NCERT Solutions
Class 10 Maths
Constructions
Ex.11.1 Q.5
Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°.
Then construct a triangle whose sides are 
of the corresponding sides of the triangle ABC.
Give the justification of the construction.
A ∆A'BC' whose sides are 
of the corresponding sides of ∆ABC can be drawn as follows.
Step 1. Draw a ∆ABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°.
Step 2. Draw a ray BX making an acute angle with BC on the opposite side of vertex A.
Step 3. Locate 4 points (as 4 is greater in 3 and 4), B1, B2, B3, B4, on line segment BX.
Step 4. Join B4 C and draw a line through B3, parallel to B4C intersecting BC at C'.
Step 5. Draw a line through C' parallel to AC intersecting AB at A'.
∆A'BC' is the required triangle.
Justification:
The construction can be justified by proving
A’B = AB, BC’ = BC, A’C’ = AC
In ∆A'BC' and ∆ABC,
∠A'C'B = ∠ACB
(Corresponding angles)
∠A'BC' = ∠ABC
(Common)
So, ∆A'BC' ∼ ∆ABC
(AA similarity criterion)
A’B ÷ AB = BC’ ÷ BC = A’C’ ÷ AC …….... (1)
In ∆BB3C' and ∆BB4C,
∠B3BC' = ∠B4BC
(Common)
∠BB3C' = ∠BB4C
(Corresponding angles)
So, ∆BB3C' ∼ ∆BB4C
(AA similarity criterion)
BC’ ÷ BC = BB3 ÷ AA4
=> BC’ ÷ BC = …………. (2)
From equations (1) and (2), we obtain
=> A’B ÷ AB = BC’ ÷ BC = A’C’ ÷ AC =
=> A’B = AB, BC’ = BC, A’C’ = AC
This justifies the construction.