ABC is a triangle and PQ is a straight line meeting AB in P and AC in Q. If AP= 1 cm, PB = 3 cm, AQ = 1.5 cm, QC= 4.5 m, prove that area of APQis 1/6 of area of ABC
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AP = 1 cm , PB = 3cm , AQ= 1.5 cm , QC = 4.5cm ( given),
we have to prove, (APQ) = 1/16 ar (ABC)
proof: AP = 1 cm and PB = 3 cm then AB = 4 cm
since the sides are in proportion then the line PQ is parallel to BC
in triangle APQ and triangle ABC
angle A = angle A (common)
angle APQ = angle B (alt. int. angles)
angle AQP = angle C (alt. int. angles)
therefore, triangle APQ ~ triangle ABC
ar(APQ)/ ar(ABC) = AP2 / AB2 (area of two similar triangles is equal to the square of their coressponding side)
ar(APQ) / ar(ABC) = 12/ 42
ar (APQ) / ar (ABC) = 1/16
therefore ar(APQ) = 1/ 16 ar(ABC)
hence proved.
Manjula29:
Ignore the above answer. See below.
Firstly, it is supposed to be 1/16 instead of 1/6. [Solution] GIVEN :- AP = 1 cm , PB = 3cm , AQ= 1.5 cm , QC = 4.5cm . TO PROVE :- ar(APQ) = 1/16 ar(ABC). PROOF:- AP = 1 cm and PB = 3 cm then AB = 4 cm.
Since the sides are in proportion, then PQ || BC. In ∆ APQ and ∆ ABC —> (1) angle A = angle A (common)
(2) angle APQ = angle B (alt. int. angles)
(3) angle AQP = angle C (alt. int. angles). Therefore, ∆ APQ ~ triangle ABC. And, ar(APQ)/ ar(ABC) = AP² / AB² (area of two similar triangles is equal to the square of their corresponding side). So, ar(APQ) / ar(ABC) = 1²/ 4² = ar(APQ) / ar(ABC) = 1/16 . Therefore ar(APQ) = 1/ 16 ar(ABC) . Hence proved.
(2) angle APQ = angle B (alt. int. angles)
(3) angle AQP = angle C (alt. int. angles). Therefore, ∆ APQ ~ triangle ABC. And, ar(APQ)/ ar(ABC) = AP² / AB² (area of two similar triangles is equal to the square of their corresponding side). So, ar(APQ) / ar(ABC) = 1²/ 4² = ar(APQ) / ar(ABC) = 1/16 . Therefore ar(APQ) = 1/ 16 ar(ABC) . Hence proved.
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