Draw a right triangle in which the sides (other than hypotenuse) are of lengths 8 cm and 6 cm. Then construct another triangle whose sides are 3/4 times the corresponding sides of the given triangle.

Given,

AB = 8 cm,

BC = 6 cm and

Ratio = 3/4 of corresponding sides

Steps for construction;

(i) Draw a line BC of 8 cm.

(ii) Taking point B as center draw an arc with any radius

(iii) With the same radius draw an arc on the previous arc and mark the point Q.

(iv) Now taking Q as center with the same radius draw an arc again on previous arc and mark the point R.

(v) By taking the distance between Q and R as radius draw an arc from the point Q and with the same radius draw an arc from R point.

(vi) Draw a line from the point where both the arc intersect each other.

(vii) An angle of 90° is formed.

(viii) Draw a ray BX at an acute angle from point B.

(ix) Plot 4 points on BX so that B₁ = B₂ = B₃ = B₄

(x) Join B₄ to point B.

(xi) Draw a line from B₄ to point C and it is parallel to B₃.

(xii) B₃ meets BC at C’.

(xiii) So, BC’ parallel to BC.

Justification:

….. (i)

In ∆ ABC and ∆A’BC’

∠ABC = ∠A’BC’ (Common angle)

∠BCA = ∠BC’A’ (Coresponding angle)

So,

∆ ABC similar to ∆A’BC’ by AA similarity

Therefore,

BC’/BC = B₃/B₄ = 3/4 ………. (ii)

On compairing (i) and (ii)

A’B/AB = BC’/BC = 3/4 (Hence Proved)

So, we get the desired ∆ A’BC’

And ratio is 3/4 times of corresponding sides.

∆A’BC’ is the required triangle.