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ABCD is a parallelogram. BC is produced to Q such that BC = CQ. Then1). area ($\triangle BCP$) = area ($\triangle DPQ$)2). area ($\triangle BCP$) > area ($\triangle DPQ$)3). area ($\triangle BCP$) < area ($\triangle DPQ$)4). area ($\triangle BCP$) + area ($\triangle DPQ$) = area ($\triangle BCD$) |
Answer» I am not SURE, may be area ($\triangle BCP$) = area ($\triangle DPQ$) is correct |
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