# Area of circle

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^{2}, you could count the total number of squares to get the area of this circle. Thus, if there were a total of 17 squares, the area of this circle would be 17 cm

^{2}However, it is easier to use one of the following formulas:

The region inside the circle is called its area. The formula to find the area is Area = π x r

^{2}; where the value of π is taken as either 3.14 or 22 / 7.

The unit of area is m

^{2}or cm

^{2}.

**Some solved examples**

1) Find the area of a circle of radius 4.2 cm.

**Solution :**Area of circle = π x r

^{2}

Area = 3.14 x (4.2)

^{2}

= 55.44 cm

^{2}

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2) A copper wire, when bent in the form of a square, encloses an area of 484 cm

^{2}. If the same wire is bent in the form of a circle, find the area enclosed by it.

**Solution :**Area of square = 484 cm

^{2}

⇒ (side )

^{2}= 22

^{2}

⇒ Side = 22 cm

So, the perimeter of square = 4 x side = 4 x 22 = 88 cm

Circumference of a circle = Perimeter of square

⇒ 2 x π x r = 88

⇒ 2 x 3.14 x r = 88

⇒ r = 14 cm

Area of circle = π x r

^{2}= 3.14 x 14

^{2}

Area = 616 cm

^{2}

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3) PQRS is a diameter of a circle of radius 6 cm. The lengths PQ,QR and RS are equal. Semicircles are drawn on PQ and QS as diameters.Find the area of the shaded region.

**Solution :**PS = 12 cm

As PQ = QR =RS

∴ PQ =QR =RS = 1/3 x PS = 1/3 x 12 = 4 cm.

QS = 2 PQ

QS = 2 x 4 = 8 cm

∴ Area of shaded region = Area of semicircle with PS as diameter + Area of semicircle with PQ as diameter – Area of semicircle with QS as diameter.

= ½ [ 3.14 x 6

^{2}+ 3.14 x 2

^{2}- 3.14 x 4

^{2}]

= ½ [ 3.14 x 36 + 3.14 x 4 – 3.14 x 16 ]

= ½ [ 3.14 ( 36 + 4 – 16)]

= ½ ( 3.14 x 24 ) = ½ x 75.36

∴ Area of shaded region = 37.68 cm

^{2}

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**Mensuration : Area and Perimeter**

• Circumference of Circle

• Area of Circle

• Circumference of Circle

• Area of Circle

7th grade math

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