step 9: Join line segment AC.

Hence, △ABC is the required triangle.

## RD Sharma Solutions for Class 9 Maths Chapter 17 Construction

In the 17th chapter of Class 9 RD Sharma Solutions students will study important concepts listed below:

- Construction Introduction
- Construction of Bisector of a Line Segment
- Construction of the Bisector of a Given Angle
- Constructions of Triangles and many more