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question on right triangles

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question on right triangles

by queenisabella » Sun Feb 26, 2012 5:24 pm
why can't you assume that this is a 3:4:5 (or 9:16:25) triangle? I know that's the trick to avoid, but want to understand why you can't assume this is a 3:4:5 triangle.

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?
A) 72
B) 96
C) 108
D) 150
E) 200

OA is D not A. Would've been A if you fell into the trap assuming this was a 3:4:5 triangle, which I did.
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by pemdas » Sun Feb 26, 2012 5:44 pm
why you should *assume* 3-4-5 sides of right triangle when you are not said so and not to consider the similar triangles sharing two angles 90` at PSQ and RSQ and side QS ???
PS/QS=QS/SR or 16/QS=QS/9 or QS^2=16*9 or QS^2=(4*3)^2
to find the area of PQR you need QS (height) and PR (PS+SR) which are 12 and (16+9)=25
hence S(PQR)=1/2 *25*12=150
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by Anurag@Gurome » Mon Feb 27, 2012 2:56 am
It is nowhere given in the question that the triangle sides are in the 9 : 16 : 25.
We can do it by the method explained by Pemdas or by using the Pythagoras Theorem:
In triangle PQR, PR² = PQ² + QR²
(16 + 9)² = PQ² + QR²
625 = PQ² + QR² ...Equation (1)

In triangle QSR, QR² = SQ² + SR²
QR² = SQ² + 9² ...Equation (2)

In triangle PQS, PQ² = SQ² + SP²
PQ² = SQ² + 16² ...Equation (3)

From equations (2) and (3), put the value of QR² and PQ² in equation 1,
625 = SQ² + 16² + SQ² + 9²
2SQ² = 625 - 256 - 81
2SQ² = 288
SQ² = 144
SQ = 12
Area of triangle PQR = (1/2) * 25 * 12 = 25 * 6 = 150 sq units

The correct answer is D.
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by GMATGuruNY » Mon Feb 27, 2012 4:30 am
queenisabella wrote:why can't you assume that this is a 3:4:5 (or 9:16:25) triangle?
Image

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.

Your instincts are good: we should always be on the lookout for special triangles.
Each of the 3 triangles here is a multiple of a 3:4:5 triangle.
In triangle PQS, the sides are 12, 16, and 20.
In triangle QRS, the sides are 9, 12, and 15.
In triangle PQR, the sides are 15, 20 and 25.
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