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Solution - Triangles on the Same Base and Between the Same Parallels

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ConceptTriangles on the Same Base and Between the Same Parallels  

Question

The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q and then parallelogram PBQR is completed (see the following figure). Show that

ar (ABCD) = ar (PBQR).

[Hint: Join AC and PQ. Now compare area (ACQ) and area (APQ)]

Solution

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Solution for concept: Triangles on the Same Base and Between the Same Parallels. For the course 8th-10th CBSE
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