Question

In the given figure, PA - 12 cm.
BQ-14 cm and CR-16 cm
Area of POR is
(A) 336 cm
(B) 441 cm
(C) 484 cm
(D) 529​

Answer 1

Given : PA - 12 cm.  BQ-14 cm and CR-16 cm

To Find : Area of PQR

(A) 336 cm²

(B) 441 cm²

(C) 484 cm²

(D) 529​  cm²

Solution:

PC = PA  = 12 cm  Equal Tangent

QA  = QB = 14 cm  Equal Tangent

RB = RC = 16 cm

PQ = PA + QA = 12 + 14 = 26

QR = QB + RB = 14 + 16  = 30 cm

PR = PC + RC = 12 + 16 = 28 cm

26 , 28 & 30 cm

s = (26 + 28 + 30)/2 = 42

Area of ΔPQR

= √42(42 - 26)(42-28)(42-30)

= √42(16)(14)(12)

= √3*14(4²)(14)(2²*3)

= 4 * 2 * 3 * 14

= 336 cm²

option A is correct

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Answer 2

Answer:

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