CBSE Class 9 Answered
In the figure, PQ||DC||AB. Prove that ar(ACP) = ar(BDQ)
Asked by Topperlearning User | 04 Jun, 2014, 01:23: PM
Expert Answer
PQ||DC
ar(PDC) = ar(QCD) ...(I)
[Triangle between same parallels and on the same base]
ACD and CDB are on same base CD and between same parallels, CD||AB
ar(ACD) = ar(CDB) …(II)
Adding (I) and (II),
ar(PDC) + ar(ACD) = ar(QCD) + ar(CDB)
ar(ACP) = ar(BDQ)
Answered by | 04 Jun, 2014, 03:23: PM
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