**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.

**Justification**

Here,

(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

Now,

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.