In the given figure, the side AB of ∥ gm ABCD is produced to a point P. A line through A drawn parallel to CP meets CB produced in Q and the parallelogram PBQR is completed. Prove that : ar (∥ gm ABCD) = ar (∥ gm BPRQ).
Theorems on Area
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Question

In the given figure, the side AB of ∥ gm ABCD is produced to a point P. A line through A drawn parallel to CP meets CB produced in Q and the parallelogram PBQR is completed. Prove that : ar (∥ gm ABCD) = ar (∥ gm BPRQ).

Theorems on Area

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KnowledgeBoat · Accepted Answer

Δ AQC and Δ AQP shares the base AQ lies between the same parallels AQ and CP. ar (ΔAQC) = ar (ΔAQP) ......(1) Both triangles share a common part, which is Δ AQB. Subtract this area from both sides of Equation (1): ar (ΔAQC) - ar (ΔAQB) = ar (ΔAQP) - ar (ΔAQB) ar (ΔABC) = ar (ΔQPB) .......(2) We know that a diagonal of a parallelogram divides it into two triangles of equal area. In parallelogram ABCD, AC is the diagonal, ar(Δ ABC) = ar(∥ gm ABCD) In parallelogram BPRQ, QP is the diagonal, ar(Δ QPB) = ar(∥ gm BPRQ) Substituting the values in equation (2), we get : ar(∥ gm ABCD) = ar(∥ gm BPRQ) ar(∥ gm ABCD) = ar(∥ gm BPRQ). Hence, proved that ar(∥ gm ABCD) = ar(∥ gm BPRQ).

Mathematics

In the given figure, the side AB of ∥ gm ABCD is produced to a point P. A line through A drawn parallel to CP meets CB produced in Q and the parallelogram PBQR is completed. Prove that : ar (∥ gm ABCD) = ar (∥ gm BPRQ).

Theorems on Area

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