SOLVE IT PLZ
Asked by | 30th Jan, 2009, 06:23: PM
Draw the triangle as mentioned.
DE is paralle to BC
triangle ADE can be shown similar to triangle ABC by AA rule.
we know that the ratio of the areas of two similar triangles is equal to theratio of squares of their corresponding sides.
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 | 30th Jan, 2009, 11:30: PM
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