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by Anurag@Gurome » Thu May 24, 2012 10:12 pm
GmatKiss wrote:MGMAT CAT
By 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 dlencz » Mon May 28, 2012 1:03 pm
If you're running short on time or don't know how to solve you can easily make this a 50/50 problem.

We know the base is 25 and area is (.5)(b)(h), so two times the area must be a multiple of 25. D and E are the only ones that match the criteria.

To solve I considered common Pythagorean triples and saw 3-4-5 would in PQS and SQR both led to a height of 12.
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by ankita1709 » Mon May 28, 2012 7:31 pm
GmatKiss wrote:MGMAT CAT
Using similar triangle property you can easily work out that QS^2 = PS*SR
No need for such heavy calculation
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by 1947 » Mon May 28, 2012 10:21 pm
Can you please explain how did you reach at QS^2=PS*SR using similar triangle.
ankita1709 wrote:
GmatKiss wrote:MGMAT CAT
Using similar triangle property you can easily work out that QS^2 = PS*SR
No need for such heavy calculation
If my post helped you- let me know by pushing the thanks button. Thanks
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by aneesh.kg » Mon May 28, 2012 10:53 pm
Ankita is right. The concept of Similar Triangles is very helpful here.

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