CBSE Class 10 Maths Chapter 12: Areas 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 $C_3 = 2\pi r$.

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

$C_3 = C_1 + C_2$

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

$$2\backslash pi\; r\; =\; 56\backslash pi$$

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

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

$$r\; =\; 28$$

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 $A_3 = \pi r^2$.

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

$A_3 = A_1 + A_2$

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

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

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

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

Now, we take the square root of both sides:

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

$$r\; =\; \backslash pm\; 10$$

Since the radius of a circle cannot be negative, we take the positive value.

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 subsequent band (Red, Blue, Black, White) is 10.5 cm. We need to find the area of each of the five scoring regions using $\pi = \frac{22}{7}$.

1. Gold Region:

Diameter of Gold region = 21 cm.

Radius of Gold region ($r_G$) = $\frac{21}{2} = 10.5$ cm.

Area of Gold region = $\pi r_G^2 = \frac{22}{7} \times (10.5)^2 = \frac{22}{7} \times 110.25 = 22 \times 15.75 = 346.5$ cm².

2. Red Region:

The Red region is a band around the Gold region. The radius of the circle including Gold and Red is the radius of Gold plus the width of the Red band.

Radius of circle (Gold + Red) ($r_{RG}$) = $10.5$ cm (radius of Gold) + $10.5$ cm (width of Red band) = $21$ cm.

Area of circle (Gold + Red) = $\pi r_{RG}^2 = \frac{22}{7} \times (21)^2 = \frac{22}{7} \times 441 = 22 \times 63 = 1386$ cm².

Area of Red region = Area of (Gold + Red) - Area of Gold

Area of Red region = $1386 - 346.5 = 1039.5$ cm².

3. Blue Region:

The Blue region is a band around the Gold and Red regions. The radius of the circle including Gold, Red, and Blue is the radius of (Gold + Red) plus the width of the Blue band.

Radius of circle (Gold + Red + Blue) ($r_{RGB}$) = $21$ cm (radius of Gold + Red) + $10.5$ cm (width of Blue band) = $31.5$ cm.

Area of circle (Gold + Red + Blue) = $\pi r_{RGB}^2 = \frac{22}{7} \times (31.5)^2 = \frac{22}{7} \times 992.25 = 22 \times 141.75 = 3118.5$ cm².

Area of Blue region = Area of (Gold + Red + Blue) - Area of (Gold + Red)

Area of Blue region = $3118.5 - 1386 = 1732.5$ cm².

4. Black Region:

The Black region is a band around the Gold, Red, and Blue regions. The radius of the circle including these three regions plus the Black band is the radius of (Gold + Red + Blue) plus the width of the Black band.

Radius of circle (Gold + Red + Blue + Black) ($r_{RGBB}$) = $31.5$ cm (radius of Gold + Red + Blue) + $10.5$ cm (width of Black band) = $42$ cm.

Area of circle (Gold + Red + Blue + Black) = $\pi r_{RGBB}^2 = \frac{22}{7} \times (42)^2 = \frac{22}{7} \times 1764 = 22 \times 252 = 5544$ cm².

Area of Black region = Area of (Gold + Red + Blue + Black) - Area of (Gold + Red + Blue)

Area of Black region = $5544 - 3118.5 = 2425.5$ cm².

5. White Region:

The White region is the outermost band. The radius of the entire target (including all five regions) is the radius of (Gold + Red + Blue + Black) plus the width of the White band.

Radius of the entire target ($r_{Total}$) = $42$ cm (radius of Gold + Red + Blue + Black) + $10.5$ cm (width of White band) = $52.5$ cm.

Area of the entire target = $\pi r_{Total}^2 = \frac{22}{7} \times (52.5)^2 = \frac{22}{7} \times 2756.25 = 22 \times 393.75 = 8662.5$ cm².

Area of White region = Area of the entire target - Area of (Gold + Red + Blue + Black)

Area of White region = $8662.5 - 5544 = 3118.5$ cm².

Summary of Areas:

• Area of Gold region = 346.5 cm²
• Area of Red region = 1039.5 cm²
• Area of Blue region = 1732.5 cm²
• Area of Black region = 2425.5 cm²
• Area of White region = 3118.5 cm²