Question

43. The diameter of two circles are 36 cm and 20 cm respectively. Find the area of the circ
which has circumference equal to the sum of the circumferences of the two circles. Al
find ratio of area of this big circle and the sum of area of two small circle.
on one circle. I​

Answer 1

Diameter of circle = 36 cm

Diameter of circle ⭕ = 20 cm

Radius of circle = 18 cm

Radius of circle ⭕ = 10 cm

Sum of circumference of circle & circle ⭕

→ 2πR + 2πr

→ 2π(R + r)

→ 2π (16 + 20)

→ 2π(36)

Therefore,

Radius of bigger circle whose circumference is sum of circumference of circle & circle ⭕ = 36 cm

So,

Area of this bigger circle = πr²

$= \frac{22}{7 } \times ({36)}^{2} \\ \\ = \frac{22}{7} \times 1296 \\ \\ = \tt1296\pi \: \: cm^{2}$

Sum of areas of circle & circle ⭕

$\rm = \pi {(18)}^{2} + \pi {(10)}^{2} \\ \rm = \pi \{ {(18)}^{2} + {(10)}^{2} \} \\ \\ = \rm\pi \{ 324 + 100\} \\ \\ \tt = 424\pi \: \: {cm}^{2}$

Now,

$\text{Required Ratio =} \frac{1296 \cancel\pi}{424 \cancel\pi} \\ \\ = \frac{162}{53} \\ \\ = 162 :53$

Answer 2

Answer:

2464 cm²

98:53

Step-by-step explanation:

Let the radius of the required circle be 'R'.

Radius of 1st circle = d/2 = 36/2 = 18 cm

Radius of 2nd circle = 20/2 = 10 cm

As given, circumference of biggest circle = circumference of 1st circle + circumference of 2nd circle.

=> 2π(18) + 2π(10) = 2πR

=> 2π[18 + 10] = 2πR

=> 18 + 10 = R

=> 28 = R

Area of biggest circle= πR² = π(28)² = π(784) = 2464 cm²

Sum of area of 1st + 2nd circle = π(18)² + π(10)² = π(18² + 10²) = π(424)

Required ratio = π(784)/ π(424)

Required ratio = 784/424 = 98/53