NCERT Exemplar Solutions - Class 10 - Mathematics - Unit 11: Area Related To Circles

Exercise 11.1

Step 1: Recall the formula for the area of a circle.
Area of a circle = \(\pi r^2\), where \(r\) is the radius.

Step 2: Find the area of the first circle.
Radius = \(R_1\). So, Area = \(\pi R_1^2\).

Step 3: Find the area of the second circle.
Radius = \(R_2\). So, Area = \(\pi R_2^2\).

Step 4: Add the two areas.
Total area = \(\pi R_1^2 + \pi R_2^2\).

Step 5: According to the question, this total area is equal to the area of a bigger circle with radius \(R\).
Area of bigger circle = \(\pi R^2\).

Step 6: Write the equation.
\(\pi R_1^2 + \pi R_2^2 = \pi R^2\).

Step 7: Cancel \(\pi\) from both sides (since it is common and non-zero).
\(R_1^2 + R_2^2 = R^2\).

Final Answer: The correct option is (B).

Step 1: Recall the formula for the circumference of a circle. Circumference (C) = \(2 \pi r\), where \(r\) is the radius (measured in metres, SI unit).

Step 2: For the first circle with radius \(R_1\), circumference = \(2 \pi R_1\).

Step 3: For the second circle with radius \(R_2\), circumference = \(2 \pi R_2\).

Step 4: The problem says the sum of these two circumferences is equal to the circumference of another circle with radius \(R\). So, \[ 2\pi R_1 + 2\pi R_2 = 2\pi R \]

Step 5: Take \(2 \pi\) common on the left side: \[ 2\pi (R_1 + R_2) = 2\pi R \]

Step 6: Cancel \(2 \pi\) from both sides (since \(2 \pi \neq 0\)): \[ R_1 + R_2 = R \]

Final Answer: The sum of the radii of the two circles is equal to the radius of the third circle. Hence, option (A) is correct.

Step 1: Let the radius of the circle be r (in metres).

Then, circumference of the circle = \(2 \pi r\) metres.

Step 2: Let the side of the square be a (in metres).

Then, perimeter of the square = \(4a\) metres.

Step 3: It is given that circumference of the circle = perimeter of the square.

So, \(2 \pi r = 4a\).

Step 4: Solve for a:

\(a = \dfrac{2 \pi r}{4} = \dfrac{\pi r}{2}\).

Step 5: Find the area of the circle:

Area of circle = \(\pi r^2\) (square metres).

Step 6: Find the area of the square:

Area of square = \(a^2 = \left(\dfrac{\pi r}{2}\right)^2 = \dfrac{\pi^2}{4}r^2\) (square metres).

Step 7: Compare the two areas:

• Circle: \(\pi r^2\)
• Square: \(\dfrac{\pi^2}{4} r^2\)

Step 8: Numerical values:

\(\pi \approx 3.14\).

\(\dfrac{\pi^2}{4} \approx \dfrac{9.87}{4} = 2.47\).

So, \(3.14 r^2 > 2.47 r^2\).

Final Step: The area of the circle is greater than the area of the square.

Correct Option: (B)

Step 1: When a triangle is drawn inside a semicircle with the diameter as its base, the triangle will always be a right-angled triangle (this is a property of a semicircle).

Step 2: The diameter of the semicircle is \(2r\). This diameter becomes the hypotenuse of the right triangle.

Step 3: Let the other two sides of the right triangle (the legs) be \(a\) and \(b\). By Pythagoras’ theorem:

\(a^2 + b^2 = (2r)^2 = 4r^2.\)

Step 4: The area of a right triangle is given by:

\( \text{Area} = \dfrac{1}{2} \times a \times b. \)

Step 5: For a fixed value of \(a^2 + b^2\), the product \(ab\) is maximum when \(a = b\). (This is a standard result from algebra or you can check by AM–GM inequality).

Step 6: If \(a = b\), then:

\(a^2 + b^2 = 2a^2 = 4r^2 \implies a^2 = 2r^2 \implies a = b = \sqrt{2}r.\)

Step 7: Now calculate the area:

\( \text{Area} = \dfrac{1}{2} \times a \times b = \dfrac{1}{2} (\sqrt{2}r)(\sqrt{2}r).\)

\( = \dfrac{1}{2} (2r^2) = r^2. \)

Final Answer: The largest area = \(r^2\) sq. units. So the correct option is (A).

Step 1: Write the formula for perimeter (circumference) of a circle.

Perimeter of circle = \(2 \pi r\), where \(r\) is the radius (SI unit: metre).

Step 2: Write the formula for perimeter of a square.

Perimeter of square = \(4a\), where \(a\) is the side length (SI unit: metre).

Step 3: Since both perimeters are equal:

\(2 \pi r = 4a\)

Divide both sides by 4:

\(a = \dfrac{\pi r}{2}\)

Step 4: Write the formula for area of a circle.

Area of circle = \(\pi r^2\) (SI unit: m²).

Step 5: Write the formula for area of a square.

Area of square = \(a^2\).

Step 6: Substitute \(a = \dfrac{\pi r}{2}\) into the area of the square.

Area of square = \(\left( \dfrac{\pi r}{2} \right)^2 = \dfrac{\pi^2 r^2}{4}\).

Step 7: Now find the ratio of the areas (circle : square).

\(\dfrac{\text{Area of circle}}{\text{Area of square}} = \dfrac{\pi r^2}{\dfrac{\pi^2 r^2}{4}}\)

Simplify:

\(= \dfrac{\pi r^2 \times 4}{\pi^2 r^2} = \dfrac{4}{\pi}\)

Step 8: Approximate the value of \(\pi\).

\(\pi \approx 3.14\), so \(\dfrac{4}{\pi} \approx \dfrac{4}{3.14} \approx 1.27\).

As a ratio, \(1.27 \approx \dfrac{14}{11}\).

Final Answer: The ratio of their areas is \(14:11\). Hence option (B).

Step 1: The diameters of the two smaller parks are given as 16 m and 12 m.

Radius is half of diameter, so:

• Radius of first park = \(\tfrac{16}{2} = 8\, \text{m}\)
• Radius of second park = \(\tfrac{12}{2} = 6\, \text{m}\)

Step 2: Area of a circle is given by the formula:

\( A = \pi r^2 \)

Step 3: Calculate the area of both small parks:

• First park area = \( \pi (8^2) = \pi (64) = 64\pi\, \text{m}^2 \)
• Second park area = \( \pi (6^2) = \pi (36) = 36\pi\, \text{m}^2 \)

Step 4: Add the two areas:

Total area = \(64\pi + 36\pi = 100\pi\, \text{m}^2\)

Step 5: Let the radius of the new (bigger) park be \(R\). Its area is:

\( A = \pi R^2 \)

Since the new park’s area must equal the sum of the smaller parks:

\( \pi R^2 = 100\pi \)

Step 6: Cancel \(\pi\) on both sides:

\( R^2 = 100 \)

Step 7: Take square root:

\( R = \sqrt{100} = 10\, \text{m} \)

Final Answer: The radius of the new park is 10 m. Hence, option (A).

Step 1: A circle inscribed in a square touches all four sides of the square. That means the diameter of the circle is equal to the side of the square.

Step 2: Side of the square is given as \(6\,\text{cm}\). So, diameter of the circle = \(6\,\text{cm}\).

Step 3: Radius is half of the diameter. Radius \(r = \tfrac{6}{2} = 3\,\text{cm}\).

Step 4: Formula for the area of a circle is \(A = \pi r^2\).

Step 5: Substitute \(r = 3\,\text{cm}\): \(A = \pi \times 3^2 = 9\pi\,\text{cm}^2\).

Final Answer: The area of the circle is \(9\pi\,\text{cm}^2\). Therefore, the correct option is (D).

Step 1: The circle has radius \(r = 8\,\text{cm}\). So, the diameter of the circle is:

\( \text{Diameter} = 2r = 2 \times 8 = 16\,\text{cm} \).

Step 2: A square inscribed in a circle means that the circle passes through all four corners of the square. In this case, the diagonal of the square is exactly equal to the diameter of the circle.

So, \( \text{Diagonal of square} = 16\,\text{cm} \).

Step 3: For a square, the relation between diagonal (d) and side (a) is:

\( d = a\sqrt{2} \).

Step 4: Substituting the diagonal value:

\( a\sqrt{2} = 16 \).

Step 5: Solve for side (a):

\( a = \dfrac{16}{\sqrt{2}} = 8\sqrt{2}\,\text{cm} \).

Step 6: The area of a square is given by:

\( \text{Area} = a^2 \).

Step 7: Substitute side length:

\( \text{Area} = (8\sqrt{2})^2 = 64 \times 2 = 128\,\text{cm}^2 \).

Final Answer: The area of the inscribed square is 128 cm². So, the correct option is (B).

Step 1: Write the given data.

First circle has diameter = 36 cm = 0.36 m. Radius \(r_1 = \tfrac{0.36}{2} = 0.18\, \text{m}\).

Second circle has diameter = 20 cm = 0.20 m. Radius \(r_2 = \tfrac{0.20}{2} = 0.10\, \text{m}\).

Step 2: Find circumferences of the two circles.

Formula: Circumference of a circle = \(2 \pi r\).

So, \(C_1 = 2 \pi r_1 = 2 \pi (0.18) = 0.36 \pi\, \text{m}\).

\(C_2 = 2 \pi r_2 = 2 \pi (0.10) = 0.20 \pi\, \text{m}\).

Step 3: Add the two circumferences.

Total circumference = \(C_1 + C_2 = 0.36 \pi + 0.20 \pi = 0.56 \pi\, \text{m}\).

Step 4: Let the required circle have radius \(R\,\text{m}\). Then its circumference = \(2 \pi R\).

We are told this is equal to the total circumference: \(2 \pi R = 0.56 \pi\).

Step 5: Solve for \(R\).

Divide both sides by \(2\pi\): \(R = \tfrac{0.56}{2} = 0.28\, \text{m}\).

Step 6: Convert back into cm.

\(0.28\, \text{m} = 28\, \text{cm}\).

Final Answer: The radius of the required circle is 28 cm. So the correct option is (C) 28 cm.

Step 1: Formula for the area of a circle is \(A = \pi r^2\), where \(r\) is the radius (in cm) and area is in square centimetres (cm²).

Step 2: First circle has radius \(24\,\text{cm}\). Its area is:

\(A_1 = \pi (24)^2 = \pi \times 576 = 576\pi\,\text{cm}^2\)

Step 3: Second circle has radius \(7\,\text{cm}\). Its area is:

\(A_2 = \pi (7)^2 = \pi \times 49 = 49\pi\,\text{cm}^2\)

Step 4: Add the two areas to get the total area:

\(A_{\text{total}} = A_1 + A_2 = 576\pi + 49\pi = 625\pi\,\text{cm}^2\)

Step 5: Let the radius of the required circle be \(R\,\text{cm}\). Its area should also be equal to \(625\pi\,\text{cm}^2\).

So, \(\pi R^2 = 625\pi\)

Step 6: Cancel \(\pi\) on both sides:

\(R^2 = 625\)

Step 7: Find \(R\) by taking square root:

\(R = \sqrt{625} = 25\,\text{cm}\)

Step 8: Diameter of a circle is twice the radius:

\(D = 2R = 2 \times 25 = 50\,\text{cm}\)

Final Answer: The diameter of the circle is 50 cm. So, the correct option is (D).