CBSE Class 10 Mathematics Chapter 12: Area Related to Circles NCERT Solutions

Let the radius of the first circle be $r_1$ and the radius of the second circle be $r_2$. We are given:

$r_1 = 19$ cm

$r_2 = 9$ cm

The circumference of the first circle is $C_1 = 2\pi r_1 = 2\pi(19) = 38\pi$ cm.

The circumference of the second circle is $C_2 = 2\pi r_2 = 2\pi(9) = 18\pi$ cm.

Let the radius of the third circle be $r$. Its circumference is $C = 2\pi r$.

According to the problem, the circumference of the third circle is equal to the sum of the circumferences of the first two circles:

$C = C_1 + C_2$

$2\pi r = 38\pi + 18\pi$

$2\pi r = 56\pi$

To find the radius $r$, we divide both sides by $2\pi$:

$r = \frac{56\pi}{2\pi}$

$r = 28$ cm

Therefore, the radius of the circle whose circumference is the sum of the circumferences of the given two circles is 28 cm.

Let the radius of the first circle be $r_1$ and the radius of the second circle be $r_2$. We are given:

$r_1 = 8$ cm

$r_2 = 6$ cm

The area of the first circle is $A_1 = \pi r_1^2 = \pi (8)^2 = 64\pi$ cm².

The area of the second circle is $A_2 = \pi r_2^2 = \pi (6)^2 = 36\pi$ cm².

Let the radius of the third circle be $r$. Its area is $A = \pi r^2$.

According to the problem, the area of the third circle is equal to the sum of the areas of the first two circles:

$A = A_1 + A_2$

$$\backslash pi\; r^2\; =\; 64\backslash pi\; +\; 36\backslash pi$$

$$\backslash pi\; r^2\; =\; 100\backslash pi$$

To find $r^2$, we divide both sides by $\pi$:

$$\backslash nr^2\; =\; 100$$

Taking the square root of both sides:

$$\backslash nr\; =\; \backslash pm\; \backslash sqrt\{100\}$$

$$\backslash nr\; =\; \backslash pm\; 10$$

Since the radius of a circle must be a positive value, we take the positive root.

$r = 10$ cm

Therefore, the radius of the circle having an area equal to the sum of the areas of the two given circles is 10 cm.

We are given the diameter of the Gold region is 21 cm. The width of each band is 10.5 cm.

1. Area of the Gold region:

The radius of the Gold region ($r_{Gold}$) is half of its diameter: $r_{Gold} = \frac{21}{2} = 10.5$ cm.

Area of Gold = $\pi r_{Gold}^2 = \frac{22}{7} \times (10.5)^2 = \frac{22}{7} \times 10.5 \times 10.5 = \frac{22}{7} \times 110.25$ cm².

Area of Gold = $22 \times 15.75 = 346.5$ cm².

2. Area of the Red region:

The Red region is a band around the Gold region. The outer radius of the Red region ($r_{Red\_outer}$) is the radius of the Gold region plus the width of the Red band: $r_{Red\_outer} = 10.5 + 10.5 = 21$ cm.

The area of the Red region is the area of the circle with radius $r_{Red\_outer}$ minus the area of the Gold region.

Area of Red = $\pi r_{Red\_outer}^2 - \pi r_{Gold}^2 = \frac{22}{7} \times (21)^2 - 346.5$

Area of Red = $\frac{22}{7} \times 441 - 346.5 = 22 \times 63 - 346.5 = 1386 - 346.5 = 1039.5$ cm².

3. Area of the Blue region:

The outer radius of the Blue region ($r_{Blue\_outer}$) is the outer radius of the Red region plus the width of the Blue band: $r_{Blue\_outer} = 21 + 10.5 = 31.5$ cm.

The area of the Blue region is the area of the circle with radius $r_{Blue\_outer}$ minus the area of the circle with radius $r_{Red\_outer}$ (which is the outer boundary of the Red region).

Area of Blue = $\pi r_{Blue\_outer}^2 - \pi r_{Red\_outer}^2 = \frac{22}{7} \times (31.5)^2 - 1386$

Area of Blue = $\frac{22}{7} \times 992.25 - 1386 = 22 \times 141.75 - 1386 = 3118.5 - 1386 = 1732.5$ cm².

4. Area of the Black region:

The outer radius of the Black region ($r_{Black\_outer}$) is the outer radius of the Blue region plus the width of the Black band: $r_{Black\_outer} = 31.5 + 10.5 = 42$ cm.

The area of the Black region is the area of the circle with radius $r_{Black\_outer}$ minus the area of the circle with radius $r_{Blue\_outer}$.

Area of Black = $\pi r_{Black\_outer}^2 - \pi r_{Blue\_outer}^2 = \frac{22}{7} \times (42)^2 - 3118.5$

Area of Black = $\frac{22}{7} \times 1764 - 3118.5 = 22 \times 252 - 3118.5 = 5544 - 3118.5 = 2425.5$ cm².

5. Area of the White region:

The outer radius of the White region ($r_{White\_outer}$) is the outer radius of the Black region plus the width of the White band: $r_{White\_outer} = 42 + 10.5 = 52.5$ cm.

The area of the White region is the area of the circle with radius $r_{White\_outer}$ minus the area of the circle with radius $r_{Black\_outer}$.

Area of White = $\pi r_{White\_outer}^2 - \pi r_{Black\_outer}^2 = \frac{22}{7} \times (52.5)^2 - 5544$

Area of White = $\frac{22}{7} \times 2756.25 - 5544 = 22 \times 393.75 - 5544 = 8662.5 - 5544 = 3118.5$ cm².

Summary of Areas:

• Gold: 346.5 cm²
• Red: 1039.5 cm²
• Blue: 1732.5 cm²
• Black: 2425.5 cm²
• White: 3118.5 cm²