NCERT Solutions
Class 10 Maths
Constructions

Ex.11.1 Q.3
Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are
of the corresponding sides of the first triangle. Give the justification of the construction.
Step 1. Draw a line segment AB of 5 cm.
Taking A and B as centre, draw arcs of 6 cm and 5 cm radius respectively.
Let these arcs intersect each other at point C.
∆ABC is the required triangle having length of sides as 5 cm, 6 cm, and 7 cm respectively.
Step 2. Draw a ray AX making acute angle with line AB on the opposite side of vertex C.
Step 3. Locate 7 points, A1, A2, A3, A4, A5, A6, A7 (as 7 is greater between 5and 7), o
n line AX such that AA1 = A1 A2 = A2 A3 = A3 A4 = A4 A5 = A5 A6 = A6 A7.
Step 4. Join B5 and draw a line through A7 parallel to BA5 to intersect extended line segment AB at point B'.
Step 5. Draw a line through B' parallel to BC intersecting the extended line segment AC at C'.
Now, ∆AB'C' is the required triangle.
Justification:
The construction can be justified by proving that
AB’ = AB, B’C’ =
BC, AC’ =
AC
In ∆ABC and ∆AB'C',
∠ABC = ∠AB'C'
(Corresponding angles)
∠BAC = ∠B'AC'
(Common)
So, ∆ABC ∼ ∆AB'C'
(AA similarity criterion)
=> AB ÷ AB’ = BC ÷ B’C’ = AC ÷ AC’ …….... (1)
In ∆AA5B and ∆AA7B',
∠A5 AB = ∠A7 AB' (Common)
∠AA5 B = ∠AA7 B'
(Corresponding angles)
So, ∆AA5B ∼ ∆AA7B'
(AA similarity criterion)
=> AB ÷ AB’ = AA5 ÷ AA7
=> AB ÷ AB’ = …………. (2)
On comparing equations (1) and (2), we obtain
AB ÷ AB’ = BC ÷ B’C’ = AC ÷ AC’ =
AB’ = AB, B’C’ =
BC, AC’ =
AC
This justifies the construction.