Given a quadrilateral ABCD, BE is drawn parallel to AC meeting DC produced at E.Also, ar(triangle ADC)=10 cm², ar(triangle ABC)=7 cm². Then ar(triangle ADE)
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ar(Δ ADE) = 17 cm²
Step-by-step explanation:
ar(Δ ADC)=10 cm²
ar(Δ ABC)=7 cm²
let say AE intersect BC at P
ar(Δ ADE) = ar(Δ ADC) + ar(Δ ACE)
ar(Δ ACE) = ar(Δ ACB) ( as Common Base AC & between Parallell lines AC & BE)
=> ar(Δ ACE) = 7 cm²
ar(Δ ADE) = ar(Δ ADC) + ar(Δ ACE)
=> ar(Δ ADE) = 10 + 7
=> ar(Δ ADE) = 17 cm²
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