Question

In the given figure ΔABC and ΔAMP  are right angled at B and M respectively.
Given AC = 10 cm, AP = 15 cm and PM = 12 cm.
(i) Prove ΔABC ∼ Δ AMP.
(ii) Find AB and BC.

Answer 1

In ΔABC and ΔAMP
i)

∠ABC = ∠AMP   ...(90° each)
∠BAC = ∠PAM           ...(common angles)
∴ ΔABC ∼ ΔAMP ...(by AA similarity)
Hence proved.
(ii) ΔABC ∼ ΔAMP
⇒ `"AB"/"AM" = "BC"/"PM" = "AC"/"AP"`
⇒ `"BC"/"PM" = "AC"/"AP"`
⇒ `"BC"/(12) = (10)/(15)`
⇒ BC = `(10)/(15) xx 12`
BC = 8
Now, AB² = AC² - BC²
= 10² - 8²
= 100 - 64 = 36
AB = 6 cm.