A field is in the shape of a trapezium whose parallel sides are $25 m$ and $13 m$ . The non-parallel sides are $10 m$ and $10 m$ . Find the area of the field.

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Solution

Let $A B C D$ be the given trapezium in which $A B = 25 m$ , $D C = 13 m$ , $B C = 10 m$ and $A D = 10 m$ .
Through $C$ , draw $C E ∥ A D$ , meeting $A B$ at $E$ .
Also, draw $C F ⊥ A B$ .
Now, $E B = (A B - A E) = (A B - D C)$
$= (25 - 13) m = 12 m;$
$C E = A D = 10 m$ ; $A E = D C = 13 m .$

Now, in $△ E B C$ , we have $C E = B C = 10 m .$
So, it is an isosceles triangle.
Also, $C F ⊥ A B$
So, $F$ is the midpoint of $E B$ .

$∴$ $E F = \frac{1}{2} \times E B = 6 m .$

Thus, in right-angled $△ C F E$ , we have $C E = 10 m, E F = 6 m .$
By Pythagoras theorem, we have
$C F = \sqrt{C E^{2} - E F^{2}}$
$= 8 m .$
Thus, the distance between the parallel sides is $8 m$ .
Area of trapezium $A B C D = \frac{1}{2} \times (sum of parallel sides) \times (distance between them)$
$= 0.5 \times (25 + 13) \times 8 m^{2}$
$= 152 m^{2}$

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A field is in the shape of a trapezium whose parallel sides are 25m and 13m. The non-parallel sides are 10m and 10m. Find the area of the field.

A field is in the shape of a trapezium whose parallel sides are $25 m$ and $13 m$ . The non-parallel sides are $10 m$ and $10 m$ . Find the area of the field.