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 

BB₁ = B₁B₂ = B₂B₃.

Step 5: Further connect B₃C.

Step 6: Construct B’C’ parallel to B₃C and C’A’ parallel to CA 

Therefore,

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 = B₁B₂ = B₂B₃ = B₃B₄ = B₄B_s = B₅B₆ = B₆B₇

Step 5: Now connect B₇ and C

Step 6: Construct B₅C’ parallel to B₇C 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 BB₁ = B₁B₂ = B₂B₂.

Step 4: Connect B₃C.

Step 5: From B₂, Construct B₂C’ parallel to B₃C 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 BB₁=  B₁B₂ = B₂B₃ = B₃B₄.

Step 5: Connect B₄ and C.

Step 6: From B₃C Construct C₃C’ parallel to B₄C 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 

BB₁ = B₁B₂ = B₂B₃ = B₃B₄ = B₄B₅ = B₅B₆ = B₆B₇.

Step 5: Connect B₅ and C.

Step 6: From B7, Construct B₇C’ parallel to B₅C 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 

AA₁ = A₁A₂ = A₂A₃ = A₃A₄ = A₄A₅

Step 5: Connect A₄ and B.

Step 6: From 45, Construct 45B’ parallel to  A₄B 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 

BB₁ = B₁B₂ = B₂B₃ = B₃B₄ = B₄B₅.

Step 5: Connect B₃ and C.

Step 6: From B₅, Construct B₅C’ parallel to B₃C 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 

DD₁ = D₁D₂ = D₂D₃ = D₃D₄.

Step 4: Connect D₂.

Step 5: Construct D₃A’ parallel to D₂A 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 

BB₁= B₁B₂ = B₂B₃=B₃B₄.

Step 5: Connect B₄ and C.

Step 6: From B₃, Construct B₃C’ parallel to B₄C 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 

AA₁ = A₁A₂ = A₂A₃ = A₃A₄=A₄A₅.

Step 6: Connect A₄ and B.

Step 7: From A₅, ConstructA₅B’ parallel to A₄B and B’C’ parallel to BC.

Therefore,

We have the required triangle ΔC’AB’.