P and Q are points on the sides AB and AC of a ABC such that BP=CQ=x, PA= 6 cm,AQ= 20 cm, BC=25 cm. If PAQ and quadrilateral BPQC have equal areas, then the value of x
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Dear student

We know that,ar(APQ)+ar(BPQC)=ar(ABC)Therefore, 2(APQ)=ABC    ar(APQ)=ar(BPQC)Applying the sine formula, this is equivalent to 212×20×6×sinA=12x+20x+6sinA240sinA=x+20x+6sinA240=x2+26x+120x2+26x-120=0x2+30x-4x-120=0xx+30-4x+30=0x+30x-4=0And since x>0 the only solution is x=4.
Regards

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