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(2) InABC, A-P-B and A-Q-C such thatseg PQ İ side BC and seg PQ divides Δ ABC in twoBPparts whose areas are equal. Find the value of AB |
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Answer» Given : PQ is parallel to BC and PQ divides triangle ABC into two parts.To find : BP/AB Proof : InΔ APQΔ ABC, ∠ APQ =∠ ABC (As PQ is parallel to BC) ∠ PAQ =∠ BAC (Common angles) ⇒Δ APQ ~Δ ABC (BY AA similarity) Therefore, ar(Δ APQ)/ar(Δ ABC) = AP²/AB² ⇒ ar(Δ APQ)/2ar(Δ APQ) = AP²/AB² ⇒ 1/2 = AP²/AB² ⇒ AP/AB = 1/√2 ⇒ (AB - BP)/AB = 1/√2 ⇒ AB/AB - BP/AB = 1/√2 ⇒ 1 - BP/AB = 1/√2 ⇒ BP/AB = 1 - 1/√2 ⇒ BP/AB =√2 - 1/√2 |
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