Test: Area Related to Circles - 3 - Class 10 MCQ

Therefore, the correct answer is Option 1: 22 cm.

According to question, Diameter (BD) = 2 x radius = 2 x 12 = 24 cm
∴ Area of square

Let the radius of the circle be rr and the side of the square be a.a. Then according to the question, 

  Now, Ratio of their perimeters

Area of the sector

⇒ Area of the sector 
= 462 sq. cm

Given : Diameter (d) = 4.2 cm
∴ Circumference  

Since if a circle inscribed a square. then the radius of the circle is half of to side of square.
∴  Radius = 10/2 = 5 cm
∴ Area of the circle = πr² = π(5)² = 25π sq. cm 

Given: Radius (r) = 10 cm
Let side of the square be x cm
Now. using Pythagoras theorem.
x² + x² = (2r)² ⇒ 2x² = (20)²
⇒ 2x² = 400 ⇒ x² = 200 sq. cm
Therefore. area of the square = 200 sq. cm. 

Time duration between 4.00 om and 4.25 pm = 25 minutes
∵ Angle described by minute hand in 60 minutes = 360°
∴ Angle described by minute hand in 25 minutes  

Given: Length of wire = 66 m = 6600 cm and Circumference = 13.2 cm

Let the radius of the circle be rr and side of the square be a. Then, according to question,

Now, ratio of their areas, 

Let radius of the circle berunits.
∴ Area of the circle = Circumference of the circle 

Here Side of square ABCD = AB = 3 x diameter of circular design
⇒ AB = 3 x (2 x 7) = 42 cm
∴ Area of square = 42 x 42 = 1764 cm²
And Area of one circular design   
And Area of one circular design
∴ Area of 9 circular designs = 154 x 9 = 1386 sq. cm
∴ Area of remaining portion of handkerchief = 1764 - 1386 = 378 sq. cm

Area of the shaded region
Area of the shaded region 
Area of the shaded region = 154 sq. cm 

Let radius of the protractor be r 
∴ Perimeter of protractor = Perimeter of semicircle + Diameter of semicircle ⇒ Perimeter of protractor = πr+2r

Let required radius be R. Then according to the question,

Area of the circle with radius r =πr²
Now, New Radius = r + 100% of r = r + r = 2r
∴ New Area = π(2r)² - 4πr²
∴ Increased Area = 4πr² -  πr² = 3πr²
And Area increased in

Here, Area of circle (A) = πr² and Circumference of circle (C) = 2πr
 
Putting value of C in A = πr², we get

Area of the sector
⇒ Area of the sector
⇒ Area of the shaded region = 462 sq. cm

Let the radius of the circle be r and side of the square be a.
Then, according to question,
2πr = 4a

Now, ratio of their areas,

Therefore, Area of the circle > Area of the square 

The distance around the circle is called its circumference