In triangle PQR, PQ=17 in, PR=10 in, QR=21 in. Find the altitude from P to the side QR.?

　
Drop perpendicular from P to point S on QR
PS = x
QS = y
SR = 21−y

x² + y² = 17² = 289
x² + (21−y)² = 10² = 100

100−(21−y)² = 289−y²
100−y²+42y−441 = 289−y²
42y−341 = 289
42y = 630
y = 15

x² = 289−y² = 289−225 = 64
x = 8

Altitude = 8 in

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Alternate method using law of cosines.

In △PQR:

PR² = PQ² + QR² − 2(PQ)(QR) cosQ
100 = 289 + 441 − 714 cosQ
cosQ = (289 + 441 − 100) / 714 = 630/714 = 15/17

sinQ = √(1−(15/17)²) = √(1−225/289) √(64/289) = 8/17

In △PQS:

sinQ = PS/PQ
8/17 = x/17
x = 8

Altitude = 8 in

s
= (1/2) × (PQ + PR + QR)
= (1/2) × (17 + 10 + 21) in
= 24 in

By Heron formula, area of ΔPQR
= √[s (s - PQ) (s - PR) (s - QR)]
=√[24 × (24 - 17) × (24 - 10) × (24 - 21)] in²
= 84 in²

Let h in be the altitude from P to the side QR.

Area of ΔPQR (in²) :
(1/2) × h × QR = 84
(1/2) × h × 21 = 84
h = 84 × 2 / 21
h = 8

The altitude from P to the side QR = 8 in