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CBSE Class 10 Answered

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Asked by | 30 Jan, 2009, 06:23: PM
answered-by-expert Expert Answer

Draw  the triangle as mentioned.

Now,

DE is paralle to BC

So,

triangle ADE can be shown similar to triangle ABC by AA rule.

Now,

we know that the ratio of the areas of two similar triangles is equal to theratio of squares of their corresponding sides.

So

Area(triangle ABC)/Area(triangle ADE)=49/25..(...as BC:DE=7:5)

Subtracting 1 from both sides we get,

[Area(triangle ABC)/Area(triangle ADE)] -1=[49/25]-1

So we get

Area (triangle ABC)-Area(triangle ADE)/Area(triangle ADE)=49-25/25

But the numerator of the LHS will come out to be area of trapezium BCED according to the figure

So, we have got,

Area(trapezium BCED)/Area(triangle ADE)=24/25

 

Answered by | 30 Jan, 2009, 11:30: PM
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