Advertisements
Advertisements
Question
A steel wire, when bent in the form of a square, encloses an area of 121 cm2. The same wire is bent in the form of a circle. Find area the circle.
Advertisements
Solution
Area of a square = ( Side )2
⇒ 121 = ( Side )2
⇒ Side of a square = 11 cm
Now,
The perimeter of a square = Perimeter of a circle
⇒ 4 x Side = Perimeter of a circle
⇒ 4 x 11 = Perimeter of a circle
⇒ Perimeter of a circle = 44 cm
⇒ 2πr = 44 ....( r is radius of a circle )
⇒ r = `44/[2π] = 44/[ 2 xx 22/7]` = 7 cm
∴ Area of a circle = πr2 = `22/7 xx 7 xx 7 = 154 "cm"^2`
APPEARS IN
RELATED QUESTIONS
Find the area and perimeter of an isosceles right angled triangle, each of whose equal sides measure 10cm.
A rectangular park 358 m long and 18 m wide is to be covered with grass, leaving 2.5 m uncovered all around it. Find the area to be laid with grass.
Find the area of a parallelogram with base equal to 25 cm and the corresponding height measuring 16.8 cm.
A circle is inscribed in an equilateral triangle ABC is side 12 cm, touching its sides (the following figure). Find the radius of the inscribed circle and the area of the shaded part.

In the given figure, ABCD is a trapezium with AB || DC, AB = 18 cm DC = 32 cm and the distance between AB and DC is 14 cm. Circles of equal radii 7 cm with centres A, B, C and D have been drawn. Then find the area of the shaded region.
(Use \[\pi = \frac{22}{7}\]

The area of a circle whose area and circumference are numerically equal, is
Area of the largest triangle that can be inscribed in a semi-circle of radius r units is ______.
In the given figure, PQRS represents a flower bed. If OP = 21 m and OR = 14 m, find the area of the flower bed.

The length of the minute hand of a clock is 21 cm. The area swept by the minute hand in 10 minutes is
Find the area of the shaded region:

