A field is in the shape of a trapezium whose parallel sides are 25m and 10m. The non-parallel sides are 13m and 14m. What is the area of the field ?

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CW Share
Aug 10, 2017

$A r e a = 196 {\text{ m}}^{2}$

Explanation:

Draw a line $A E$, which is parallel to $B C$
$\implies A B C E$ is a parallelogram
$\implies$ Area of trapezium $=$ Area $\Delta A D E$ + Area $A B C E$
In $\Delta A D E$, as the lengths of the three sides are $13 , 14 \mathmr{and} 15$, we can use Heron's Formula to find its area.
Heron's formula : $A = \sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}$
where $s = \frac{\text{perimeter}}{2}$, and $a , b , c$ are the three known sides.
$\implies s = \frac{13 + 14 + 15}{2} = 21$
$\implies$ Area of DeltaADE = sqrt(21(21-13)(21-14)(21-15)
$= \sqrt{\left(21 \cdot 8 \cdot 7 \cdot 6\right)} = \sqrt{7056} = 84$
$84 = \frac{1}{2} \cdot 15 \cdot h$
$\implies h = \frac{84 \cdot 2}{15} = 11.2$
Area $A B C E = 10 \times h = 10 \times 11.2 = 112$
Hence Area of trapezium $A B C D =$ Area $\Delta A D E$ + Area $A B C E$
$= 84 + 112 = 196 {\text{ m}}^{2}$

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dk_ch Share
Aug 10, 2017

Let $A B C D$ be the trapezium whose parallel sides are $A B = 10 m \mathmr{and} C D = 25 m$ The non parallel sides are $B C = 14 m \mathmr{and} D A = 13 m$

$B E$ is drawn parallel to $A D$.It intersects $C D$ at E.$B F$ is the perpendicular drawn from $B$ to $C E$ . Hence $B E$ is the height of the triangle $B C E$ and trapezium $A B C D$

Now for triangle $B C E$

$B E = 13 m , B C = 14 m \mathmr{and} C E = 15 m$

So semi perimeter of $\Delta B C E$ is $s = \frac{1}{2} \left(13 + 14 + 15\right) = 21 m$

Hence area of $\Delta B C E$

$= \sqrt{21 \times \left(21 - 13\right) \times \left(21 - 14\right) \times \left(21 - 15\right)} {m}^{2}$

$= \sqrt{21 \times 8 \times 7 \times 6} {m}^{2} = 84 {m}^{2}$

Now
$\frac{1}{2} \times C E \times B F = 84$

$\implies \frac{1}{2} \times 15 \times B F = 84$

$\implies B F = \frac{84 \times 2}{15} = 11.2 m$

So area of the trapezium $A B C D$

$= \frac{1}{2} \times \left(A B + C D\right) \times B F$

$= \frac{1}{2} \times \left(10 + 25\right) \times 11.2 {m}^{2}$

$= 196 {m}^{2}$

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