Ex 11.1, 7
Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. Then construct another triangle whose sides are 5/3 times the corresponding sides of the given triangle.
Given sides other than hypotenuse in a right angled triangle.
∴ Both sides will be perpendicular to each other
Steps of construction
- Draw a line segment AB = 4 cm. Draw a ray RA making 90° with it.
- Taking A as center and 3 cm as radius, draw an arc intersecting the ray RA at C.
- Join BC. ΔABC is the required triangle.
- Locate 5 points (as 5 is greater in 5 and 3), A_1, A_2, A_3, A_4, A_5, on line segment AX such that 〖AA〗_1 = A_1 A_2 = A_2 A_3 = A_3 A_4= A_4 A_5.
- Join A_3B. Draw a line through A_5 parallel to A_3B intersecting extended line segment AB at B'.
- Through B', draw a line parallel to BC intersecting extended line segment AC at C'.
ΔAB'C' is the required triangle.
(AB^′)/AB=(AA_5)/(AA_3 ) = 5/3
Also, A’C’ is parallel to AC
So, the will make the same angle with line BC
∴ ∠ AB’C’ = ∠ ABC
In Δ AB’C’ and Δ ABC
∠ A = ∠ A
∠ AB’C’ = ∠ ABC
Δ A’BC’ ∼ Δ ABC
Since corresponding sides of
similar triangles are in the same ratio
AB/AB′=BC/(B^′ C^′ )=AC/(AC^′ )
So, AB/AB′=BC/(B^′ C^′ )=AC/(AC^′ )=5/3
This justifies the construction.