Question

Construct a $\Delta ABC$ with BC = 7 cm, $\angle B={60}^{\circ }$ and AB = 6 cm. Construct another triangle whose sides are $\frac{3}{4}$ times the corresponding sides of $\Delta ABC.$

Answer 1

Following steps will be followed to draw a triangle A'BC' whose sides are of corresponding sides of ABC.

1. Draw a line segment BC of 7 cm. Draw an arc of any radius while taking B as centre. Let it intersect line BC at point O. Now taking O as centre draw another arc to cut the previous arc at point O'. Joint BO' which is the ray making 60° with line BC.

2. Now draw an arc of 6 cm. radius, while taking, B as centre, intersecting extended line segment BO' at point A. Join AC. Triangle ABC is having AB = 6 cm.
BC = 7 cm and angle ABC = 60°.

3. Draw a ray BX making an acute angle with BC on opposite side of vertex A.

4. Locate 4 points (as 4 is greater in 3 and 4). B₁, B₂, B₃, B₄ on line segment BX.

5. Join B₄C and draw a line through B₃, parallel to B₄C intersecting BC at C'.

6. Draw a line through C' parallel to AC intersecting AB at A'. Triangle A'BC' is the required triangle.