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

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

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