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प्रश्न
Parallelogram ABCD and rectangle ABEF are on the same base AB and have equal areas. Show that the perimeter of the parallelogram is greater than that of the rectangle.
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उत्तर
As the parallelogram and the rectangle have the same base and equal area, therefore, these will also lie between the same parallels.
Consider the parallelogram ABCD and rectangle ABEF as follows.
Here, it can be observed that parallelogram ABCD and rectangle ABEF are between the same parallels AB and CF.
We know that opposite sides of a parallelogram or a rectangle are of equal lengths. Therefore,
AB = EF (For rectangle)
AB = CD (For parallelogram)
∴ CD = EF
⇒ AB + CD = AB + EF ... (1)
Of all the line segments that can be drawn to a given line from a point not lying on it, the perpendicular line segment is the shortest.
∴ AF < AD
And similarly, BE < BC
∴ AF + BE < AD + BC ... (2)
From equations (1) and (2), we obtain
AB + EF + AF + BE < AD + BC + AB + CD
Perimeter of rectangle ABEF < Perimeter of parallelogram ABCD
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In the given figure, P is a point in the interior of a parallelogram ABCD. Show that
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In the given below fig. ABCD, ABFE and CDEF are parallelograms. Prove that ar (ΔADE)
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ABCD is a parallelogram, G is the point on AB such that AG = 2 GB, E is a point of DC
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(4) ar (ΔEBG) = ar (ΔEFC)
(5)ΔFind what portion of the area of parallelogram is the area of EFG.
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