Question

use the following figure, find its area​

Answer 1

To Find:-

The area of the figure.

Construction:-

Draw a line perpendicular to AB which joins point C.

Solution:-

In the figure,

AB || CD

AE || CD

AD || CE

Now,

AE = CD = 23m

CE = AD = 9m

So now we can clearly see that AECD is a rectangle, Whose dimensions are:-

Length = 23 m

Breadth = 9 m

Therefore Area of the rectangle = $\sf{Length\times Breadth\: sq.units}$

= $\sf{Area = 23\times 9}$

= $\sf{Area = 207\:m^2}$

Therefore area of the rectangle = 207 m²

Now,

In ∆BEC,

BC = 15 m

CE = 9 m

According to Pythagoras Theorem,

$\sf{(Hypotenuse)^2 = (Base)^2 + (Perpendicular)^2}$

$\sf{\implies (Base)^2 = (Hypotenuse)^2 - (Perpendicular)^2}$

= $\sf{(BE)^2 = (BC)^2 - (CE)^2}$

= $\sf{BE = \sqrt{(15)^2 - (9)^2}}$

= $\sf{BE = \sqrt{225-81}}$

= $\sf{BE = \sqrt{144}}$

= $\sf{BE = 12\:m}$

Now let us find the Area of ∆BEC

We know,

Area of triangle = $\sf{\dfrac{1}{2}\times base\times perpendicular}$ sq.units

$\sf{\dfrac{1}{2}\times 12\times 9}$

= $\sf{54\:m^2}$

Therefore area of the triangle is 54 m²

Now,

Area of the figure = Area of rectangle + Area of triangle

= $\sf{Area\:of\:figure = 207 + 54\:m^2}$

= $\sf{Area\:of\:the\:figure = 261\:m^2}$

Therefore the area of the figure is 261 m².

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