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Answer 1

Given :-

A rectangle ABCD .

To Find :-

The area of the remaining piece of rectangle.

Used Concepts :-

• Area of Quadrant i.e πr²/4
• Area of semi circle i.e πr²/2
• Area of rectangle i.e l × b .

Solution :-

Here , AB = DC = 14 cm

BC = AD = 7 cm

At first , it is clearly understandable that the side BC is the radius of the Quadrant i.e 7cm . Now we knows that all radius are same of a Circle or parts of circle so EC = 7 cm .

Now , Diameter of DGE = DC - EC

=> 14 - 7

=> 7 cm

Let,. Length of ABCD = l = 14 cm

Breadth of ABCD = b = 7 cm

Radius of Quadrant BFEC = r = 7 cm

Radius of DGE = d = 7/2 cm

=> Area of the remaining piece of rectangle = Area of ABCD - Area of BFEC - Area of DGE

=> l × b - πr²/4 - πd²

=> 14 × 7 - 22 × 7 × 7/4 × 7 - 22 × 7 × 7 / 2 × 2 × 2 × 7

=> 98 - 77/2 - 77/4

=> 98 - ( 77/2 + 77/4 )

=> 98 - ( 154 + 77/4 )

=> 98 - 231/4

=> 392 - 231/4

=> 161/4 = 40.25 cm²

Henceforth , Our required answer is 40.25 cm² .

Answer 2

$\\ \bigstar \: \large \underline{ \underline{\bold{Given :-}}}$

• □ABCD is a rectangle.
• Length of AB = 14 cm and BC = 7 cm

$\\ \bigstar \: \large \underline{ \underline{\bold{To \: find :-}}}$

• We have to calculate the area of the remaining piece of the rectangle.

$\\ \bigstar \: \large \underline{ \underline{\bold{Solution:-}}}$

At first, We have to find the area of Rectangle

So,

$\\ \: \: \: \: \: \sf \: \: Area \: of \: Rectangle = length \times breadth$

$\\ \: \: \: \: \: \sf \: \: \: \: \: \: \: \: \ \: \: \: \: \: \: \: \: \: \: \: \: \:\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = AB × BC$

$\\ \: \: \: \: \: \sf \: \: \: \: \: \: \: \: \ \: \: \: \: \: \: \: \: \: \: \: \: \:\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 14 \times 7$

$\\ \: \: \: \: \: \sf \: \: \: \: \: \: \: \: \ \: \: \: \: \: \: \: \: \: \: \: \: \:\: \: \: \: \: \: \: \: \: \: \: \: \: \: \green{\: = 98 {cm}^{2} }$

Now, we have to find the area of quarter circle And Area of semi-circle

So,

$\\ \sf \: Area \: of \: quarter \: circle( BFES)= \frac{1}{4} \pi {(BC)}^{2}$

$\\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \: \: \: \: \: \: \: \: \: = \frac{1}{4} \pi {(BC)}^{2}$

$\\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \: \: \: \: \: \: \: \: \: = \frac{1}{4} \pi{(7)}^{2}$

$\\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \: \: \: \: \: \: \: \: \: = \frac{49}{4} \pi$

And ,

$\\ \sf \: Area \: of \: semi \: circle( DGE)= \frac{1}{2} \pi { \bigg( \frac{7}{2} \bigg)}^{2}$

$\\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \: \: \: \: \: \: \: \: \: = \frac{1}{2} \times { \bigg( \frac{49}{4} \bigg)} \pi$

So,

Area of remaining piece = Area of Rectangle - (Area of quarter circle + Area of semi circle )

$\\ \sf \: \: \: \: \: \implies \: 98 - \bigg( \frac{49}{4} \: \pi + \frac{1 }{2} \times \frac{49}{4} \pi\bigg)$

$\\ \sf \: \: \: \: \: \implies \: 98 - \frac{49}{4} \pi\bigg( 1 + \frac{1 }{2} \bigg)$

$\\ \sf \: \: \: \: \: \implies \: 98 - \frac{49}{4} \pi\bigg( \frac{1 + 2 }{2} \bigg)$

$\\ \sf \: \: \: \: \: \implies \: 98 - \frac{49}{4} \times \pi \times \frac{3 }{2}$

$\\ \sf \: \: \: \: \: \implies \: 98 - \frac{49}{4} \times \frac{22}{7} \times \frac{3 }{2}$

$\\ \sf \: \: \: \: \: \implies \: 98 - \frac{231}{4}$

$\\ \sf \: \: \: \: \: \implies \: 98 - 57.75$

$\\ \sf \: \: \: \: \: \implies \frak{\: \pink{ 40.25 {cm}^{2}} }$

Hence, The area of the remaining piece of the rectangle is 40.24 cm².