Question

Find the area of a trapezium shaped field having parallel sides 18cm and 10cm.
if the non-parallel sides are each of length 5 cm., find the area of the trapezium.​

Answer 1

$\huge\mathbb{SOLUTION:-}$

$\bold{Given}\begin{cases}\sf{parallel\:sides\:of\: trapezium=18 and 10cm}\\ \sf{Non\: parallel\:sides\:are\:equal=5cm}\end{cases}$

• Find the height of trapezium

$\sf Height {}^{2}=5{}^{2}-4{}^{2}\\ \\ \sf\leadsto h{}^{2}=25-16\\ \\ \sf\leadsto h=\sqrt{9}\\ \\ \sf\implies height=3cm$

• The height of trapezium is 3cm

Now Area of trapezium

$\boxed{\sf{Area\:of\: trapezium=\frac{1}{2}\times (sum\:of\: parallel\:sides)\times height}}$

$\sf Area=\frac{1}{2}\times (18+10)\times 3\\ \\ \sf\implies Area=\frac{1}{\cancel2}\times \cancel{28}\times 3\\ \\ \sf\implies Area=14\times 3\\ \\ \sf\implies Area=42cm{}^{2}$

$\boxed{\sf{Area\:of\: trapezium=42cm{}^{2}}}$

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Answer 2

$\underline \mathbb{SOLUTION}$

$\underline \mathbb{GIVEN}$

PARELLEL SIDES OF TRAPEZIUM = 18 cm and 10 cm

NON PARALLEL SIDES = 5 cm

$\underline \mathbb{TO\:FIND}$

$= > {height}^{2} = {5}^{2} - {4}^{2}$

$= > {h}^{2} = 25 - 16$

$= > h = \sqrt{9}$

$= > h = 3cm$

HENCE, HEIGHT OF TRAPEZIUM = 3 cm

$area \: of \: trapezium = \frac{1}{2} \times (sum \: of \: parallel \: sides) \times height$

$= > area = \frac{1}{2} \times (18 + 10) \times 3$

$= > area = \frac{1}{2} \times 28 \times 3$

$= > area = 14 \times 3$

$= > area = {42cm}^{2}$

$area \: of \: trapezium = 42 {cm}^{2}$

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