mgmat pythagorus

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mgmat pythagorus

by rommysingh » Fri Sep 11, 2015 11:26 am
In the diagram to the right, triangle PQR has a right angle at Q and line segment QS is perpendicular to PR. If line segment PS has a length of 16 and line segment SR has a length of 9, what is the area of triangle PQR?



72


96


108


150


200

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by GMATGuruNY » Fri Sep 11, 2015 12:04 pm
rommysingh wrote:In the diagram to the right, triangle PQR has a right angle at Q and line segment QS is perpendicular to PR. If line segment PS has a length of 16 and line segment SR has a length of 9, what is the area of triangle PQR?



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QS is a height drawn through the right angle of triangle PQR.

A height drawn through the right angle of a triangle forms THREE SIMILAR TRIANGLES.

In the figure above:
The angles of triangle PQS are x-y-90.
The angles of triangle QRS are x-y-90.
The angles of triangle PQR are x-y-90.
Since each triangle has the same combination of angles, all 3 triangles are similar.

Since triangle PQS is similar to triangle QRS, the ratio of the legs in each triangle must be the same.
In triangle PQS:
(side opposite x)/(side opposite y) = h/16.
In triangle QRS:
(side opposite x)/(side opposite y) = 9/h.
Since the ratios must be equal:
h/16 = 9/h
h² = 9*16
h = 3*4 = 12.

Area of PQR = (1/2)bh = (1/2)(12)(25) = 150.

The correct answer is D.
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