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# In the figure shown, QS is an altitude of right triangle PQR, PS = 16,

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Math Expert
Joined: 02 Sep 2009
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In the figure shown, QS is an altitude of right triangle PQR, PS = 16, [#permalink]

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26 Mar 2018, 00:17
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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. $$32\sqrt{5}$$

[Reveal] Spoiler:
Attachment:

2018-03-26_1113.png [ 9.37 KiB | Viewed 339 times ]
[Reveal] Spoiler: OA

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Re: In the figure shown, QS is an altitude of right triangle PQR, PS = 16, [#permalink]

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26 Mar 2018, 02:34
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 $$16 + 4 + 4\sqrt{5} + 8\sqrt{5} = 20 +12\sqrt{5}$$

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In the figure shown, QS is an altitude of right triangle PQR, PS = 16, [#permalink]

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26 Mar 2018, 02:53
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Attachment:

2018-03-26_1113.png [ 11.74 KiB | Viewed 252 times ]

In Triangle PQS, $$PQ^2 = PS^2 + QS^2$$ -> $$PQ^2 = 16^2 + x^2 = 256 + x^2$$

In Triangle RQS, $$RQ^2 = SR^2 + QS^2$$ -> $$RQ^2 = 4^2 + x^2 = 16 + x^2$$

In Triangle PQR, $$PR^2 = PQ^2 + RQ^2$$ -> $$20^2 = 256 + x^2 + 16 + x^2$$ -> $$2x^2 = 400 - 272 = 128$$ -> $$x^2 = 64$$ -> $$x = 8$$

Perimeter = $$20 + \sqrt{256 + 64} + \sqrt{16 + 64} = 20 + \sqrt{320} + \sqrt{80} = 20 + 2\sqrt{80} + \sqrt{80} = 20 + 3\sqrt{2^4*5} = 20 + 12\sqrt{5}$$

Therefore, the perimeter of triangle PQR is $$20 + 12\sqrt{5}$$(Option D)
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In the figure shown, QS is an altitude of right triangle PQR, PS = 16,   [#permalink] 26 Mar 2018, 02:53
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