Parallelograms and rectangles on same base and between same parallels are equal in area.
In the adjoining figure, parallelogram ABCD and rectangle ABEF are on the same base BC and between the same parallels AB and l.
Therefore, area of parallelogram ABCD = Area of rectangle ABEF
= AB × AF
AF is also the altitude of the parallelogram.
Hence, area of the parallelogram = base × height
= AB × AF
Solved examples for the rectangles and parallelograms on same base and between same parallels:
1. Parallelogram ABCD and rectangle ABFE have some base AB, and the length and breadth of the rectangle are as 7 cm and 4 cm. find the area of the parallelogram.
Solution:
Length of rectangle = 7 cm
Breadth of rectangle = 4 cm
Therefore, area of rectangle = 7 × 4 cm^{2}Since, rectangle ABFE and parallelogram ABCD are on the same base AB.
Therefore, area of parallelogram = Area of rectangle
Therefore, area of parallelogram ABCD = 28 cm^{2}2. In the adjacent figure ABCD is a parallelogram and EFCD is a rectangle.
Also, AL ⊥ DC. Prove that
(a) Area (ABCD) = Area (EFCD)
(b) Area (ABCD) = DC × AL
Solution:
(a) As a rectangle is also a parallelogram,
Therefore, Area (ABCD) = Area (EFCD)
(b) From above result,
Area (ABCD) = DC × FC (Area of the rectangle = length × breadth) ……………… (i)
As AL ⊥ DC, therefore, AFCL is also a rectangle
So, AL ⊥ FC ……………… (ii)
Therefore, from (i) and (ii) we get
Area (ABCD) = DC × AL
From the above result we see that, area of a parallelogram is the product of it’s any side and the corresponding altitude.
`Figure on Same Base and between Same Parallels
Parallelograms on Same Base and between Same Parallels
Parallelograms and Rectangles on Same Base and between Same Parallels
Triangle and Parallelogram on Same Base and between Same Parallels
Triangle on Same Base and between Same Parallels
8th Grade Math Practice
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