# Class 10 RD Sharma Solutions – Chapter 11 Constructions – Exercise 11.2 | Set 1

### Question 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:**

Follow these steps for the construction:

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

Step 2: From the centre B take the radius of 4 cm and from the centre C take the radius 6 cm, construct arcs bisecting each other at the point A.

Step 3: Connect the lines AB and AC.

After that we will have ABC as the triangle.

Step 4: Construct a ray BX creating an acute angle with the line BC and break off 3 equal parts creating

BB

_{1}= B_{1}B_{2 }= B_{2}B_{3}.Step 5: Further connect B

_{3}C.Step 6: Construct B’C’ parallel to B

_{3}C and C’A’ parallel to CATherefore,

We have the required triangle ΔA’BC’.

### Question 2. Construct a triangle similar to a given ΔABC such that each of its sides is (5/7)^{th }of the corresponding sides of ΔABC. It is given that AB = 5 cm, BC = 7 cm and ∠ABC = 50°.

**Solution:**

Follow these steps for the construction:

Step 1: Construct a line segment BC of 7 cm.

Step 2: Construct a ray BX an angle of 50° and break off BA = 5 cm.

Step 3: Connect AC.

After that we have ABC is the triangle.

Step 4: Construct a ray BY creating an acute angle with BC and break off 7 equal parts creating

BB = B

_{1}B_{2 }= B_{2}B_{3 }= B_{3}B_{4 }= B_{4}B_{s }= B_{5}B_{6 }= B_{6}B_{7}Step 5: Now connect B

_{7}and CStep 6: Construct B

_{5}C’ parallel to B_{7}C and C’A’ parallel to CA.Therefore,

We have the required triangle ΔA’BC’.

### Question 3. Construct a triangle similar to a given ΔABC such that each of its sides is (2/3)^{rd} of the corresponding sides of ΔABC. It is given that BC = 6 cm, ∠B = 50° and ∠C = 60°.

**Solution:**

Follow these steps for the construction:

Step 1: Construct a line segment BC of 6 cm.

Step 2: Construct a ray BX creating an angle of 50° and CY creating 60° Along BC which bisect each other at A. After that, ABC is the triangle.

Step 3: From B, Construct one more ray BZ creating an acute angle below BC and intersect 3 equal parts, creating BB

_{1}= B_{1}B_{2}= B_{2}B_{2}.Step 4: Connect B

_{3}C.Step 5: From B

_{2}, Construct B_{2}C’ parallel to B_{3}C and C’A’ parallel to CA.Therefore,

We have the required triangle ΔA’BC’.

### Question 4. Construct a ΔABC in which BC = 6 cm, AB = 4 cm and AC = 5 cm. Draw a triangle similar to ΔABC with its sides equal to 3/4^{th} of the corresponding sides of ΔABC.

**Solution:**

Follow these steps for the construction:

Step 1: Construct a line segment BC of 6 cm.

Step 2: Along centre B and radius 4 cm and Along centre C and radius 5 cm,

Construct arcs’ bisecting each other at A.

Step 3: Connect AB and AC.

After that ABC is the triangle,

Step 4: Construct a ray BX creating an acute angle along BC and break off 4 equal parts creating BB

_{1}= B_{1}B_{2}= B_{2}B_{3}= B_{3}B_{4}.Step 5: Connect B

_{4}and C.Step 6: From B

_{3}C Construct C_{3}C’ parallel to B_{4}C and from C’,Step 7: Construct C’A’ parallel to CA.

Therefore,

We have the required triangle ΔA’BC’.

### Question 5. Construct a triangle Along sides 5 cm, 6 cm, and 7 cm and then another triangle whose sides are (7/5)^{th} of the corresponding sides of the first triangle. Give the justification of the construction.

**Solution:**

Follow these steps for the construction:

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

Step 2: Along centre B and radius 6 cm and Along centre C and radius 7 cm,

Construct arcs bisecting each other at A.

Step 3: Connect AB and AC.

After that ABC is the triangle.

Step 4: Construct a ray BX creating an acute angle along BC and break off 7 equal parts creating

BB

_{1}= B_{1}B_{2}= B_{2}B_{3}= B_{3}B_{4}= B_{4}B_{5}= B_{5}B_{6}= B_{6}B_{7}.Step 5: Connect B

_{5}and C.Step 6: From B7, Construct B

_{7}C’ parallel to B_{5}C and C’A’ parallel CA.Therefore,

We have the required triangle ΔA’BC’.

### Question 6. Construct a right triangle ABC in which AC = AB = 4.5 cm and ∠A = 90°. Draw a triangle similar to ΔABC with its sides equal to (5/4)^{th} of the corresponding sides of ΔABC.

**Solution:**

Follow these steps for the construction:

Step 1: Construct a line segment AB of 4.5 cm.

Step 2: At A, Construct a ray AX perpendicular to AB and break off AC = AB = 4.5 cm.

Step 3: Connect BC.

After that ABC is the triangle.

Step 4: Construct a ray AY creating an acute angle along AB and break off 5 equal parts creating

AA

_{1}= A_{1}A_{2}= A_{2}A_{3}= A_{3}A_{4}= A_{4}A_{5}Step 5: Connect A

_{4}and B.Step 6: From 45, Construct 45B’ parallel to A

_{4}B and B’C’ parallel to BC.Therefore,

We have the required triangle ΔAB’C’.

### Question 7. Construct a right triangle in which the sides (other than hypotenuse) are of lengths 5 cm and 4 cm. Then 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 line segment BC of 5 cm.

Step 2: At B, Construct perpendicular BX and break off BA = 4 cm.

Step 3: Connect AC ,

After that ABC is the triangle

Step 4: Construct a ray BY creating an acute angle along BC, and break off 5 equal parts creating

BB

_{1}= B_{1}B_{2}= B_{2}B_{3}= B_{3}B_{4}= B_{4}B_{5}.Step 5: Connect B

_{3}and C.Step 6: From B

_{5}, Construct B_{5}C’ parallel to B_{3}C and C’A’ parallel to CA.Therefore,

We have the required triangle ΔA’BC’.

### Question 8. Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are 3/2 times the corresponding sides of the isosceles triangle.

**Solution:**

Follow these steps for the construction:

Step 1: Construct a line segment BC of 8 cm.

Step 2: Construct its perpendicular bisector DX and break off DA = 4 cm.

Step 3: Connect AB and AC.

After that ABC is the triangle.

Step 4: Construct a ray DY creating an acute angle along OA and break off 3 equal parts creating

DD

_{1}= D_{1}D_{2}= D_{2}D_{3}= D_{3}D_{4}.Step 4: Connect D

_{2}.Step 5: Construct D

_{3}A’ parallel to D_{2}A and A’B’ parallel to AB meeting BC at C’ and B’ respectively.Therefore,

We have the required triangle ΔB’A’C’.

### Question 9. Draw a ΔABC with BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct a triangle whose sides are (3/4)^{th} of the corresponding sides of the ΔABC.

**Solution:**

Follow these steps for the construction:

Step 1: Construct a line segment BC of 6 cm.

Step 2: At B, Construct a ray BX creating an angle of 60° Along BC and break off BA of 5 cm.

Step 3: Connect AC. After that ABC is the triangle.

Step 4: Construct a ray BY creating an acute angle along BC and break off 4 equal parts creating

BB

_{1}= B_{1}B_{2}= B_{2}B_{3}=B_{3}B_{4}.Step 5: Connect B

_{4}and C.Step 6: From B

_{3}, Construct B_{3}C’ parallel to B_{4}C and C’A’ parallel to CA.Therefore,

We have the required triangle ΔA’BC’.

### Question 10. Construct a triangle similar to ΔABC in which AB = 4.6 cm, BC = 5.1 cm, ∠A = 60° Along scale factor 4 : 5.

**Solution:**

Follow these steps for the construction:

Step 1: Construct a line segment AB of 4.6 cm.

Step 2: At A, Construct a ray AX creating an angle of 60°.

Step 3: Along centre B and radius 5.1 cm.

Construct an arc bisecting AX at C.

Step 4: Connect BC.

After that ABC is the triangle.

Step 5: From A, Construct a ray AX creating an acute angle along AB and break off 5 equal parts creating

AA

_{1}= A_{1}A_{2}= A_{2}A_{3}= A_{3}A_{4}=A_{4}A_{5}.Step 6: Connect A

_{4}and B.Step 7: From A

_{5}, ConstructA_{5}B’ parallel to A_{4}B and B’C’ parallel to BC.Therefore,

We have the required triangle ΔC’AB’.

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