Students can understand the methods of constructing a triangle similar to a given triangle in this exercise. The RD Sharma Solutions Class 10 has all the answers prepared by experts at BYJUâ€™s for clearing doubts and giving the correct methodologies to solve a problem. Students can be greatly benefited from this resource. The RD Sharma Solutions for Class 10 Maths Chapter 11 Constructions Exercise 11.2 PDF is also provided below for students to help them with the right steps in solving questions.

## RD Sharma Solutions for Class 10 Chapter 11 Constructions Exercise 11.2 Download PDF

### Access RD Sharma Solutions for Class 10 Chapter 11 Constructions Exercise 11.2

**1. Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are (2/3) of the corresponding sides of it.
Solution:**

Steps of construction:

1. Draw a line segment BC = 5 cm.

2. With centre as B and radius 4 cm and with centre C and radius 6 cm, draw arcs from both points to intersect each other at A.

3. Now, join AB and AC. Then ABC is the triangle.

4. Draw a ray BX making an acute angle with BC and cut off 3 equal parts making BB_{1}Â = B_{1}B_{2}= B_{2}B_{3}.

5. Join B_{3}C.

6. Draw Bâ€™Câ€™ parallel to B_{3}C and Câ€™Aâ€™ parallel to CA.

Then, Î”Aâ€™BCâ€™ is the required triangle.

**2. Construct a triangle similar to a given Î”ABC such that each of its sides is (5/7) ^{thÂ }of the corresponding sides of Î”ABC. It is given that AB = 5 cm, BC = 7 cm and âˆ ABC = 50Â°.
Solution:**

Steps of construction:

1. Draw a line segment BC = 7 cm.

2. Draw a ray BX making an angle of 50Â° and cut off BA = 5 cm.

3. Join AC. Then ABC is the triangle.

4. Draw a ray BY making an acute angle with BC and cut off 7 equal parts making BB_{1} = B_{1}B_{2 }= B_{2}B_{3 }= B_{3}B_{4 }= B_{4}B_{s }= B_{5}B_{6 }= B_{6}B_{7
}

5. Now, join B_{7}Â and C

6. Draw B_{5}Câ€™ parallel to B_{7}C and Câ€™Aâ€™ parallel to CA.

Then, Î”Aâ€™BCâ€™ is the required triangle.

**3. Construct a triangle similar to a given Î”ABC such that each of its sides isÂ (2/3) ^{rd} of the corresponding sides of Î”ABC. It is given that BC = 6 cm, âˆ B = 50Â° and âˆ C = 60Â°.
Solution:**

Steps of construction:

1. Draw a line segment BC = 6 cm.

2. Draw a ray BX making an angle of 50Â° and CY making 60Â° with BC which intersect each other at A. Then, ABC is the triangle.

3. From B, draw another ray BZ making an acute angle below BC and then cut off 3 equal parts making BB_{1}Â = B_{1}B_{2}Â = B_{2}B_{2}

4. Now, join B_{3}C.

5. From B_{2}, draw B_{2}Câ€™ parallel to B_{3}C and Câ€™Aâ€™ parallel to CA.

Then Î”Aâ€™BCâ€™ is the required triangle.

**4. Draw a Î”ABC in which BC = 6 cm, AB = 4 cm and AC = 5 cm. Draw a triangle similar to Î”ABC with its sides equal toÂ (3/4) ^{th} of the corresponding sides of Î”ABC.
Solution:**

Steps of construction:

1. Draw a line segment BC = 6 cm.

2. With centre as B and radius 4 cm and with C as centre and radius 5 cm, draw arcs intersecting each other at A.

3. Join AB and AC. Then, ABC is the triangle.

4. Draw a ray BX making an acute angle with BC and cut off 4 equal parts making BB_{1 }=Â B_{1}B_{2Â }= B_{2}B_{3}Â = B_{3}B_{4}.

5. Join B_{4}Â and Câ€™.

6. From B_{3}C draw C_{3}Câ€™ parallel to B_{4}C and from Câ€™, draw Câ€™Aâ€™ parallel to CA.

Then Î”Aâ€™BCâ€™ is the required triangle.

**5. Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides areÂ (7/5) ^{th} of theÂ corresponding sides of the first triangle.
Solution:**

Steps of construction:

1. Draw a line segment BC = 5 cm.

2. With B as centre and radius 6 cm and with C as centre and radius 7 cm, draw arcs intersecting each other at A.

3. Now, join AB and AC. Then, ABC is the triangle.

4. Draw a ray BX making an acute angle with BC and cut off 7 equal parts making BB_{1}Â = B_{1}B_{2}Â = B_{2}B_{3}Â = B_{3}B_{4}Â = B_{4}B_{5}Â = B_{5}B_{6}Â = B_{6}B_{7}.

5. Join B_{5}Â and C.

6. From B_{7}, draw B_{7}Câ€™ parallel to B_{5}C and Câ€™Aâ€™ parallel CA.

Then, Î”Aâ€™BCâ€™ is the required triangle.

**6. Draw a right triangle ABC in which AC = AB = 4.5 cm and âˆ A = 90Â°. Draw a triangle similar to Î”ABC with its sides equal to (5/4) ^{th} of the corresponding sides of Î”ABC.
Solution:**

Steps of construction:

1. Draw a line segment AB = 4.5 cm.

2. At A, draw a ray AX perpendicular to AB and cut off AC = AB = 4.5 cm.

3. Now, join BC. Then, ABC is the triangle.

4. Draw a ray AY making an acute angle with AB and cut off 5 equal parts making AA_{1}Â = A_{1}A_{2}Â = A_{2}A_{3}Â = A_{3}A_{4}Â = A_{4}A_{5
}

5. Join A_{4}Â and B.

6. From A_{5}, draw A_{5}Bâ€™ parallel toÂ A_{4}B and Â Bâ€™Câ€™ parallel to BC.

Then, Î”ABâ€™Câ€™ is the required triangle.

**7. Draw a right triangle in which the sides (other than hypotenuse) are of lengths 5 cm and 4 cm. Then construct another triangle whose sides areÂ 5/3 times the corresponding sides of the given triangle.
Solution:**

Steps of construction:

1. Draw a line segment BC = 5 cm.

2. At B, draw perpendicular BX and cut off BA = 4 cm.

3. Now, join AC. Then, ABC is the triangle

4. Draw a ray BY making an acute angle with BC and cut off 5 equal parts making BB_{1}Â = B_{1}B_{2}Â = B_{2}B_{3}Â = B_{3}B_{4}Â = B_{4}B_{5}

5. Join B_{3}Â and C.

6. From B_{5}, draw B_{5}Câ€™ parallel to B_{3}C and Câ€™Aâ€™ parallel to CA.

Then, Î”Aâ€™BCâ€™ is the required triangle.