CBSE Class 10 Maths NCERT Solutions: Area Related To Circles

The problem states that the sum of the areas of two circles with radii <math>R_1</math> and <math>R_2</math> is equal to the area of a circle with radius R.

The area of the first circle is <math>π R_1^2</math>.

The area of the second circle is <math>π R_2^2</math>.

The area of the third circle is <math>π R^2</math>.

According to the problem statement, we can write the equation:

$$\pi \; R\_1^2\; +\; \pi \; R\_2^2\; =\; \pi \; R^2$$

To simplify this equation, we can divide all terms by <math>π</math> (since <math>π</math> is a non-zero constant):

$$R\_1^2\; +\; R\_2^2\; =\; R^2$$

This equation represents the relationship between the radii. Therefore, option (B) is the correct answer.

The problem states that the sum of the circumferences of two circles with radii <math>R_1</math> and <math>R_2</math> is equal to the circumference of a circle with radius R.

The circumference of the first circle is <math>2\pi R_1</math>.

The circumference of the second circle is <math>2\pi R_2</math>.

The circumference of the third circle is <math>2\pi R</math>.

According to the problem statement, we can write the equation:

$$2\backslash pi\; R\_1\; +\; 2\backslash pi\; R\_2\; =\; 2\backslash pi\; R$$

To simplify this equation, we can divide all terms by <math>2\pi</math> (since <math>2\pi</math> is a non-zero constant):

$$R\_1\; +\; R\_2\; =\; R$$

This equation represents the relationship between the radii. Therefore, option (A) is the correct answer.

Let the radius of the circle be <math>r</math> and the side length of the square be <math>a</math>.

We are given that the circumference of the circle is equal to the perimeter of the square.

Circumference of the circle = <math>2\pi r</math>

Perimeter of the square = <math>4a</math>

Setting them equal:

$$2\backslash pi\; r\; =\; 4a$$

We can express <math>a</math> in terms of <math>r</math> (using <math>\pi = \frac{22}{7}</math>):

$$2\; \backslash times\; \backslash frac\{22\}\{7\}\; \backslash times\; r\; =\; 4a$$

$$\backslash frac\{44\}\{7\}\; r\; =\; 4a$$

Dividing both sides by 4:

$$a\; =\; \backslash frac\{11\}\{7\}\; r$$

Now, let's find the areas of the circle and the square.

Area of the circle (<math>A_{circle}</math>) = <math>\pi r^2</math>

Area of the square (<math>A_{square}</math>) = <math>a^2</math>

Substitute the expression for <math>a</math> into the area of the square formula:

$$A\_\{square\}\; =\; \backslash left(\backslash frac\{11\}\{7\}\; r\backslash right)^2\; =\; \backslash frac\{121\}\{49\}\; r^2$$

Now, let's express the area of the circle in terms of <math>r^2</math> and compare it with the area of the square.

<math>A_{circle} = \pi r^2 = \frac{22}{7} r^2</math>

To compare <math>\frac{22}{7} r^2</math> and <math>\frac{121}{49} r^2</math>, we can find a common denominator or compare the coefficients.

Let's rewrite <math>A_{circle}</math> with a denominator of 49:

$$A\_\{circle\}\; =\; \backslash frac\{22\}\{7\}\; r^2\; =\; \backslash frac\{22\; \backslash times\; 7\}\{7\; \backslash times\; 7\}\; r^2\; =\; \backslash frac\{154\}\{49\}\; r^2$$

Now we compare <math>\frac{154}{49} r^2</math> with <math>\frac{121}{49} r^2</math>.

Since <math>154 > 121</math>, it follows that:

$$A\_\{circle\}\; >\; A\_\{square\}$$

Therefore, the area of the circle is greater than the area of the square when their perimeters are equal. Option (B) is the correct answer.