Ex 11.1, 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 sides of Δ ABC.
In Δ ABC
∠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°
Steps of construction
- Draw a ΔABC with side BC = 7 cm, ∠B = 45°, ∠C = 30°.
- Draw a ray BX making an acute angle with BC on the opposite side of vertex A.
- Locate 4 points (as 4 is greater in 4 and 3), B_1, B_2, B_3, B_4, on BX.
- Join B_3C. Draw a line through B_4 parallel to B_3C intersecting extended BC at C'.
- Through C', draw a line parallel to AC intersecting extended line segment at C'.
ΔA'BC' is the required triangle.
(BC^′)/BC=(BB_4)/(BB_3 ) = 4/3
Also, A’C’ is parallel to AC
So, the will make the same angle with line BC
∴ ∠ A’C’B = ∠ ACB
In Δ A’BC’ and ABC
∠ B = ∠ B
∠ A’C’B = ∠ ACB
Δ A’BC’ ∼ Δ ABC
Since corresponding sides of
similar triangles are in the same ratio
(A^′ B)/AB=(A^′ C^′)/AC=(BC^′)/BC
So, (A^′ B)/AB=(A^′ C^′)/AC=(BC^′)/BC =4/3.
This justifies the construction.