Area Related to Circles- T1

If the addition of the areas of two circles having radii R₁ and R₂ is equal as the Area of a circle of radius R, then

• R = R₁ + R₂

• R₁² + R₂² = R²

• R₁ + R₂ < R

• R₁² + R₂² < R₂²

Answer explanation

The area of a circle is given by A = πR². For two circles with radii R1 and R2, their combined area is π(R1² + R2²). If this equals the area of a circle with radius R, then R1² + R2² = R².

When the sum of the circumferences of two circles having radii R₁ and R₂ is equal as the Circumference of the circle of radius R, then

• R₁ + R₂ = R

• R₁ + R₂ > R

• R₁ + R₂ < R

• No relation among R₁, R₂, and R.

Answer explanation

The circumference of a circle is given by C = 2πR. For two circles with radii R1 and R2, their combined circumference is 2π(R1 + R2). If this equals the circumference of a circle with radius R, then 2π(R1 + R2) = 2πR, leading to R1 + R2 = R.

When the Circumference of the circle and the perimeter of a square are equal, we get

• Area of a circle = the Area of the square

• area of a circle > the Area of the square

• Area of a circle < the Area of the square

• No relation between the areas of the circle and square.

Answer explanation

When the circumference of the circle (C = 2πr) equals the perimeter of the square (P = 4s), the radius of the circle is larger than the side of the square. Thus, the area of the circle (πr²) is greater than the area of the square (s²).

The Area of the largest triangle that can be inside in a semicircle of radius r units is

• r² sq. units

• 1/2r² sq. units

• 2r² sq. units

• √2r² sq. units

Answer explanation

The largest triangle inscribed in a semicircle is a right triangle with its hypotenuse as the diameter. The area is given by (1/2) * base * height. For a semicircle of radius r, the area simplifies to r^2 sq. units.

When the perimeter of a circle is equal to that of a square, then the ratio of their areas is

• 22: 7

• 14: 11

• 7: 22

• 11: 14

Answer explanation

Let the perimeter of the circle (C) be equal to the perimeter of the square (P). C = 2πr and P = 4s. Setting them equal gives r = 2s/π. The areas are A_circle = πr² and A_square = s². The ratio A_circle:A_square simplifies to 14:11.

It is proposed to build the single circular park equal in the Area to the sum of areas for the two circular parks of the diameters of 16 m and 12 m in a locality. The radius of the new park will be

• 10 m

• 15 m

• 20 m

• 24 m

Answer explanation

The area of a circle is A = πr². The areas of the two parks are A1 = π(8)² and A2 = π(6)². Their sum is A1 + A2 = π(64 + 36) = π(100). The new park's area is πr² = π(100), so r² = 100, giving r = 10 m.

The Area of the circle which can be inscribed in a square of side 6 cm is

• 36π cm2

• 18π cm2

• 12π cm2

• 9π cm2

Answer explanation

The diameter of the inscribed circle is equal to the side of the square, which is 6 cm. Thus, the radius is 3 cm. The area of the circle is A = πr² = π(3)² = 9π cm².

The Area of the square which can be inscribed in a circle of Radius 8 cm is

• 256 cm2

• 128 cm2

• 64√2 cm2

• 64 cm2

Answer explanation

The diameter of the circle is 16 cm (2 * 8 cm). The side of the inscribed square is equal to the diameter divided by √2, which is 16/√2 = 8√2 cm. The area is (8√2)² = 128 cm².

The Radius of the circle whose Circumference is the same as the sum of the circumferences of the two circles of the diameters of 36 cm and 20 cm is

• 56 cm

• 42 cm

• 28 cm

• 16 cm

Answer explanation

First, calculate the circumferences of the two circles: C1 = π * 36 and C2 = π * 20. Their sum is C1 + C2 = π * (36 + 20) = π * 56. For a circle with this circumference, C = 2πr, so r = 28 cm.

The diameter of a circle which has an Area is equal to the sum of the areas of the two circles of radii, 24 cm, and 7 cm, is

• 31 cm

• 25 cm

• 62 cm

• 50 cm

Answer explanation

To find the diameter of the circle, first calculate the areas of the two circles: A1 = π(24^2) and A2 = π(7^2). Their sum is A1 + A2 = π(576 + 49) = π(625). The area of the new circle is π(r^2), so r^2 = 625, giving r = 25 cm. Thus, the diameter is 50 cm.