NCERT Solutions
Class 10 Maths
Constructions

Ex.11.1 Q.6
Draw a triangle ABC with side BC = 7 cm, ∠B = 450, ∠A = 1050.
Then, construct a triangle whose sides are times the corresponding side of ∆ABC.
Give the justification of the construction.
Given, ∠B = 450, ∠A = 1050
Sum of all interior angles in a triangle is 1800.
∠A + ∠B + ∠C = 1800
=> 1050 + 450 + ∠C = 1800
=> ∠C = 1800 − 1500
=> ∠C = 300
The required triangle can be drawn as follows:
Step 1. Draw a ∆ABC with side BC = 7 cm, ∠B = 450, ∠C = 300.
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 4 and 3), B1, B2, B3, B4, on BX.
Step 4. Join B3 C. Draw a line through B4 parallel to B3 C intersecting extended BC at C'.
Step 5. Through C', draw a line parallel to AC intersecting extended line segment at C'.
Now, ∆A'BC' is the required triangle.
Justification:
The construction can be justified by proving that
A’B = AB, BC’ =
BC, A’C’ =
AC
In ∆ABC and ∆A'BC',
∠ABC = ∠A'BC' (Common)
∠ACB = ∠A'C'B (Corresponding angles)
So, ∆ABC ∼ ∆A'BC' (AA similarity criterion)
AB ÷ A’B = BC ÷ BC’ = AC ÷ A’C’ ……... (1)
In ∆BB3C and ∆BB4C',
∠B3BC = ∠B4BC'
(Common)
∠BB3C = ∠BB4C'
(Corresponding angles)
So, ∆BB3 C ∼ ∆BB4 C'
(AA similarity criterion)
BC ÷ BC’ = BB3 ÷ BB4
=> BC ÷ BC’ = …………. (2)
On comparing equations (1) and (2), we obtain
=> AB ÷ A’B = BC ÷ BC’ = AC ÷ A’C’ =
=> A’B = AB, BC’ =
BC, A’C’ =
AC
This justifies the construction.