P is a point on side BC of rectangle ABCD such that PA^2 + PD^2 = AD^2. What is the area of triangle PBD?
(1) The longer side of the rectangle ABCD measures 10, and P is such that BP:PC = 9:16.
(2) The shorter side of the rectangle ABCD measures 4.8, and P is such that BP:PC = 9:16.
Made Up!
PA^2 + PD^2 = AD^2
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- sanju09
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[spoiler]My answer is D.[/spoiler]theCodeToGMAT wrote:Is the Answer [spoiler]{E}[/spoiler]?
The mind is everything. What you think you become. -Lord Buddha
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It will feel like a Challenge Problem of the Week if one tries to finish the calculation part, which is never necessary on DS. If we can foresee where a statement can take us to, we are done.
The key catch of the problem lies in the stem itself, and that is that P must lie on the longer side of the rectangle ABCD, or the rectangle ABCD cannot be a square and BC is the longer side; and this should not take us too long by trying sophisticated algebra to realize that P is along the longer side, simple imagination is sufficient.
Those who fail to draw it right would fail to make the decision right. Further, we can note that the three right triangles ABP, APD, and PCD so formed are all similar, and also recall that triangles on the same base and under the same parallel lines are same in area. Hence, the area of triangle PBD is same as the area of triangle ABP which is ½ AP.AB.
One side and ratio can determine everything about three similar right triangles, therefore [spoiler]each statement alone is sufficient[/spoiler], hence [spoiler]D[/spoiler] is the right choice. We don't need to be on pins and needles really.
The mind is everything. What you think you become. -Lord Buddha
Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001
www.manyagroup.com
Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001
www.manyagroup.com