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प्रश्न
In ΔABC, PQ is a line segment intersecting AB at P and AC at Q such that PQ || BC and PQ divides ΔABC into two parts equal in area. Find `(BP)/(AB)`
उत्तर
We have,
PQ || BC
And ar(ΔAPQ) = ar(trap. PQCB)
⇒ ar(ΔAPQ) = ar(ΔABC) – ar(ΔAPQ)
⇒ 2ar(ΔAPQ) = ar(ΔABC) ...(i)
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))/(2"Area"(triangleAPQ)) ="AP"^2/"AB"^2` ...[By using (i)]
`rArr1/2 = "AP"^2/"AB"^2`
`rArr1/sqrt2 = "AP"/"AB"^2`
`rArr1/sqrt2 = "AP"/"AB"` ...[Taking square root]
`rArr1/sqrt2 = (AB - BP)/(AB)`
`rArr1/sqrt2 = "AB"/"AB"-"BP"/"AB"`
`rArr1/sqrt2 = 1 - "BP"/"AB"`
`rArr"BP"/"AB" = 1 - 1/sqrt2`
`rArr"BP"/"AB" = (sqrt2-1)/sqrt2`
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