Let ABCD be a trapezoid with altitude of length 6 and bases AB of length 4 and DC of length 8. If P is the point of intersection of the diagonals AC and BD then the distance from P to DC is
(a) 4
(b) 7/2
(c) 9/2
(d) 14/3
(e) there are many possible answers
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artistocrat
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Begin by drawing a very good diagram. Draw all lines and figure out what you are asked to find. Look for relationships and patterns among diagrams in the figure? Is one figure similar to another? Take it from there and see what you come up with.artistocrat wrote:How do you solve this question?
OA is A.dtweah wrote:Let ABCD be a trapezoid with altitude of length 6 and bases AB of length 4 and DC of length 8. If P is the point of intersection of the diagonals AC and BD then the distance from P to DC is
(a) 4
(b) 7/2
(c) 9/2
(d) 14/3
(e) there are many possible answers
The triangles APB and DPC are similar and thus the distance from P to DC is twice the distance from P to AB. Hence the answer is (2/3)(6) = 4.
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dumb.doofus
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yes.. good point. the triangles would be always similar with the bases as the parallel sides of a trapezium..
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