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Question

If triangle ABC is an isosceles triangle in which AB = AC = 13 cm, then find the value of area of ΔADCarea of ΔEFB. (upto two decimal places)


  1. 0.52

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Solution

The correct option is A 0.52

In ΔADC and ΔFEB,

C=B

(since base angles of isosceles triangle are equal)

ADC=EFG=90

ΔADCΔEFB (by AA similarity criterion)

area of ΔADCarea of ΔEFB=AC2BE2=AC2(BC+CE)2

(The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides)

area of ΔADCarea of ΔEFB=(13)2(10+8)2

=(13)2(18)2=169324=0.521


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