Question 6:
Draw a triangle ABC with side BC = 7 cm, ∠B = 45°, ∠A = 105°. Then, construct a triangle whose sides are 4/3 times the corresponding side of ΔABC. Give the justification of the construction.
Answer:
∠B = 45°, ∠A = 105°
Sum of all interior angles in a triangle is 180°.
∠A + ∠B + ∠C = 180°
105° + 45° + ∠C = 180°
∠C = 180° − 150°
∠C = 30°
The required triangle can be drawn as follows.
Step 1
Draw a ΔABC with side BC = 7 cm, ∠B = 45°, ∠C = 30°.
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 B3C. Draw a line through B4 parallel to B3C intersecting extended BC at C’.
Step 5
Through C’, draw a line parallel to AC intersecting extended line segment at C’.
ΔA’BC’ is the required triangle.
![](https://entranceindia.com/wp-content/uploads/2016/05/Constructions-Class-X-Ex-11.1-Q-6.1.png)
Justification
The construction can be justified by proving that
![](https://entranceindia.com/wp-content/uploads/2016/05/Constructions-Class-X-Ex-11.1-Q-6.2.png)
In ΔABC and ΔA’BC’,
∠ABC = ∠A’BC’ (Common)
∠ACB = ∠A’C’B (Corresponding angles)
∴ ΔABC ∼ ΔA’BC’ (AA similarity criterion)
⇒ AB/A’B = BC/BC’ = AC/A’C’ ……… (1)
In ΔBB3C and ΔBB4C’,
∠B3BC = ∠B4BC’ (Common)
∠BB3C = ∠BB4C’ (Corresponding angles)
∴ ΔBB3C ∼ ΔBB4C’ (AA similarity criterion)
⇒ BC/BC’ = BB3/BB4
⇒ BC/BC’ = 3/4 ………………….. (2)
On comparing equations (1) and (2), we obtain
![](https://entranceindia.com/wp-content/uploads/2016/05/Constructions-Class-X-Ex-11.1-Q-6.3.png)
This justifies the construction.