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प्रश्न
In ABC, P divides the side AB such that AP : PB = 1 : 2. Q is a point in AC such that PQ || BC. Find the ratio of the areas of ΔAPQ and trapezium BPQC.
उत्तर
We have,
PQ || BC
And `"AP"/"PB"=1/2`
In ΔAPQ and ΔABC
∠A = ∠A [Common]
∠APQ = ∠B [Corresponding angles]
Then, ΔAPQ ~ΔABC [By AA similarity]
By area of similar triangle theorem
`("area"(triangleAPQ))/("area"(triangleABC))="AP"^2/"AB"^2`
`rArr("area"(triangleAPQ))/(("area"(triangleAPQ)+"area"("trap.BPQC")))=1^2/3^2` `["AP"/"PB"=1/2]`
⇒ 9ar (APQ) = ar(ΔAPQ) + ar(trap. BPQC)
⇒ 9ar (APQ) − ar(ΔAPQ) = ar(trap. BPQC)
⇒ 8ar(APQ) = ar(trap. BPQC)
`rArr("area"(triangleAPQ))/("area"("trap.BPQC"))=1/8`
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