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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 $$\times$$ height
Therefore,
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 $$\times$$ height
Therefore,
Area of trapezium $$ABCD = \cfrac{1}{2} \times \left( 11 + 25 \right) \times 12 = 216 \; {cm}^{2}$$
Thus the area of trapezium is $$216 \; {cm}^{2}$$.

1143384_1159143_ans_fc8ea03492ab4860b43a3f8010737b7b.jpeg

Mathematics

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