Question

Find the area of following diagram​

Answer 1

Formula Used :-

$\boxed{\red{\sf\:Area_{(rectangle)} = Length \times Breadth}}$

$\boxed{\red{\sf\:Area_{(trapezium)} = \dfrac{1}{2}(sum \: of \parallel \: sides) \times \: distance}}$

Solution :-

The given figure consists of 3 figures.

Figure 1, is a trapezium whose

• Parallel sides are 8 cm and 12 cm

• Distance between parallel sides is 10 cm

So,

$\rm :\longmapsto\:Area_{(figure \: 1)} \: = \: Area_{(trapezium)}$

$\rm :\longmapsto\:Area_{(figure \: 1)} \: = \: \dfrac{1}{2} \times (8 + 12) \times 10$

$\rm :\longmapsto\:Area_{(figure \: 1)} \: = \: 20 \times 5$

$\rm :\longmapsto\:Area_{(figure \: 1)} \: = \: 100 \: {cm}^{2}$

Figure 2 is a rectangle of dimensions

• Length = 20 cm

• Breadth = 8 cm

Thus,

$\rm :\longmapsto\:Area_{(figure \: 2)} \: = \: Area_{(rectangle)}$

$\rm :\longmapsto\:Area_{(figure \: 2)} \: = \: 20 \times 8$

$\rm :\longmapsto\:Area_{(figure \: 2)} \: = \: 160 \: {cm}^{2}$

Figure 3. consist of a trapezium whose

• Parallel sides are 8 cm and 14 cm

• Distance between parallel sides is 12 cm

So,

$\rm :\longmapsto\:Area_{(figure \: 3)} \: = \: Area_{(trapezium)}$

$\rm :\longmapsto\:Area_{(figure \: 3)} \: = \: \dfrac{1}{2} \times (14 + 8) \times 12$

$\rm :\longmapsto\:Area_{(figure \: 3)} \: = \: (14 + 8) \times 6$

$\rm :\longmapsto\:Area_{(figure \: 3)} \: = \: 22\times 6$

$\rm :\longmapsto\:Area_{(figure \: 3)} \: = \: 132 \: {cm}^{2}$

Therefore,

• The area of whole figure is

$\rm :\longmapsto\:Area_{(figure)} = Area_{(fig1)}+Area_{(fig2)}+Area_{(fig3)}$

$\rm :\longmapsto\:Area_{(figure)} \: = 100 + 160 + 132$

$\bf :\longmapsto\:Area_{(figure)} \: = 392 \: {cm}^{2}$

Additional Information :-

$\boxed{\red{\sf\:Area_{(square)} = {(side)}^{2}}}$

$\boxed{\red{\sf\:Area_{(square)} = \dfrac{1}{2} {(diagonal)}^{2}}}$

$\boxed{\red{\sf\:Perimeter_{(rectangle)} = 2(Length + Breadth)}}$

$\boxed{\red{\sf\:Perimeter_{(square)} = 4 \times side}}$

$\boxed{\red{\sf\:Perimeter_{(rhombus)} = 4 \times side}}$

$\boxed{\red{\sf\:Area_{(circle)} = \pi \: {r}^{2}}}$

$\boxed{\red{\sf\:Perimeter_{(circle)} = 2\pi \: r}}$