If the perimeter of a trapezium-shaped floor is $56$ m, its non-parallel sides are $12$ m each and the altitude is $10$ m, find the area that the floor covers.

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Solution

The perimeter of floor is $56 m$

The length of non parallel sides is $12 m$

So the sum of lengths of parallel sides is $(a + b) 56 - 12 - 12 = 32 m$

The height of trapezium= $10 m$

The area of floor = $\frac{1}{2} (a + b) \times h$

= $\frac{1}{2} (32) \times 10$

$= 160 m^{2}$

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If the perimeter of a trapezium-shaped floor is 56 m, its non-parallel sides are 12 m each and the altitude is 10 m, find the area that the floor covers.

The perimeter of floor is 56mThe length of non parallel sides is 12mSo the sum of lengths of parallel sides is (a+b)56−12−12=32mThe height of trapezium=10mThe area of floor =12(a+b)×h=12(32)×10=160m2

If the perimeter of a trapezium-shaped floor is $56$ m, its non-parallel sides are $12$ m each and the altitude is $10$ m, find the area that the floor covers.

The perimeter of floor is $56 m$

The length of non parallel sides is $12 m$

So the sum of lengths of parallel sides is $(a + b) 56 - 12 - 12 = 32 m$

The height of trapezium= $10 m$

The area of floor = $\frac{1}{2} (a + b) \times h$

= $\frac{1}{2} (32) \times 10$

$= 160 m^{2}$