Question

Draw a right triangle in which the sides (other than the hypotenuse) are of lengths 4 cm and 3 cm. Now construct another triangle whose sides are \[\frac{3}{5}\] times the corresponding sides of the given triangle.

Answer 1

Construct a right triangle of sides  \[AB = 4 cm, AC = 3 cm \text { and } \angle A = 90^\circ \] and then a triangle similar to it whose sides are \[\left( \frac{3}{5} \right)^{th}\]

of the corresponding sides of ΔABC.

We follow the following steps to construct the given

Step of construction

Step: I- First of all we draw a line segment AB = 4 cm.

Step: II- With A as centre and draw an angle ∠A = 90°.

Step: III- With A as centre and radius AC = 3 cm.

Step: IV-Join BC to obtain right ΔABC.

Step: V- Below AB, makes an acute angle \[\angle BAX\].

Step: VI- Along AX, mark off five points  \[A_1 , A_2 , A_3 , A_4 \text { and } A_5\]  such that  `A A_1 = A_1A_2 = A_2 A_3 = A_3 A_4` \[= A_4 A_5\]

Step: VII- Join \[A_5 B\].

Step: VIII -Since we have to construct a triangle each of whose sides is \[\left( \frac{3}{5} \right)^{th}\] of the corresponding sides of right ΔABC.

So, we draw a line  `A_3B' `on AX from point `A_3`which is  \[A_3 B \lVert A_5 B\] and meeting AB at B’.

Step: IX- From B’ point draw B'C' || BC, and meeting AC at C’

Thus, ΔAB'C'is the required triangle, each of whose sides is \[\left( \frac{3}{5} \right)^{th}\] of the corresponding sides of ΔAB.