Question

the given figure P and Q trisect the line segment bc of triangle ABC and triangle A P Q is equal to k into area triangle ABC find the value of k

Answer 1

let height of ABC be h as P & Q trisect the base PQ =1/3 BC and area of ABC = 1/2 * BC *h , area of APQ =1/2 *PQ *h
substitute PQ with 1/3 BC so u will get area of APQ =1/2 *BC*1/3 *h =1/3area of ABC

Answer 2

k = 1/3 if  P and Q trisect the line segment bc of ΔABC and Area of Δ APQ=  k * Area of Δ ABC

Step-by-step explanation:

P and Q trisect the line segment bc of triangle ABC

BP = PQ = CQ = BC/3

Lets draw AM ⊥ BC /PQ

Area of Triangle ABC = (1/2)BC * AM

Area of Δ APQ = (1/2) PQ * AM

Area of Δ APQ  = (1/2) (BC/3) * AM

=> Area of Δ APQ  = (1/2) * BC * AM / 3

=> Area of Δ APQ  = Area of Δ ABC / 3

Comparing with

Area of Δ APQ  = k * Area of Δ ABC

=> k = 1/3

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