CBSE Class 10 Maths Chapter 12: Areas Related to Circles NCERT Solutions

Solution:

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

Let the radius of the third circle, whose circumference is the sum of the circumferences of the first two circles, be $r$.

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.

The circumference of the third circle is $C = 2\pi r$.

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

$C = C_1 + C_2$

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

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

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 which has a circumference equal to the sum of the circumferences of the given two circles is 28 cm.

Solution:

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

Let the radius of the third circle, whose area is the sum of the areas of the first two circles, be $r$.

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.

The area of the third circle is $A = \pi r^2$.

According to the problem statement, 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 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 must be a positive value, we take the positive root.

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

Solution:

The archery target has five scoring regions: Gold, Red, Blue, Black, and White. We are given the diameter of the Gold region and the width of each subsequent band.

1. Gold Region:

Diameter of Gold region = 21 cm.

Radius of Gold region ($r_1$) = Diameter / 2 = 21 cm / 2 = 10.5 cm.

Area of Gold region ($A_{Gold}$) = $\pi r_1^2 = \pi (10.5)^2$.

Using $\pi = \frac{22}{7}$:

$$A\_\{Gold\}\; =\; \backslash frac\{22\}\{7\}\; \backslash times\; (10.5)^2\; =\; \backslash frac\{22\}\{7\}\; \backslash times\; 110.25\; =\; 22\; \backslash times\; 15.75\; =\; 346.5\; \backslash text\{\; cm\}^2$$

2. Red Region:

The Red region is a band around the Gold region. The width of this band is 10.5 cm.

The radius of the circle including the Gold and Red regions ($r_2$) = Radius of Gold region + Width of Red band = 10.5 cm + 10.5 cm = 21 cm.

The area of the Red region ($A_{Red}$) is the area of the circle with radius $r_2$ minus the area of the Gold region:

$A_{Red} = \pi r_2^2 - \pi r_1^2 = \pi (21)^2 - \pi (10.5)^2$

$$A\_\{Red\}\; =\; \backslash frac\{22\}\{7\}\; \backslash times\; (21)^2\; -\; 346.5\; =\; \backslash frac\{22\}\{7\}\; \backslash times\; 441\; -\; 346.5$$

$$A\_\{Red\}\; =\; 22\; \backslash times\; 63\; -\; 346.5\; =\; 1386\; -\; 346.5\; =\; 1039.5\; \backslash text\{\; cm\}^2$$

3. Blue Region:

The Blue region is a band around the Red region, with a width of 10.5 cm.

The radius of the circle including the Gold, Red, and Blue regions ($r_3$) = Radius up to Red region + Width of Blue band = 21 cm + 10.5 cm = 31.5 cm.

The area of the Blue region ($A_{Blue}$) is the area of the circle with radius $r_3$ minus the area of the circle with radius $r_2$:

$A_{Blue} = \pi r_3^2 - \pi r_2^2 = \pi (31.5)^2 - \pi (21)^2$

$$A\_\{Blue\}\; =\; \backslash frac\{22\}\{7\}\; \backslash times\; (31.5)^2\; -\; 1386\; =\; \backslash frac\{22\}\{7\}\; \backslash times\; 992.25\; -\; 1386$$

$$A\_\{Blue\}\; =\; 22\; \backslash times\; 141.75\; -\; 1386\; =\; 3118.5\; -\; 1386\; =\; 1732.5\; \backslash text\{\; cm\}^2$$

4. Black Region:

The Black region is a band around the Blue region, with a width of 10.5 cm.

The radius of the circle including the Gold, Red, Blue, and Black regions ($r_4$) = Radius up to Blue region + Width of Black band = 31.5 cm + 10.5 cm = 42 cm.

The area of the Black region ($A_{Black}$) is the area of the circle with radius $r_4$ minus the area of the circle with radius $r_3$:

$A_{Black} = \pi r_4^2 - \pi r_3^2 = \pi (42)^2 - \pi (31.5)^2$

$$A\_\{Black\}\; =\; \backslash frac\{22\}\{7\}\; \backslash times\; (42)^2\; -\; 3118.5\; =\; \backslash frac\{22\}\{7\}\; \backslash times\; 1764\; -\; 3118.5$$

$$A\_\{Black\}\; =\; 22\; \backslash times\; 252\; -\; 3118.5\; =\; 5544\; -\; 3118.5\; =\; 2425.5\; \backslash text\{\; cm\}^2$$

5. White Region:

The White region is the outermost band, with a width of 10.5 cm.

The radius of the circle including all five regions ($r_5$) = Radius up to Black region + Width of White band = 42 cm + 10.5 cm = 52.5 cm.

The area of the White region ($A_{White}$) is the area of the circle with radius $r_5$ minus the area of the circle with radius $r_4$:

$A_{White} = \pi r_5^2 - \pi r_4^2 = \pi (52.5)^2 - \pi (42)^2$

$$A\_\{White\}\; =\; \backslash frac\{22\}\{7\}\; \backslash times\; (52.5)^2\; -\; 5544\; =\; \backslash frac\{22\}\{7\}\; \backslash times\; 2756.25\; -\; 5544$$

$$A\_\{White\}\; =\; 22\; \backslash times\; 393.75\; -\; 5544\; =\; 8662.5\; -\; 5544\; =\; 3118.5\; \backslash text\{\; cm\}^2$$

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