CBSE Solutions for Class 10 Maths Chapter 12: Areas Related to Circles || CBSE Class 10 Maths Chapter 12 Areas Related to Circles solutions Ex 12.1

Radius \( \left(r_{1}\right) \) of gold region (i.e., \( 1^{\text {st }} \) circle) \( =\frac{21}{2}=10.5 \mathrm{~cm} \)
Given: Each circle is 10.5 cm wider than the previous circle.
Hence,
Radius \( \left(r_{2}\right) \) of \( 2^{\text {nd }} \) circle \( =10.5+10.5 =21 cm\)
Radius \( \left(r_{3}\right) \) of \( 3^{\text {rd }} \) circle \( =21+10.5=31.5 \mathrm{~cm} \)
Radius \( \left(r_{4}\right) \) of \( 4^{\text {th }} \) circle \( =31.5+10.5=42 \mathrm{~cm} \)
Radius \( \left(r_{5}\right) \) of \( 5^{\text {th }} \) circle \( =42+10.5=52.5 \mathrm{~cm} \)
Area of gold region \( = \) Area of \( 1^{\text {st }} \)circle
\(=\pi \mathrm{r}_{1}^{2}\)
\(=\pi(10.5)^{2}\)
\(=346.5 \mathrm{~cm}^{2}\)
Area of red region \( = \) Area of \( 2^{\text {nd }} \) circle - Area of \( 1^{\text {st }} \) circle
\(=\pi r_{2}^{2}-\pi r_{1}^{2}\)
\(=\pi(21)^{2}-\pi(10.5)^{2}\)
\(=441 \pi-110.25 \pi\)
\(=330.75 \pi\)
\(=1039.5 \mathrm{~cm}^{2}\)
Area of blue region \( = \) Area of \( 3^{\text {rd }} \) circle - Area of \( 2^{\text {nd }} \) circle
\(=\pi r_{3}^{2}-\pi r_{2}^{2}\)
\(=\pi(31.5)^{2}-\pi(21)^{2}\)
\(=992.25 \pi-441 \pi\)
\(=551.25 \pi\)
\(=1732.5 \mathrm{~cm}^{2}\)
Area of black region \( = \) Area of \( 4^{\text {th }} \) circle - Area of \( 3^{\text {rd }} \) circle
\(=\pi \mathrm{r}_{4}{ }^{2}-\pi \mathrm{r}_{3}^{2}\)
\(=\pi(42)^{2}-\pi(31.5)^{2}\)
\(=1764 \pi-992.25 \pi\)
\(=771.75 \pi\)
\(=2425.5 \mathrm{~cm}^{2}\)
Area of white region \( = \) Area of \( 5^{\text {th }} \) circle - Area of \( 4^{\text {th }} \) circle
\(=\pi r_{5}^{2}-\pi r_{4}^{2}\)
\(=\pi(52.5)^{2}-\pi(42)^{2}\)
\(=2756.25 \pi-1764 \pi\)
\(=992.25 \pi\)
\(=3118.5 \mathrm{~cm}^{2}\)
Therefore, areas of gold, red, blue, black, and white regions are \(346.5 \mathrm{cm}^{2}, 1039.5 \mathrm{~cm}^{2}, 1732.5 \mathrm{~cm}^{2}, 2425.5 \mathrm{~cm}^{2} \), and \(3118.5 \mathrm{cm}^{2} \) respectively.