Bunuel wrote:

In the figure shown, QS is an altitude of right triangle PQR, PS = 16, and RS = 4. Find the perimeter of PQR.

A. \(20 + 4\sqrt{5}\)

B. \(20 + 6\sqrt{5}\)

C. \(20 + 8\sqrt{5}\)

D. \(20 + 12\sqrt{5}\)

E. \(32\sqrt{5}\)

As the answers are close together we can't estimate and need to rely on specific geometric rules for an exact solution.

This is a Precise approach.

Since the larger triangle is right-angle then angle P and angle R sum to 90.

Since the two smaller triangle are also right-angle then angle SQP and angle P sum to 90 as do angles R and angle SRQ.

That is, angle SQP = angle R and angle SQR = angle P.

So our small triangles are similar both to each other and to the larger triangle.

Then using similarity of the two smaller triangles: SQ/SP = SR/SQ --> SQ^2 = 16*4 = 64 --> SQ = 8.

Then the hypotenuse of the small internal triangle can be calculated as \(\sqrt{8^2 + 4^2}=\sqrt{4^2(2^2+1)}=4\sqrt{5}\)

Then the hypotenuse of the large internal triangle can be calculated as the \(\sqrt{8^2 + 16^2}=\sqrt{8^2(1+2^2)}=8\sqrt{5}\)

Then our total perimeter is \(16 + 4 + 4\sqrt{5} + 8\sqrt{5} = 20 +12\sqrt{5}\)

(D) is our answer.

_________________

David

Senior tutor at examPAL

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