Question

In PQR, Q = 90° and QR is 4 cm more than
PQ. If area of PQR = 96 cm2, find the sides of
the triangle​

Answer 1

Answer:

PQ=12,QR=16,PR=20

Step-by-step explanation:

LET ASSUME PQ BE X (PQ IS ALSO HEIGHT)

THEN QR BE X+4

NOW, AREA OF TRIANGE

$= ( \frac{1}{2} ) \times b \times h$

HERE BASE IS QR => X+4

MEANS, AREA OF TRIANGLE

$= ( \frac{1}{2} ) \times x + 4 \times x$

IT IS GIVEN AREA OF TRIANGLE = 96 CM^2 MEANS,

$( \frac{1}{2} ) \times x + 4 \times x = {92cm}^{2}$

SHIFTING 2 TO R.H.S

$1 \times x \times x + 4x = 92 \times 2 \\ {x}^{2} \ + 4x = 192$

${x}^{2} + 4x - 192 = 0 \\ {x}^{2} + 16x - 12x - 192 = 0 \\ x(x + 16) - 12(x + 16) = 0 \\ (x - 12)(x + 16) = 0 \\ x = 12or - 16$

THEREFORE

PQ=12

QR=12+4=>16

NOW PR BY PYTHAGORAS

$= \sqrt{ {pq}^{2} + {qr}^{2} } = \sqrt{ {12}^{2} + {16}^{2} } \\ = \sqrt{400} = 20$

SO AFTER ALL

PQ=12

QR=16

PR=20

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