Question

ΔABC ∼ ΔPBR, BC = 8 cm, AC = 10 cm, ∠B = 90°, `"BC"/"BR" = 5/4`, then construct ΔPBR.

Answer 1

In ∆ABC, ∠B = 90°    ......[To prove]

∴ AC² = AB² + BC²    ......…[According to the Pythagorean theorem]

∴ 10² = AB² + 8²

∴ AB² = 100 – 64

∴ AB² = 36  

∴ AB = 6 सेमी ............[Taking the square root of both sides]

Steps of construction:

1. Draw a line BC of length 8 cm.
2. Take ∠B = 90° and draw an arc of length 6 cm on it. Name that point A.
3. Join line AC to get ∆ABC.
4. Draw a ray BX such that ∠CBX is the small angle.
5. Take points B₁, B₂, B₃, B₄, B₅ on ray BX such that BB₁ = B₁B₂ = B₂B₃ = B₃B₄ = B₄B₅.
6. Join points C and B₅.
7. Draw a line parallel to CB₅ from point B₄. This line intersects line BC at point R.
8. Draw a line parallel to side AC from point R. Name the intersection of this line and line AB as P. ∆PBR is an isosceles triangle similar to ∆ABC.

Construction: