ABCD is a trapezium with parallel sides AB=a cm, CD=b cm. E and F are the midpoints of non-parallel sides. The ratio of ar(ABFE) and ar(EFCD) is_______?
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Answered by
29
Suppose AB is base. Suppose h is the perpendicular distance between AB and CD.
Then the area of ABCD is (1/2)(a+b)h.
EF || AB.
EF lies half way between AB and CD.
Therefore the perpendicular distances between AB and EF and between DF and CD are each (1/2)h.
The length of EF lies half way between that of AB and that of CD, so
EF = (1/2)(a+b).
So the area of ABFE is
(1/2)(a+(1/2)(a+b))(1/2)h = (1/8)(3a+b)h
and the area of EFCD is
(1/2)((1/2)(a+b)+b)(1/2)h = (1/8)(a+3b)h.
Therefore the ratio of areas of ABFE and EFCD is (3a+b)/(a+3b).
Answered by
14
3a+b:2a+2b is the answer.... please mark as brainliest
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