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 D38×ar(ΔABC)
Let the perpendicular height of ΔABC=h Area(ΔABC)=(12)×BC×handarea(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))