Draw a ΔABC in which BC = 6 cm, AB = 4 cm and AC = 5 cm. Draw a triangle similar to ΔABC with its sides equal to (3/4)^{th} of the corresponding sides of ΔABC.

#### Solution

Given that

Construct a triangle of sides BC = 6 cm, AB = 4 cm and AC = 5 cm and then a triangle similar to it whose sides are (3/4)^{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 radius = AC = 5 cm, draw an arc.

Step: III -With *B *as centre and radius = BC = 6 cm, draw an arc, intersecting the arc drawn in step II at *C.*

Step: IV -Joins *AC *and *BC *to obtain ΔABC.

Step: V -Below *AB, *makes an acute angle ∠BAX = 60°.

Step: VI -Along *AX,* mark off four points A_{1}, A_{2}, A_{3} and A_{4} such that AA_{1} = A_{1}A_{2} = A_{2}A_{3} = A_{3}A_{4}

Step: VII -Join A_{4}B.

Step: VIII -Since we have to construct a triangle each of whose sides is (3/4)^{th }of the corresponding sides of ΔABC.

So, we take three parts out of four equal parts on *AX *from point A_{3} draw A_{3}B' || A_{4}B, and meeting *AB *at B*’.*

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

Thus, ΔAB'C' is the required triangle, each of whose sides is (3/4)^{th }of the corresponding sides of ΔABC*.*