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Question

In the figure give, D divides the side BC of ΔABC in the ratio 3:5. What is the area of ΔABD?

A
25×ar(ΔABC)
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B
35×ar(ΔABC)
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C
58×ar(ΔABC)
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D
38×ar(ΔABC)
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Solution

The correct option is D 38×ar(ΔABC)

Let the perpendicular height of ΔABC=h
Area (ΔABC)=(12)×BC×hand area (ΔABD)=(12)×BC×h
Now, D divides BC in the ratio 3:5
So, BD : DC = 3 : 5
And BD : BC = BD : (BD + DC)
= 3 : (3 + 5)
= 3 : 8
So, BD=(38)×BC
Area (ΔABD)=(12)×(38)×BC×h(38){(12)×BC×h}=(38×Area(ΔABC))

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