Question

Find the area of a trapezium whose parallel sides are $$11$$ m and $$25$$ m and the non parallel sides are $$15$$ m and $$13$$ m.

Solution

Trapezium $$ABCD$$ is divide into parallelogram $$AECD$$ and triangle $$CEB$$. Now in $$\triangle{CEB}$$, we have$$EB = 25 - 11 = 14 \; cm$$Using Heron’s theorem, we will first evaluate the semi-perimeter of $$\triangle{CEB}$$ and then evaluate its area.$$S = \cfrac{BE + EC+ BC}{2}$$$$\Rightarrow S = \cfrac{13 + 14 + 15}{2} = 21 \; cm$$Therefore,area of $$\triangle{CEB} = \sqrt{S \left( S - BE \right) \left( S - EC \right) \left( S - BC \right)}$$$$\Rightarrow$$ Area of $$\triangle{CEB} = \sqrt{21 \left( 21 - 13 \right) \left( 21 - 14 \right) \left( 21 - 15 \right)} = 84 \; {cm}^{2}$$As we know that,Area of triangle $$= \cfrac{1}{2} \times$$ Base $$\times$$ heightTherefore,Area of $$\triangle{CEB} = \cfrac{1}{2} \times 14 \times BF$$$$\Rightarrow BF = \cfrac{84 \times 2}{14} = 12 \; cm$$Again, as we know that,Area of trapezium $$=$$ Sum of parallel sides $$\times$$ heightTherefore,Area of trapezium $$ABCD = \cfrac{1}{2} \times \left( 11 + 25 \right) \times 12 = 216 \; {cm}^{2}$$Thus the area of trapezium is $$216 \; {cm}^{2}$$.Mathematics

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