# Class 10 RD Sharma Solutions – Chapter 11 Constructions – Exercise 11.2 | Set 2

### Question 11. Construct a triangle similar to a given Î”XYZ with its sides equal to (3/2)^{th} of the corresponding sides of Î”XYZ. Write the steps of construction.

**Solution:**

Follow these steps for the construction:

Step 1: Construct a triangle XYZ along some feasible data.

Step 2: Construct a ray YL creating an acute angle along XZ and break off 5 equal parts creating

YY

_{1}= Y_{1}Y_{2}= Y_{2}Y_{3}= Y_{3}Y_{4}.Step 3: Connect Y

_{4}and Z.Step 4: From Y

_{3}, construct Y_{3}Zâ€™ parallel to Y_{4}Z and Zâ€™Xâ€™ parallel to ZX.Therefore,

We have the required triangle Î”Xâ€™YZâ€™.

### Question 12. Draw a right triangle in which sides (other than the hypotenuse) are of lengths 8 cm and 6 cm. Then construct another triangle whose sides are 3/4 times the corresponding sides of the first triangle.

**Solution:**

Follow these steps for the construction

Step 1: Construct right Î”ABC right angle at B and BC of 8 cm and BA of 6 cm.

Step 2: Construct a line BY creating a cut angle along BC and break off 4 equal parts.

Step 3: Connect 4C and Construct 3Câ€™ || 4C and Câ€™Aâ€™ parallel to CA.

Therefore,

We have the required triangle Î”BCâ€™Aâ€™ is the required triangle.

### Question 13. Construct a triangle with sides 5 cm, 5.5 cm, and 6.5 cm. Now construct another triangle, whose sides are 3/5 times the corresponding sides of the given triangle.

**Solution:**

Follow these steps for the construction:

Step 1: Construct a line segment BC of 5.5 cm.

Step 2: Along centre B and radius 5 cm and along centre C and radius 6.5 cm,

Construct arcs that bisect each other at A

Step 3: Connect BA and CA. Î”ABC is the given triangle.

Step 4: At B, construct a ray BX creating an acute angle and break off 5 equal parts from BX.

Step 5: Connect C5 and Construct 3D || 5C which connects BC at D.

From D, construct DE || CA which meets AB at E.

Therefore,

We have the required triangle Î”EBD.

### Question 14. Construct a triangle PQR with side QR = 7 cm, PQ = 6 cm and âˆ PQR = 60Â°. Then construct another triangle whose sides are 3/5 of the corresponding sides of Î”PQR.

**Solution:**

Follow these steps for the construction:

Step 1: Construct a line segment QR = 7 cm.

Step 2: At Q construct a ray QX creating an angle of 60Â° and cut of PQ = 6 cm.

Connect PR.

Step 3: Construct a ray QY creating an acute angle and break off 5 equal parts.

Step 4: Connect 5, R and through 3, construct 3, S parallel to 5, R which meet QR at S.

Step 5: Through S, construct ST || RP meeting PQ at T.

Therefore,

We have the required Î”QST.

### Question 15. Construct a Î”ABC in which base BC = 6 cm, AB = 5 cm and âˆ ABC = 60Â°. Then construct another triangle whose sides are 3/4 of the corresponding sides of Î”ABC.

**Solution:**

Follow these steps for the construction:

Step 1: To construct a triangle ABC

Along side of BC of 6 cm,

AB of 5 cm and âˆ ABC = 60Â°.

Step 2: Construct a ray BX, which creates an acute angle âˆ CBX below the line BC.

Step 4: Make four points B

_{1}, B_{2}, B_{3}and B_{4}on BX such that BB_{1}= B_{1}B_{2 }= B_{2}B_{3}= B_{3}B_{4}.Step 5: Connect B

_{4}C and construct a line through B_{3}parallel to B_{4}C bisecting BC to Câ€™.Step 6: Construct a line through Câ€™ parallel to the line CA to intersect BA at Aâ€™.

Therefore,

We have the required triangle Î”ABC.

### Question 16. Construct a right triangle in which the sides (other than the hypotenuse) arc of lengths 4 cm and 3 cm. Now, construct another triangle whose sides are 5/3 times the corresponding sides of the given triangle.

**Solution:**

Follow these steps for the construction:

Step 1: Construct a right triangle ABC in which the sides (other than the hypotenuse) are

of lengths 4 cm and 3 cm. âˆ B = 90Â°.

Step 2: Construct a line BX, that creates an acute angle âˆ CBX below the line BC.

Step 3: Mark 5 points B

_{1}, B2_{2}, B_{3}, B_{4}and B_{5}on BX such that BB_{1}= B_{1}B_{2 }= B_{2}B_{3}= B_{3}B_{4}= B_{4}B_{5}.Step 4: Connect B

_{3}to C and Construct a line through B_{5}parallel to B_{3}C,bisecting the extended line segment BC at Câ€™.

Step 5: Construct a line through Câ€™ parallel to CA bisecting the extended line segment BA at Aâ€™.

Therefore,

We have the required triangle Î”A’BC’.

### Question 17. Construct a Î”ABC in which AB = 5 cm, âˆ B = 60Â°, altitude CD = 3 cm. Construct a Î”AQR similar to Î”ABC such that side of Î”AQR is 1.5 times that of the corresponding sides of Î”ACB.

**Solution:**

Follow these steps for the construction:

Step 1: Construct a line segment AB of 5 cm.

Step 2: At A, construct a perpendicular and break off AE = 3 cm.

Step 3: From E, construct EF || AB.

Step 4: From B, construct a ray creating an angle of 60 meeting EF at C.

Step 5: Connect CA. After that ABC is the triangle.

Step 6: From A, construct a ray AX creating an acute angle along AB and break

off 3 equal parts creating AA

_{1 }= A_{1}A_{2}= A_{2}A_{3}.Step 7: Connect A

_{2}and B.Step 8: From A , construct A^Bâ€™ parallel to A

_{2}B and Bâ€™Câ€™ parallel to BC.Therefore,

We have the required triangle Î”Câ€™ABâ€™.

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