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# In the figure, PS is perpendicular to QR. If PQ = PR = 26 and PS = 24

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Math Expert
Joined: 02 Sep 2009
Posts: 43804
In the figure, PS is perpendicular to QR. If PQ = PR = 26 and PS = 24 [#permalink]

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08 Feb 2018, 05:33
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In the figure, PS is perpendicular to QR. If PQ = PR = 26 and PS = 24, then QR =

(A) 14

(B) 16

(C) 18

(D) 20

(E) 22

[Reveal] Spoiler:
Attachment:

2018-02-08_1732.png [ 8.05 KiB | Viewed 395 times ]
[Reveal] Spoiler: OA

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In the figure, PS is perpendicular to QR. If PQ = PR = 26 and PS = 24 [#permalink]

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08 Feb 2018, 08:24
Bunuel wrote:

In the figure, PS is perpendicular to QR. If PQ = PR = 26 and PS = 24, then QR =

(A) 14

(B) 16

(C) 18

(D) 20

(E) 22

[Reveal] Spoiler:
Attachment:
2018-02-08_1732.png

PQR is an isosceles triangle, hence PS will bisect the base QR

so triangle PQS will be a right angle triangle with PQ as hypotenuse

using Pythagorean triplet 13-12-5 we have our sides PQ-PS-QS=26:24:10

so QS=10, hence QR=2*10=20

Option D
VP
Joined: 22 May 2016
Posts: 1329
In the figure, PS is perpendicular to QR. If PQ = PR = 26 and PS = 24 [#permalink]

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13 Feb 2018, 20:05
Bunuel wrote:

In the figure, PS is perpendicular to QR. If PQ = PR = 26 and PS = 24, then QR =

(A) 14

(B) 16

(C) 18

(D) 20

(E) 22

[Reveal] Spoiler:
Attachment:
2018-02-08_1732.png

Side PQ = PR. The triangle is isosceles.

Rule: The altitude to the base of an isosceles triangle bisects both the vertex and the base, and forms two congruent right triangles.

If you do not recognize the Pythagorean triplet 5x:12x:13x (where x=2 and sides of ONE right triangle are 10:24:26), let x = QS, and use the Pythagorean theorem:

$$x^2 + 24^2 = 26^2$$
$$x^2 + 576 = 676$$
$$x^2 = 100$$
$$x = 10 = QS$$

The base, from above, is bisected. QS = RS
QS + RS = base QR of big triangle
(10 * 2) = 20 = length of QR

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In the figure, PS is perpendicular to QR. If PQ = PR = 26 and PS = 24   [#permalink] 13 Feb 2018, 20:05
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