In $△ A B C$ , if $D$ is a point in $B C$ and divides it in the ratio $3 : 5$ i.e., if $B D : D C = 3 : 5$ then, $a r (△ A D C) : a r (△ A B C) = ?$

3 : 5

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3 : 8

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5 : 8

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8 : 3

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Solution

The correct option is C $5 : 8$
Given: $B D = 3 x, D C = 5 x$
$(\frac{A r e a o f △ A D C}{A r e a o f △ A B C}) = \frac{\frac{1}{2} \times D C \times A E}{\frac{1}{2} \times (B D + D C) \times A E}$
$= \frac{5 x \times A E}{8 x \times A E} = \frac{5}{8}$

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In △ABC, if D is a point in BC and divides it in the ratio 3:5 i.e., if BD:DC=3:5 then, ar(△ADC):ar(△ABC)=?

The correct option is C 5:8Given: BD=3x,DC=5x(Areaof△ADCAreaof△ABC)=12×DC×AE12×(BD+DC)×AE=5x×AE8x×AE=58

In $△ A B C$ , if $D$ is a point in $B C$ and divides it in the ratio $3 : 5$ i.e., if $B D : D C = 3 : 5$ then, $a r (△ A D C) : a r (△ A B C) = ?$

The correct option is C $5 : 8$
Given: $B D = 3 x, D C = 5 x$
$(\frac{A r e a o f △ A D C}{A r e a o f △ A B C}) = \frac{\frac{1}{2} \times D C \times A E}{\frac{1}{2} \times (B D + D C) \times A E}$
$= \frac{5 x \times A E}{8 x \times A E} = \frac{5}{8}$