In δpqr, pq = 17 in, pr = 10 in, qr = 21 in. find the altitude from p to the side qr

To get the altitude we proceed as follows:
Half perimeter of the triangle is:
s=(17+10+21)/2
s=24
Area of the triangle will be:
A=√s(s-PR)(s-PQ)(s-QR)
A=√(24×7×3×14)
A=84

But:
A=1/2*base*height
hence:
84=1/2×21×h
⇒h=(2×84)/21=8

If the altitude from P meets QR at S, then,

PS = Sqrt {(PQ)^2-(QS)^2} = Sqrt {(PR)^2-(RS)^2}

Therefore,
Sqrt {(PQ)^2-(QS)^2} = Sqrt {(PR)^2-(RS)^2) => (PQ)^2-(QS)^2 = (PR)^2-(RS)^2

But, PQ= 17 in, QS+RS = QR = 21 in => QS = 21-RS, PR = 10 in

Substituting,
17^2 - (21-RS)^2 = 10^2-(RS)^2 => 289-(21-RS)(21-RS) = 100 -(RS)^2
289-(441-42RS +(RS)^2) = 100- (RS)^2 => 289-441+42RS-(RS)^2 = 100 -(RS)^2 => -152+42RS = 100 => 42RS = 252 => RS = 6 in
Then, QS = 21-6 = 15 in

Finally,
PS (altitude) = Sqrt {17^2-15^2} = 8 in