#### Chapter 17 ConstructionÂ R.D. Sharma Solutions for Class 9th Exercise 17.3

**Exercise 17.3**

1.Â Construct a Î” ABC in which BC = 3.6 cm, AB + AC = 4.8 cm and âˆ B = 60Â°.

**Solution**

We are asked to construct a triangle ABC in which BC = 3.6 cm, âˆ B = 60Â° and AB + AC = 4.8 cm

We will follow the following algorithm for the construction .

Step 1: Draw BC of 3.6 cm

Step 2: Draw âˆ EBC = 60Â°.

Step 3: From ray BE, cut BD of 4.8 cm.

Step 4: Draw segment DC.

Step 5: Draw the perpendicular bisector of DC, intersecting the ray BE at A

Step 6: Draw segment AC.

Î”ABC is the required angle.

2. Construct a Î” ABC in which AB + AC = 5.6 cm, BC = 4.5 cm and âˆ B = 45Â°.

**Solution**

We are asked to construct a triangle ABC in which BC = 4.5 cm and AB + AC = 5.6 cm .

We will follow the following algorithm for the construction .

StepÂ 1: Draw BC of 4.5 cm.

Step 2: Draw âˆ EBC = 45Â°

Step 3: From ray BE, cut BD of 5.6 cm.

Step 4: Draw segment DC.

Step 5: Draw the perpendicular bisector of DC, intersecting the ray BE at A

Step 6: Draw segment AC.

Î”ABC is the required angle.

3. Construct a Î” ABC in which BC = 3.4 cm, AB âˆ’ AC = 1.5 cm and âˆ B = 45Â°.

**Solution**

We are asked to construct a triangle ABC in which BC = 3.4 cm , and âˆ B = 45Â° and AB - AC = 1.5 cmÂ Â Â Â Â Â

We will follow the following algorithm for the construction .

Steps of construction

StepÂ 1: Draw BC of 3.4 cm.

Step 2: Draw âˆ CBE = 45Â°

Step 3: From the ray BE, cut BD of 1.5 cm.

Step 4: Draw segment DC.

Step 5: Draw the perpendicular bisector of DC, intersecting the ray BE at A

Step 6: Draw AC.

Î”ABC is the required angle.

4. Using ruler and compasses only, construct a Î”ABC, given base BC = 7 cm, âˆ ABC = 60Â° and AB+AC = 12 cm.

**Solution**

We are asked to construct a triangle ABC in which BC = 7 cm, âˆ B = 60Â° and AB + AC = 12 cm

We will follow the following algorithm for the construction .

Step 1: Draw BC of length 7 cm.

Step 2: With B as a centre, and convenient radius draw an arc cutting BC at N.

Step 3: With N as a centre, and same radius draw an arc cutting the previous arc at M.

Step 4: Draw BM and produce it to P.

Step 5: From BP, cut BR = 12 cm.

Step 6: Draw segment RC.

Step 7: Draw perpendicular bisector of RC. It cuts the ray BR at A.

Step 8: Draw AC.

Î”ABC is the required triangle.

5. Construct a triangle whose perimeter is 6.4 cm, and angles at the base are 60Â° and 45Â°.

**Solution**

We are asked to draw triangle ABC whose base angles are 60Â° and 45Â° and its perimeter is 6.4 cm

We will follow the following algorithm for the construction

StepÂ 1: Draw DE of 6.4 cm.

Step 2: Draw âˆ EDP =âˆ B = 60Â° and âˆ DER = âˆ C = 45Â°.

Step 3: Draw the angle bisectors of âˆ EDP and âˆ DER and mark their point of intersection as A.

Step 4: Draw the perpendicular bisectors of AD and AE. They meet DE at point B and Point C respectively.

Step 5: Draw AB and AC to get the required Î”ABC.

6. Using ruler and compasses only, construct a Î” ABC from the following data:

AB + BC + CA = 12 cm, âˆ B = 45Â° and âˆ C = 60Â°.

**Solution**

We are asked to draw triangle ABC whose base angles are 60Â° and 45Â° and its perimeter is 12 cm

We will follow the following algorithm for the construction

Step 1: Draw DE of length 12 cm.

Step 2: Construct âˆ EDP =âˆ C = 60Â° and âˆ DER =âˆ B = 45Â°, using regular methods.

Step 3: Draw the angle bisectors of âˆ EDP and âˆ DER and mark their point of intersection as A.

Step 4: Draw the perpendicular bisectors of AE and AD. They meet DE at point B and Point C respectively.

Step 5: Draw AB and AC to get the required Î”ABC.

7. Construct a right-angled triangle whose perimeter is equal to 10 cm and one acute angle equal to 60Â°.

**Solution**

We are asked to draw triangle ABC whose base angles are 60Â° and 90Â° and its perimeter is 10 cm.

We will follow the following algorithm for the construction

Steps of construction

Step 1: Draw a line segment DE of 10 cm.

Step 2: Draw âˆ EDM = âˆ B = 90Â° and âˆ DEN = âˆ C = 60Â°.

Step 3: Draw the angle bisectors of âˆ EDM and âˆ DEN and mark their point of intersection as A.

Step 4: Draw the perpendicular bisectors of AD and AE. They meet DE at B and C respectively.

Step 5: Draw AB and AC to get the desired Î”ABC.

8. Construct a triangle ABC such that BC = 6 cm, AB = 6 cm and median AD = 4 cm.

**Solution**

We are given BC = 6 cm, and median AD = 4 cm

We have to construct the triangle ABC

We will follow the following algorithm for the construction .

StepÂ 1: Draw a line segment BC of length 6 cm.

Step 2: Draw the perpendicular bisector of line BC cutting the line at point D, such that BD = 3 cm.

Step 3: With centre B and radius = 6 cm, draw an arc.

Step 4: With centre D and radius = 4 cm, draw an arc intersecting the previous arc at A.

Step 5: Draw AC to get the desired Î”ABC.

9. Construct a right triangle ABC whose base BC is 6 cm and the sum of hypotenuse AC and other side AB is 10 cm.

**Solution**

We have to construct a right triangle ABC with base BC = 6 cm and the sum of hypotenuse and the other side is 10 cm

We will follow the following algorithm for the construction

Steps of construction

StepÂ 1: Draw a line segment BC of length 6 cm.

Step 2: Draw âˆ CBE = 90Â°

Step 3: From BE, cut BD of length 10 cm.

Step 4: Draw the line segment DC.

Step 5: Draw the perpendicular bisector of DC, intersecting the ray BE at point A

Step 6: Draw AC to get the desired Î”ABC.

Â 10. Construct a triangle XYZ in which âˆ Y = 30Â°, âˆ Z = 90Â° and XY + YZ + ZX = 11.

**Solution**

We are asked to construct a triangle XYZ such that âˆ Y = 30Â°, âˆ Z = 90Â° XY + YZ + ZX = 11 cm

We will follow the following algorithm for the construction

StepÂ 1: Draw a line segment AB of length 11 cm.

Step 2: Draw âˆ BAC = âˆ Y = 30 and âˆ ABD = âˆ Z = 90Â°

Step 3: Bisect âˆ BAC and âˆ ABD and mark the point of intersection of these angle bisectors by X.

Step 4: Draw perpendicular bisectors of AX and XB. They meet AB at point Y and Z, respectively.

Step 5: Draw XY and XZ to get the desired Î”XYZ.