Question

find the area of the each of the following polygons
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Answer 1

Answer:

Hey mate here is your answer

Step-by-step explanation:

22m + 9m + 9m + 6m + 12m + 4m = 62 m is the area.

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Answer 2

$\large\underline{\sf{Solution-}}$

Construction :-

• Produce AG to H intersecting DE at H.

• Through B, draw BI perpendicular to DE intersects DE at H.

Calculations

Now,

Dimensions of sides :-

• DE = 22 m

• EH = FG = 9 m

• BI = AH = AG + GH = 9 + 6 = 15 m

• AB = HI = 12 m

• DI = DE - EH - HI = 21 - 12 - 9 = 1 m

Now,

Area of figure EFGH

EFGH is a square with sides

• EF = FG = GH = HE = 9 m

$\rm :\longmapsto\:Area_{(square)} = {(side)}^{2}$

$\rm :\longmapsto\:Area_{(square)} = {(9)}^{2}$

$\bf :\longmapsto\:Area_{(square)} = 81 \: {m}^{2}$

Area of figure ABIH

ABIH is a rectangle with sides,

• Breadth, AB = 12 m

• Length, BI = 15 m

$\rm :\longmapsto\:Area_{(rectangle)} = length \times breadth$

$\rm :\longmapsto\:Area_{(rectangle)} = 15 \times 12$

$\bf :\longmapsto\:Area_{(rectangle)} = 180 \: {m}^{2}$

Area of figure DIBC

DIBC is a trapezium with DC || BI and

• DC = 4 m

• BI = 15 m

• DI = 1 m

$\rm :\longmapsto\:Area_{(trapezium)} = \dfrac{1}{2}(sum \: of \parallel \: sides) \times distance$

$\rm :\longmapsto\:Area_{(trapezium)} = \dfrac{1}{2}(4 + 15) \times1$

$\rm :\longmapsto\:Area_{(trapezium)} = \dfrac{1}{2}(19)$

$\bf :\longmapsto\:Area_{(trapezium)} = 9.5 \: {m}^{2}$

Thus,

$\bf :\longmapsto\:Area_{(polygon)}$

$\: \: \rm \: = \: \:Area_{(square)} + Area_{(rectangle)} + Area_{(trapezium)}$

$\: \: \rm \: = \: \:81 + 180 + 9.5$

$\: \: \rm \: = \: \:270.5 \: {m}^{2}$

$\bf :\implies\:Area_{(polygon)} = 270.5 \: {m}^{2}$

Additional Information :-

$\rm :\longmapsto\:Perimeter_{(rectangle)} = 2(l + b)$

$\rm :\longmapsto\:Perimeter_{(square)} = 4 \times side$

$\rm :\longmapsto\:Perimeter_{(rhombus)} = 4 \times side$

$\rm :\longmapsto\:Perimeter_{(circle)} = 2\pi \: r$

$\rm :\longmapsto\:Area_{(circle)} = \pi \: {r}^{2}$