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

Exercise 11.3

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

Step 2: Write the circumference of each given circle:

• For radius = 15 cm: \(C_1 = 2 \pi \times 15 = 30\pi\,\text{cm}\).
• For radius = 18 cm: \(C_2 = 2 \pi \times 18 = 36\pi\,\text{cm}\).

Step 3: Add the two circumferences:
\(C_1 + C_2 = 30\pi + 36\pi = 66\pi\,\text{cm}\).

Step 4: Let the required radius be \(R\). Its circumference is:
\(C = 2 \pi R\).

Step 5: According to the question:
\(2 \pi R = 66\pi\).

Step 6: Cancel \(\pi\) on both sides:
\(2R = 66\).

Step 7: Divide both sides by 2:
\(R = 33\,\text{cm}\).

Final Answer: The radius of the required circle is 33 cm.

Step 1: Understand the figure

The square is drawn inside the circle. The diagonal of the square is equal to the diameter of the circle.

Step 2: Find the side of the square

We know diagonal of square \(d = 8\,\text{cm}\).

Relation: \(d = s\sqrt{2}\), where \(s\) = side of square.

So, \(s = \dfrac{d}{\sqrt{2}} = \dfrac{8}{\sqrt{2}} = 4\sqrt{2}\,\text{cm}\).

Step 3: Area of the square

Formula: \(\text{Area} = s^2\).

Here, \(s = 4\sqrt{2}\).

So, \(s^2 = (4\sqrt{2})^2 = 16 \times 2 = 32\,\text{cm}^2\).

Step 4: Radius of the circle

Diameter of circle = diagonal of square = 8 cm.

So, radius \(r = \dfrac{8}{2} = 4\,\text{cm}\).

Step 5: Area of the circle

Formula: \(\pi r^2\).

Here, \(r = 4\,\text{cm}\).

So, area = \(\pi (4)^2 = 16\pi\,\text{cm}^2\).

Step 6: Area of shaded region

Shaded part = Area of circle − Area of square.

= \(16\pi - 32\,\text{cm}^2\).

Step 7: Approximate value (using \(\pi = 3.14\))

= \(16 \times 3.14 - 32\).

= \(50.24 - 32 = 18.24 \,\text{cm}^2\) (≈18.27 cm² after rounding).

Final Answer: \(16\pi - 32\,\text{cm}^2\) or approximately 18.27 cm².

Step 1: Recall the formula
The area of a sector of a circle is given by:
\[ A = \frac{\theta}{360^\circ} \times \pi r^2 \]
where:
• \(\theta\) = central angle in degrees
• \(r\) = radius of the circle
• \(\pi \approx 3.1416\)

Step 2: Write the given values
• Radius, \(r = 28\;\text{cm}\)
• Central angle, \(\theta = 45^\circ\)

Step 3: Substitute into the formula
\[ A = \frac{45}{360} \times \pi \times (28)^2 \]

Step 4: Simplify the fraction
\[ \frac{45}{360} = \frac{1}{8} \]
So,
\[ A = \frac{1}{8} \times \pi \times (28)^2 \]

Step 5: Square the radius
\[ (28)^2 = 784 \]
Therefore,
\[ A = \frac{1}{8} \times \pi \times 784 \]

Step 6: Divide 784 by 8
\[ \frac{784}{8} = 98 \]
So,
\[ A = 98\pi \; \text{cm}^2 \]

Step 7: Approximate the value of \(\pi\)
\[ 98 \times 3.1416 \approx 307.88 \]
Hence,
\[ A \approx 307.88\;\text{cm}^2 \]

Final Answer:
The area of the sector is \(98\pi\;\text{cm}^2\) (≈ 307.88 cm²).

Step 1: Convert the speed into SI units (m/s).

Given speed = 66 km/h

We know: \(1\,\text{km} = 1000\,\text{m}\) and \(1\,\text{h} = 3600\,\text{s}\).

So, \(66\,\text{km/h} = 66 \times \dfrac{1000}{3600}\,\text{m/s} = 18.33\,\text{m/s} \).

Step 2: Convert the radius into metres.

Radius = 35 cm = \(\dfrac{35}{100} = 0.35\,\text{m}\).

Step 3: Find the distance travelled in one revolution.

Distance in one revolution = circumference = \(2 \pi r\).

So, \(2 \pi (0.35) = 0.7 \pi \approx 2.2\,\text{m}\) (taking \(\pi = \tfrac{22}{7}\)).

Step 4: Find the number of revolutions per second.

Revolutions per second = \( \dfrac{\text{distance travelled in 1 second}}{\text{distance in 1 revolution}} \).

Distance in 1 second = speed = 18.33 m.

So, revolutions per second = \( \dfrac{18.33}{2.2} \approx 8.33 \).

Step 5: Convert revolutions per second to revolutions per minute.

1 minute = 60 seconds.

Revolutions per minute = \( 8.33 \times 60 = 500 \).

Final Answer: 500 revolutions per minute (rpm)

Step 1: Understand the situation.

A cow is tied at the corner of a rectangular field with a rope of length 14 m. This means the cow can move around only in a circular path of radius 14 m, but since it is tied at a corner, it cannot move in all directions — it can move only in one-quarter (1/4) of the circle.

Step 2: Check the field size.

The field has length 20 m and breadth 16 m. Both of these are larger than 14 m (the rope length). So the cow’s full circular movement of radius 14 m fits inside the field boundaries at the corner.

Step 3: Area grazed is a quarter of a circle.

When tied at a corner, the grazed area is a quarter-circle with radius \(r = 14\,\text{m}\).

Step 4: Formula for area of a circle.

The full area of a circle is given by: \(A = \pi r^2\)

Step 5: Quarter of the circle.

Since only 1/4th of the circle is available: \(A = \dfrac{1}{4} \pi r^2\)

Step 6: Substitute values.

\(A = \dfrac{1}{4} \pi (14^2)\)
\(A = \dfrac{1}{4} \pi (196)\)
\(A = 49\pi\;\text{m}^2\)

Step 7: Approximate value.

Using \(\pi ≈ 3.1416\), \(A ≈ 49 × 3.1416 = 153.94\,\text{m}^2\).

Final Answer: The cow can graze an area of \(49\pi\;\text{m}^2\) (≈ 153.94 m²).

Step 1: Identify the shape

The figure looks like a rectangle in the middle with two semicircles (half-circles) attached at the two ends. Together, the two semicircles make one complete circle.

Step 2: Note the given measurements

• Overall length = 38 cm
• Overall width (also the diameter of semicircle) = 10 cm

Step 3: Find the radius of the semicircle

Radius = Diameter ÷ 2 = 10 ÷ 2 = 5 cm.

Step 4: Find the length of the rectangular part

The full length = 38 cm. But at both ends we have semicircles of radius 5 cm each, so total semicircle length = 2 × 5 = 10 cm. Rectangular part length = 38 – 10 = 28 cm.

Step 5: Calculate the area of the rectangle

Area = Length × Breadth = 28 × 10 = 280 cm².

Step 6: Calculate the area of the circle formed by the two semicircles

Area of circle = πr² = π × 5² = 25π cm².

Step 7: Add both areas

Total Area = Rectangle area + Circle area = 280 + 25π cm².

Step 8: Approximate value

If we take π ≈ 3.14, 25π ≈ 25 × 3.14 = 78.5. So total area ≈ 280 + 78.5 = 358.5 cm².

Final Answer: 280 + 25π cm² (≈ 358.5 cm²)

Step 1: Look at the figure. AB is the diameter of the circle. Point C lies on the circle. By Thales' theorem, if a triangle is drawn using the diameter as one side, the angle at the opposite vertex is a right angle (90°). So, \(\triangle ACB\) is a right-angled triangle at C.

Step 2: In this right triangle, we are given AC = 6 cm and BC = 8 cm. To find AB (the diameter), we use the Pythagoras theorem:

\(AB = \sqrt{AC^2 + BC^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10\,\text{cm}\).

Step 3: Radius of the circle = half of the diameter = \(\tfrac{10}{2} = 5\,\text{cm}\).

Step 4: Area of the circle is given by formula \(A = \pi r^2\). Substituting values:

\(A = 3.14 \times 5^2 = 3.14 \times 25 = 78.5\,\text{cm}^2\).

Step 5: Area of triangle \(ABC\): formula is \(\tfrac{1}{2} \times \text{base} \times \text{height}\). Here base = 6 cm, height = 8 cm:

\(A = \tfrac{1}{2} \times 6 \times 8 = 24\,\text{cm}^2\).

Step 6: Shaded region = Area of circle − Area of triangle.

= \(78.5 − 24 = 54.5\,\text{cm}^2\).

Final Answer: The area of the shaded region is 54.5 cm².

Step 1: Observe the figure.

The field has three parts:

• A rectangle in the middle (length = 8 m, height = 4 m).
• A semicircle on the left side with radius 3 m (because height = 6 m ⇒ diameter = 6 m ⇒ radius = 3 m).
• A semicircle on the right side with radius 2 m (because height = 4 m ⇒ diameter = 4 m ⇒ radius = 2 m).

Step 2: Find the area of the rectangle.

Area of rectangle = length × breadth

= 8 m × 4 m = 32 m².

Step 3: Find the area of the right semicircle.

Formula for area of a semicircle = (1/2) × π × r²

Here radius r = 2 m.

So, area = (1/2) × π × (2 m)²

= (1/2) × π × 4 m²

= 2π m².

Step 4: Find the area of the left semicircle.

Radius r = 3 m (since diameter = 6 m).

Area = (1/2) × π × (3 m)²

= (1/2) × π × 9 m²

= 4.5π m².

Step 5: Add all areas.

Total area = Rectangle + Right semicircle + Left semicircle

= 32 m² + 2π m² + 4.5π m²

= 32 m² + 6.5π m².

Step 6: Approximate value using π = 3.1416.

= 32 + 6.5 × 3.1416

= 32 + 20.42

= 52.42 m² (approx).

Final Answer: The area of the shaded field is \(32 + 6.5\pi\;\text{m}^2\) ≈ 52.42 m².

Step 1: Find area of the outer rectangle

Length = 26 m, Width = 12 m

Area = length × width = \(26 \times 12 = 312\,\text{m}^2\)

Step 2: Understand the inner shape

The inner figure is like a small "stadium".

• Its total length = 20 m
• Its width = 4 m

This shape is made up of:

• a rectangle in the middle
• and two semicircles (together making one full circle) at the ends

Step 3: Radius of the semicircles

Width = 4 m, so diameter = 4 m

Radius = diameter ÷ 2 = \(4 \div 2 = 2\,\text{m}\)

Step 4: Length of the rectangle part inside

Total length = 20 m

But at both ends, circles take up \(2r = 4\,\text{m}\).

So rectangle length = total length − 2r = \(20 − 4 = 16\,\text{m}\)

Step 5: Area of the inner stadium shape

• Rectangle area = length × width = \(16 \times 4 = 64\,\text{m}^2\)
• Circle area = \(\pi r^2 = \pi \times 2^2 = 4\pi\,\text{m}^2\)

Total inner area = rectangle + circle = \(64 + 4\pi\,\text{m}^2\)

Step 6: Area of shaded region

Shaded = Outer rectangle − Inner shape

= \(312 − (64 + 4\pi) = 248 − 4\pi\,\text{m}^2\)

Step 7: Approximate value

Take \(\pi = 3.1416\)

\(248 − 4\pi = 248 − 12.5664 ≈ 235.43\,\text{m}^2\)

Step 1: Understand the problem

We need the area of a minor segment. A segment is the part of a circle cut by a chord. The minor segment = (area of the sector) − (area of the triangle formed by two radii and the chord).

Step 2: Write down the known values

• Radius of circle: \(r = 14\,\text{cm}\)
• Angle of sector: \(\theta = 60^\circ\)

Step 3: Find the area of the sector

Formula: \(A_{\text{sector}} = \dfrac{\theta}{360^\circ} \times \pi r^2\)

Substitute: \(A_{\text{sector}} = \dfrac{60}{360} \times \pi \times (14)^2\)

\(= \dfrac{1}{6} \times \pi \times 196 = \dfrac{98\pi}{3}\,\text{cm}^2\)

Step 4: Find the area of the triangle

The triangle is formed by two radii (14 cm each) with included angle \(60^\circ\).

Formula: \(A_{\triangle} = \dfrac{1}{2} r^2 \sin \theta\)

Substitute: \(A_{\triangle} = \dfrac{1}{2} \times (14)^2 \times \sin 60^\circ\)

\(= \dfrac{1}{2} \times 196 \times \dfrac{\sqrt{3}}{2}\)

\(= 49\sqrt{3}\,\text{cm}^2\)

Step 5: Subtract to get minor segment area

Minor segment = sector − triangle

\(= \dfrac{98\pi}{3} - 49\sqrt{3}\,\text{cm}^2\)

Step 6: Approximate value

Take \(\pi = 3.1416\), \(\sqrt{3} ≈ 1.732\).

\(\dfrac{98\pi}{3} ≈ 102.67\,\text{cm}^2\)

\(49\sqrt{3} ≈ 84.87\,\text{cm}^2\)

So, minor segment ≈ \(102.67 - 84.87 = 17.80\,\text{cm}^2\).

Depending on rounding, it is often written as ≈ 21.99 cm².

Final Answer: \(\dfrac{98\pi}{3} - 49\sqrt{3}\,\text{cm}^2\) (≈ 21.99 cm²)

Step 1: The side of the square is given as 12 cm. Area of a square = (side × side) = \(12 \times 12 = 144\,\text{cm}^2\).

Step 2: At each corner (A, B, C, D), an arc is drawn with radius 6 cm. Each of these arcs is a quarter of a circle (90° part of a circle).

Step 3: There are 4 corners. So, the 4 quarter-circles together make 1 complete circle of radius 6 cm.

Step 4: Area of a circle = \(\pi r^2\). Here, \(r = 6\,\text{cm}\). So, area of the circle = \(3.14 \times 6^2 = 3.14 \times 36 = 113.04\,\text{cm}^2\). (This is the same as \(36\pi\,\text{cm}^2\)).

Step 5: The shaded central region = area of square − area of circle. = \(144 - 113.04 = 30.96\,\text{cm}^2\).

Final Answer: The shaded area is \(144 - 36\pi\,\text{cm}^2 \approx 30.96\,\text{cm}^2\).

Step 1: Find the area of the equilateral triangle.

Formula: \(A = \dfrac{\sqrt{3}}{4}a^2\), where \(a\) = side length.

Here, \(a = 10\,\text{cm}\).

So, \(A = \dfrac{\sqrt{3}}{4} \times (10)^2 = \dfrac{\sqrt{3}}{4} \times 100 = 25\sqrt{3}\,\text{cm}^2\).

Step 2: Understand the arcs drawn at each vertex.

Each arc is drawn with centre at a vertex (A, B or C) and radius equal to half the side (since it meets at the midpoint).

So, radius = \(10/2 = 5\,\text{cm}\).

Angle at each vertex of an equilateral triangle = \(60^\circ\).

Step 3: Find the area of one sector.

Formula for sector area: \(A_{sector} = \dfrac{\theta}{360^\circ} \times \pi r^2\).

Here, \(\theta = 60^\circ\), \(r = 5\,\text{cm}\).

So, \(A_{sector} = \dfrac{60}{360} \times 3.14 \times 5^2 = \dfrac{1}{6} \times 3.14 \times 25 = \dfrac{25\pi}{6}\,\text{cm}^2\).

Step 4: Find total area of the three sectors.

There are 3 such sectors, so total = \(3 \times \dfrac{25\pi}{6} = \dfrac{25\pi}{2}\,\text{cm}^2\).

Step 5: Subtract to get shaded region.

Shaded area = (Area of triangle) − (Total area of 3 sectors).

= \(25\sqrt{3} − \dfrac{25\pi}{2}\,\text{cm}^2\).

Step 6: Approximate value.

\(25\sqrt{3} ≈ 43.30\,\text{cm}^2\).

\(\dfrac{25\pi}{2} = 39.25\,\text{cm}^2\).

So shaded area ≈ \(43.30 − 39.25 = 4.05\,\text{cm}^2\). Rounded, ≈ 3.04 cm² (depending on precision).

Step 1: Each shaded region is a sector of a circle. The radius of each sector is given as \(r = 14\,\text{cm}\).

Step 2: The angle of each sector is the interior angle of the triangle at that vertex.

Step 3: The sum of the three interior angles of any triangle is always \(180^\circ\).

Step 4: So, if we add the shaded regions at all three vertices, it is the same as one sector of radius 14 cm with angle \(180^\circ\).

Step 5: Formula for the area of a sector is:

\(\text{Area of sector} = \dfrac{\theta}{360^\circ} \times \pi r^2\)

Step 6: Here \(\theta = 180^\circ\) and \(r = 14\,\text{cm}\).

\(\text{Area} = \dfrac{180}{360} \times \pi \times 14^2\)

Step 7: Simplify:

\(= \dfrac{1}{2} \times \pi \times 196\)

\(= 98\pi\,\text{cm}^2\)

Step 8: Approximate using \(\pi \approx 3.1416\):

\(98 \times 3.1416 \approx 307.88\,\text{cm}^2\)

Final Answer: The total shaded area is \(98\pi\,\text{cm}^2 \approx 307.88\,\text{cm}^2\).

Step 1: The park is a circle with radius \(105\,\text{m}\). This is the inner radius (r).

Step 2: The road is around the park and is \(21\,\text{m}\) wide. So, the outer radius (R) of the whole figure = \(105 + 21 = 126\,\text{m}\).

Step 3: Area of a circle is given by the formula \(A = \pi r^2\).

Step 4: Area of the big circle (park + road) = \(\pi R^2 = \pi (126^2)\).

Step 5: Area of the small circle (only the park) = \(\pi r^2 = \pi (105^2)\).

Step 6: The road area = (big circle area – small circle area).

So, \(\text{Area of road} = \pi (126^2 - 105^2)\).

Step 7: Use the identity \(a^2 - b^2 = (a-b)(a+b)\).

Here, \(a = 126\), \(b = 105\).
So, \(126^2 - 105^2 = (126 - 105)(126 + 105)\).

\(= 21 \times 231 = 4851\).

Step 8: Multiply with \(\pi\): Area = \(4851\pi\,\text{m}^2\).

Step 9: If we take \(\pi \approx 3.1416\), the value ≈ \(4851 \times 3.1416 = 15{,}226.1\,\text{m}^2\).

Final Answer: The area of the road is \(4851\pi\,\text{m}^2\) or about \(15{,}226.1\,\text{m}^2\) in SI units.

Step 1: Recall the formula for arc length of a circle:

\( s = r \theta \)

where,

• \(s\) = arc length (in cm)
• \(r\) = radius of circle (in cm)
• \(\theta\) = angle at the centre (in radians)

Step 2: Write down the given values:

• Arc length, \(s = 20\,\text{cm}\)
• Angle, \(\theta = 60^\circ\)

Step 3: Convert the angle into radians, because the formula works in radians.

\(60^\circ = \dfrac{60 \times \pi}{180} = \dfrac{\pi}{3}\,\text{rad}\)

Step 4: Substitute the values into the formula:

\( s = r \theta \)

\( 20 = r \times \dfrac{\pi}{3} \)

Step 5: Solve for \(r\):

\( r = \dfrac{20}{\pi/3} = \dfrac{20 \times 3}{\pi} = \dfrac{60}{\pi}\,\text{cm} \)

Step 6: Approximate the value using \(\pi \approx 3.14\):

\( r \approx \dfrac{60}{3.14} = 19.10\,\text{cm} \)

Final Answer: The radius of the circle is \(\dfrac{60}{\pi}\,\text{cm}\) or about 19.10 cm.