Question

ΔABC ~ ΔPBR, BC = 8 cm, AC = 10 cm , ∠B = 90°, `"BC"/"BR" = 5/4` then construct ∆ABC and ΔPBR

Answer 1

Analysis: In ∆ABC, ∠B = 90°    ......[Given]

∴ AC² = AB² + BC²    ......[Pythagoras theorem]

∴ 10² = AB² + 8²

∴ AB² = 100 – 64

∴ AB² = 36  

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

Steps of construction:

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