Question

In India triangle ABC A-P-B and A-Q-C SUCH THAT seg PQ ||side BC and seg PQ divides triangle ABC in two parts whose areas are equal find the value of BP/AB

Answer 1

Answer:

Step-by-step explanation:

Given : PQ is parallel to BC and PQ divides triangle ABC into two parts.To find : BP/AB

Proof : In Δ APQ Δ 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 Answer.

Answer 2

Answer:

Step-by-step explanation:Secondary SchoolMath 5+3 pts

In India triangle ABC A-P-B and A-Q-C SUCH THAT seg PQ ||side BC and seg PQ divides triangle ABC in two parts whose areas are equal find the value of BP/AB

Ask for details FollowReport by Febhasanthos.

Answer:

Step-by-step explanation:

Given : PQ is parallel to BC and PQ divides triangle ABC into two parts.To find : BP/AB

Proof : In Δ APQ Δ 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 Answer.

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