Question

Construct an isosceles ∆ABC with base BC = 6 cm and altitude = 4 cm. Now, construct a triangle similar to ∆ABC, each of whose sides is $\frac{3}{2}$ times the corresponding sides of ∆ABC.

Answer 1

Steps of Construction :

Step 1. Draw a line segment BC = 6 cm.
Step 2. With B as centre, draw an arc each above and below BC.
Step 3. With C as centre, draw an arc each above and below BC.
Step 4. Join their points of intersection and thus, we obtain the perpendicular bisector of BC. Let it intersect BC at M.
Step 5. From D, cut an arc of radius 4 cm and mark the point as A.
Step 6. Join AB and AC.
Thus, △ABC is obtained .
Step 7. Extend BC to D, such that BD = $\frac{3}{2}$BC = $\frac{3}{2}$(6) cm = 9 cm.
Step 8. Draw DE $\parallel AC$, cutting AB produced to E.


Thus, △EBD is the required triangle, each of whose sides is $\frac{3}{2}$ the corresponding sides of ∆ABC.