Question

1
In Fig. 10.11, ABC is a triangle right angled at B. P is a point on AB
such that AP = 10 cm. From P, PM is drawn perpendicular to AC. If
MP = 6 cm and BC = 18 cm, find
(1) AM
(ii) BP
(iii) MC
Fig. 10.11​

Answer 1

AM = 8  cm .  BP = 14 cm  . MC = 22 cm

Step-by-step explanation:

Comparing Δ  AMP & ΔABC

∠A = ∠A  common

∠AMP = ∠ABC = 90°

=> Δ  AMP ≈ ΔABC

=> AM/AB = MP/BC = AP/AC

=> AM/AB = 6/18 = 10/AC

=> AM/AB = 1/3  = 10/AC

=> AC = 30

    AB = 3AM

AB² = AC² - BC²

=> AB² = 30² - 18²

=> AB² = (30 + 18)(30 - 18)

=> AB² = 48 * 12

=> AB² = 2 * 2  * 12 * 12

=> AB = 2 * 12

=> AB = 24 cm

24 = 3AM

=> AM = 8  cm

BP = AB - AP  = 24 - 10  = 14 cm

BP = 14 cm

MC = AC - AM

=>  MC = 30 - 8

=> MC = 22 cm

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