Question

In figure, D is a point on side BC such that B-D-C
find A(ABD)/A(ADC)​

Answer 1

Area of triangle = 1/2 X base X height

so, A(ABD)=1/2 X 3 X AD ...(1)

A(ADC)=1/2 X 5 X AD ...(2)

DIVIDING EQU (1) BY (2)

A(ABD)/A(ADC)=1/2 X 3 X AD/1/2 X 5 X AD

=>3/5

OR 0.6

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Answer 2

Given:

D is the point on BC

BD = 3cm

DC = 5 cm

To Find:

$\frac{ar(ABD)}{ar(ADC)}$

Solution:

Area of triangle = 1/2×base×height

Now,

In ΔABD,

area of ΔABD = 1/2×3×AD ..(i)

In ΔADC,

area of ΔADC = 1/2×5×AD ..(ii)

Then, dividing equation (i) and (ii),

⇒ ar(ΔABD)/ar(ΔADC)

⇒ 1/2×3×AD/1/2×5×AD

⇒ 3/5 = 0.6

Therefore, ar(ABD)/ar(ADC) = 0.6 cm