NCERT Solutions
Class 10 Maths
Constructions
Ex.11.1 Q.2
Construct a triangle of sides 4 cm, 5cm and 6cm and then a triangle similar to it whose sides are 
of the corresponding sides of the first triangle. Give the justification of the construction.
Step 1. Draw a line segment AB = 4 cm. Taking point A as centre, draw an arc of 5 cm radius.
Similarly, taking point B as its centre, draw an arc of 6 cm radius.
These arcs will intersect each other at point C. Now, AC = 5 cm and BC = 6 cm and ∆ABC is the required triangle.
Step 2. Draw a ray AX making an acute angle with line AB on the opposite side of vertex C.
Step 3. Locate 3 points A1, A2, A3 (as 3 is greater between 2 and 3) on line AX such that: AA1 = AA2 = AA3
Step 4. Join BA3 and draw a line through A2 parallel to BA3 to intersect AB at point B'.
Step 5. Draw a line through B' parallel to the line BC to intersect AC at C'.
Now, ∆AB'C' is the required triangle.
Justification:
The construction can be justified by proving that
AB’ = 
B, B’C’ = 
BC, AC’ = AC
By construction, we have B’C’ || BC
So, ∠A = ∠ABC
(Corresponding angles)
In ∆AB'C' and ∆ABC,
∠ AB’C’ = ∠ABC
(Proved above)
∠ B’AC’ = ∠BAC
(Common)
So, ∆ AB’C’ ∼ ∆ABC
(AA similarity criterion)
=> AB’ ÷ AB = BC’ ÷ BC = AC’÷ AC ……...... (1)
In ∆AA2 B' and ∆AA3B,
∠A2AB' = ∠A3 AB
(Common)
∠AA2 B' = ∠AA3 B
(Corresponding angles)
So, ∆AA2 B' ∼ ∆AA3 B
(AA similarity criterion)
=> AB’÷ AB = AA2 ÷ AA3
=> AB’÷ AB = ………... (2)
From equations (1) and (2), we obtain
AB’ ÷ AB = B’C’ ÷ BC = AC’ ÷ AC =
=> AB’ = AB, B’C’ = BC, AC’ = AC
This justifies the construction.
