Question

Two parallel side s of a trapezium are 58cm and 42cm . The other two sides are of equal length which is 17cm . Find the area of trapezium.

Answer 1

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Answer 2

Answer:

The area of trapezium is $\bold{750\ \mathrm{cm}^{2}}$.

Solution:

We know that the area of trapezium

$\bold{=\frac{1}{2} \times(\text {sum of the parallel sides } a+b) \times \text {height of trapezium } h}$

$=\frac{1}{2}(a+b) h$

Let us draw a perpendicular AE and BF from the vertex A and B onto the opposite side CD shown in the figure.

Since, ACE is a right-angled triangle, Pythagoras theorem can be applied to find the height of the trapezium.

As AB and CD are parallel, when perpendiculars are drawn, AB = EF = 42

Then, the remaining CE and FD will be $\frac{58-42}{2}=\frac{16}{2}=8$.

Now, $A C^{2}=A E^{2}+C E^{2}$

$\Rightarrow 17^{2}=A E^{2}+8^{2}$

$\Rightarrow A E^{2}=289-64=225$

$\Rightarrow A E=\sqrt{225}=15$

Similarly, from right angled triangle BDF, $B D^{2}=B F^{2}+F D^{2}$

$\Rightarrow B F^{2}=17^{2}-8^{2} \Rightarrow B F=\sqrt{225}=15$  

Area of the trapezium ABCD $=\frac{1}{2} \times(52+48) \times 15$

$=\frac{1}{2} \times 100 \times 15$

$=750\ \mathrm{cm}^{2}$