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

Exercise 11.4

Step 1: Recall the formula for the area of a circle

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

Here, \( A = 22176\,\text{m}^2 \). We need to find the radius \( r \).

Step 2: Put the values into the formula

\( 22176 = \dfrac{22}{7} \times r^2 \)

Step 3: Solve for \( r^2 \)

\( r^2 = \dfrac{22176 \times 7}{22} \)

\( r^2 = 7056 \)

Step 4: Take the square root

\( r = \sqrt{7056} = 84\,\text{m} \)

So, the radius of the playground is 84 metres.

Step 5: Find the circumference (the boundary length)

Formula: \( C = 2\pi r \)

\( C = 2 \times \dfrac{22}{7} \times 84 \)

\( C = 528\,\text{m} \)

So, the fencing length needed is 528 metres.

Step 6: Find the cost of fencing

Rate of fencing = Rs 50 per metre

Total Cost = \( 528 \times 50 = 26,400 \)

Final Answer: Rs 26,400

Step 1: Convert all measurements into the same unit (SI unit: metre).

Diameter of front wheel = 80 cm = \(80 \div 100 = 0.8\,\text{m}\).

Diameter of rear wheel = 2 m (already in metres).

Step 2: Find the circumference of each wheel.

The circumference of a wheel is given by the formula: \(C = \pi \times d\).

Front wheel circumference = \(\pi \times 0.8 = 0.8\pi\,\text{m}\).

Rear wheel circumference = \(\pi \times 2 = 2\pi\,\text{m}\).

Step 3: Find the total distance travelled by the front wheel in 1400 revolutions.

Total distance = Number of revolutions \(\times\) circumference.

\(= 1400 \times 0.8\pi = 1120\pi\,\text{m}\).

Step 4: Find how many revolutions the rear wheel makes to cover the same distance.

Revolutions = \(\dfrac{\text{Total distance}}{\text{Circumference of rear wheel}}\).

\(= \dfrac{1120\pi}{2\pi} = 560\).

Final Answer: The rear wheel makes 560 revolutions.

Step 1: Find the area of the triangular field

The sides are 15 m, 16 m, and 17 m. To find the area of a triangle when all three sides are given, we use Heron’s formula:

\[ Delta = sqrt{s(s-a)(s-b)(s-c)} \]

Here, \(s\) is the semi-perimeter = half of the sum of sides.

\( s = \dfrac{15 + 16 + 17}{2} = \dfrac{48}{2} = 24 \, \text{m} \)

Now substitute into the formula:

\( Delta = sqrt{24(24-15)(24-16)(24-17)} \)

\( = sqrt{24 imes 9 imes 8 imes 7} \)

\( = sqrt{12096} \)

\( = 24\sqrt{21} \, \text{m}^2 \)

This is the total area of the triangular field.

Step 2: Find the total area grazed by the animals

Each animal is tied with a rope of length 7 m. So, each animal can graze in a circular sector of radius 7 m around the corner where it is tied.

The angle of the sector is equal to the angle of the triangle at that corner. The sum of all three angles of a triangle is \(180^\circ = \pi \, \text{radians}\).

The formula for the area of a sector is:

\( ext{Sector area} = \dfrac{1}{2} r^2 heta \)

Adding all three sectors together:

\( ext{Total grazed area} = \dfrac{1}{2} r^2 (\alpha + \beta + \gamma) \)

But \( \alpha + \beta + \gamma = \pi \).

So, \( ext{Total grazed area} = \dfrac{1}{2} imes 7^2 imes pi = \dfrac{49\pi}{2} \, \text{m}^2 \)

Step 3: Find the ungrazed area

The part of the field that cannot be grazed = (Total area of triangle) – (Total grazed area).

\( ext{Ungrazed area} = 24\sqrt{21} - \dfrac{49\pi}{2} \, \text{m}^2 \)

Now approximate the value:

\( 24\sqrt{21} \approx 109.9 \, \text{m}^2 \)

\( \dfrac{49\pi}{2} \approx 76.96 \, \text{m}^2 \)

So, \( ext{Ungrazed area} \approx 109.9 - 76.96 = 33.0 \, \text{m}^2 \)

Final Answer: The ungrazed area is about 33.0 m².

Step 1: Write the formula for the area of a sector.

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

Here, radius \(r = 12\,\text{cm}\), angle \(\theta = 60^\circ\), and \(\pi = 3.14\).

\(A_{sector} = \dfrac{60}{360} \times 3.14 \times (12)^2\)

\(= \dfrac{1}{6} \times 3.14 \times 144\)

\(= 75.36\,\text{cm}^2\)

Step 2: Write the formula for the area of an isosceles triangle inside the sector.

That triangle is formed by two radii (each 12 cm) and the included angle \(60^\circ\).

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

\(= \dfrac{1}{2} (12)^2 \sin 60^\circ\)

\(= \dfrac{1}{2} \times 144 \times \dfrac{\sqrt{3}}{2}\)

\(= 36\sqrt{3}\,\text{cm}^2\)

Now approximate \(\sqrt{3} \approx 1.732\).

So, \(A_{triangle} = 36 \times 1.732 \approx 62.35\,\text{cm}^2\).

Step 3: Find the area of the segment.

\(A_{segment} = A_{sector} - A_{triangle}\)

\(= 75.36 - 62.35\)

\(\approx 13.01\,\text{cm}^2\)

Final Answer: Area of the segment = \(13.01\,\text{cm}^2\)

Step 1: Write down the given information.

• Diameter of pond = 17.5 m
• Width of path = 2 m
• Cost of construction = Rs 25 per m²

Step 2: Find the radius of the pond (inner radius).

Radius \(r = \frac{\text{diameter}}{2} = \frac{17.5}{2} = 8.75\,\text{m}.\)

Step 3: Find the outer radius (pond + path).

Outer radius \(R = r + \text{width of path} = 8.75 + 2 = 10.75\,\text{m}.\)

Step 4: Area of path = Area of bigger circle – Area of pond.

\( \text{Area} = \pi R^2 - \pi r^2 = \pi(10.75^2 - 8.75^2). \)

Step 5: Simplify the calculation.

\( 10.75^2 = 115.5625, \; 8.75^2 = 76.5625. \)

Difference = \(115.5625 - 76.5625 = 39. \)

So, Area = \( 39\pi \; m^2. \)

Step 6: Find the cost.

Cost = Area × Rate = \(39\pi × 25 = 975\pi\; Rs. \)

Approximating with \(\pi \approx 3.1416,\)

\(975 × 3.1416 ≈ 3,064\; Rs.\)

Final Answer: Cost of constructing the path = Rs 3,064 (approx).

Step 1: Formula for trapezium area

The area of a trapezium is given by:

\( \text{Area} = \dfrac{1}{2} \times (\text{sum of parallel sides}) \times \text{height} \)

Step 2: Substitute values

Here, \(AB = 18\,\text{cm}\), \(DC = 32\,\text{cm}\), and height = \(14\,\text{cm}\).

So, \( \text{Area} = \dfrac{1}{2} (18 + 32) \times 14 \)

= \( \dfrac{1}{2} (50) \times 14 \)

= \( 25 \times 14 = 350\,\text{cm}^2 \).

Step 3: Find area of circular parts

At each corner A, B, C, D, a quarter-circle is drawn with radius \(7\,\text{cm}\).

Four quarter-circles together make one complete circle.

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

Here, \(r = 7\,\text{cm}\).

So, area of the full circle = \( \pi \times 7^2 = 49\pi\,\text{cm}^2 \).

Step 4: Subtract to get shaded area

Shaded region = area of trapezium – area of circle

= \(350 - 49\pi\,\text{cm}^2 \).

Step 5: Approximate value

Take \(\pi \approx 3.14\).

So, \(49\pi \approx 49 \times 3.14 = 153.9\,\text{cm}^2 \).

Therefore, shaded region = \(350 - 153.9 = 196.1\,\text{cm}^2 \).

Step 1: Mark the centers of the three circles as \(A, B, C.\) Since each circle touches the other two externally, the distance between any two centers is equal to twice the radius.

So, \(AB = BC = CA = 2r = 2 \times 3.5 = 7\,\text{cm}.\)

Step 2: The three centers form an equilateral triangle of side \(7\,\text{cm}.\)

Step 3: Area of an equilateral triangle of side \(a\) is:

\[ A_{\triangle} = \dfrac{\sqrt{3}}{4} a^2 \]

Substitute \(a = 7\,\text{cm}:\)

\[ A_{\triangle} = \dfrac{\sqrt{3}}{4} (7)^2 = \dfrac{49\sqrt{3}}{4}\,\text{cm}^2. \]

Step 4: Now look at the circular parts inside the triangle. At each corner, there is a sector of angle \(60^\circ\) (because angles of an equilateral triangle are \(60^\circ\)).

Area of one sector of radius \(r = 3.5\,\text{cm}\) and angle \(60^\circ\) is:

\[ A_{\text{sector}} = \dfrac{60}{360}\pi r^2 = \dfrac{1}{6} \pi (3.5)^2 \]

\[ A_{\text{sector}} = \dfrac{1}{6} \pi (12.25) = \dfrac{12.25\pi}{6}\,\text{cm}^2. \]

Since there are 3 such sectors, total area of sectors is:

\[ A_{\text{sectors}} = 3 \times \dfrac{12.25\pi}{6} = \dfrac{36.75\pi}{6} = 6.125\pi = \dfrac{49\pi}{8}\,\text{cm}^2. \]

Step 5: The required shaded area (enclosed between the circles) is the triangle area minus the three sectors:

\[ A = A_{\triangle} - A_{\text{sectors}} \]

\[ A = \dfrac{49\sqrt{3}}{4} - \dfrac{49\pi}{8} \;\,\text{cm}^2. \]

Step 6: Approximate value:

\( \dfrac{49\sqrt{3}}{4} \approx 21.22\,\text{cm}^2, \quad \dfrac{49\pi}{8} \approx 19.25\,\text{cm}^2. \)

So, \( A \approx 21.22 - 19.25 = 1.97\,\text{cm}^2. \)

Final Answer: The required area = \(1.97\,\text{cm}^2\).

Step 1: Recall the formula for the area of a sector when the arc length is given:

\[ A = \tfrac{1}{2} \times r \times l \]

Here, \(r\) = radius of the circle, and \(l\) = arc length.

Step 2: Write down the values from the question:

• Radius, \(r = 5\,\text{cm}\)
• Arc length, \(l = 3.5\,\text{cm}\)

Step 3: Substitute these values into the formula:

\[ A = \tfrac{1}{2} \times 5 \times 3.5 \]

Step 4: First multiply radius and arc length:

\[ 5 \times 3.5 = 17.5 \]

Step 5: Now divide by 2:

\[ \tfrac{17.5}{2} = 8.75 \]

Final Answer: \(A = 8.75\,\text{cm}^2\)

So, the area of the sector is 8.75 square centimetres.

Step 1: Understand the figure.
Each cardboard is a circle of radius \(7\,\text{cm}\). When they are placed so that each touches the other two, their centers form the corners of a square. The side of this square equals twice the radius of one circle, i.e., \(14\,\text{cm}\).

Step 2: Find the area of the square.
The formula for the area of a square is:
\(A_{square} = (\text{side})^2\)
Here, side = \(14\,\text{cm}\).
So, \(A_{square} = 14^2 = 196\,\text{cm}^2\).

Step 3: Find the area occupied by circular parts inside the square.
At each corner of the square, there is a quarter of a circle (because only one-fourth of each circle lies inside). Four quarters together make one full circle of radius \(7\,\text{cm}\).
Area of a full circle is:
\(A_{circle} = \pi r^2 = \pi (7^2) = 49\pi\,\text{cm}^2\).

Step 4: Calculate the required enclosed area.
Required area = Area of the square – Area of one full circle
= \(196 - 49\pi\,\text{cm}^2\).

Step 5: Approximate value.
Using \(\pi \approx 3.1416\):
\(49\pi \approx 153.94\).
So, required area = \(196 - 153.94 = 42.06\,\text{cm}^2\).

Final Answer: The enclosed area is about \(42.06\,\text{cm}^2\).

Step 1: The square cardboard has an area of \(784\,\text{cm}^2\).

To find the side length of the square, take the square root of the area:

\(\text{Side of square} = \sqrt{784} = 28\,\text{cm}\).

Step 2: The four circular plates are placed inside the square so that each one touches two sides of the square and also touches two other circles. This means the circles are as large as possible within the square.

Step 3: Look at half of the side of the square: \(28/2 = 14\,\text{cm}\). Each circle will fit exactly in one corner, so the diameter of each circle = \(14\,\text{cm}\).

Step 4: Radius of each circle = half of diameter = \(14/2 = 7\,\text{cm}\).

Step 5: Area of one circle = \(\pi r^2 = \pi (7)^2 = 49\pi\,\text{cm}^2\).

Step 6: There are 4 circles. Total area of 4 circles = \(4 \times 49\pi = 196\pi\,\text{cm}^2\).

Step 7: Area of the square sheet not covered by the circles = Area of square – Total area of circles.

\(= 784 - 196\pi\,\text{cm}^2\)

Step 8: Approximate using \(\pi \approx 3.14\):

\(196 \times 3.14 \approx 615.75\)

Uncovered area = \(784 - 615.75 = 168.25\,\text{cm}^2\).

Final Answer: The uncovered area is \(168.25\,\text{cm}^2\).

Step 1: Convert all dimensions to the same unit.

The floor is given as \(5\,\text{m} \times 4\,\text{m}\). Since the tiles are in cm, convert the floor to cm:

\(5\,\text{m} = 500\,\text{cm}, \; 4\,\text{m} = 400\,\text{cm}.\)

Step 2: Find the area of the floor.

Area of rectangle \(= \text{length} \times \text{breadth}\).

\(500 \times 400 = 200000\,\text{cm}^2.\)

Step 3: Work out how many tiles fit.

Diameter of one tile = 50 cm ⇒ radius = 25 cm.

Along the length: \(500/50 = 10\) tiles fit.

Along the breadth: \(400/50 = 8\) tiles fit.

Total number of tiles = \(10 \times 8 = 80.\)

Step 4: Area of one circular tile.

Area = \(\pi r^2 = \pi (25)^2 = 625\pi\,\text{cm}^2.\)

Step 5: Total area of 80 tiles.

\(80 \times 625\pi = 50000\pi\,\text{cm}^2.\)

Step 6: Uncovered area.

Uncovered area = Floor area – Tile area

= \(200000 - 50000\pi\,\text{cm}^2\).

Step 7: Approximate value.

Take \(\pi \approx 3.1416\).

\(50000\pi \approx 157080.\)

Uncovered area \(= 200000 - 157080 = 42920\,\text{cm}^2.\)

Step 8: Convert back to m² (SI unit).

\(1\,\text{m}^2 = 10000\,\text{cm}^2.\)

So, \(42920\,\text{cm}^2 = 4.292\,\text{m}^2.\)

Step 1: Understand the property
If all vertices of a rhombus lie on a circle, then that rhombus must be a square (special property of cyclic quadrilaterals).

Step 2: Write the area formula of a circle
The area of a circle is given by: \[ A = \pi R^2 \] where \(R\) is the radius of the circle.

Step 3: Substitute the given values
\(1256 = 3.14 \times R^2\)

Step 4: Solve for \(R^2\)
Divide both sides by 3.14: \[ R^2 = \frac{1256}{3.14} = 400 \]

Step 5: Find the radius
\[ R = \sqrt{400} = 20\,\text{cm} \]

Step 6: Relating radius to square
In a square inscribed in a circle, the diagonal of the square equals the diameter of the circle. So, diagonal of square = \(2R = 40\,\text{cm}\).

Step 7: Find the side of the square
Diagonal of square = \( \sqrt{2} \times \text{side} \). \[ \text{side} = \frac{40}{\sqrt{2}} = 20\sqrt{2}\,\text{cm} \]

Step 8: Find the area of the square
Area = (side)² \[ = (20\sqrt{2})^2 = 400 \times 2 = 800\,\text{cm}^2 \]

Final Answer: The area of the rhombus is \(\boxed{800\,\text{cm}^2}\).

Step 1: The diameters are in the ratio \(1:2:3\).

That means the radii are also in the ratio \(1:2:3\), because radius = diameter ÷ 2.

Step 2: The area of a circle depends on the square of its radius:

\( A = \pi r^2 \)

Step 3: Since the radii are in ratio \(1:2:3\), the areas of the full circles will be in the ratio:

\(1^2 : 2^2 : 3^2 = 1 : 4 : 9\).

Step 4: Now find the areas of each region (ring):

• Innermost circle area = \(1\) unit².
• Middle ring area = (area of radius 2 circle) − (area of radius 1 circle) = \(4 − 1 = 3\) unit².
• Outer ring area = (area of radius 3 circle) − (area of radius 2 circle) = \(9 − 4 = 5\) unit².

Step 5: So the ratio of the areas of the three regions is:

\(1 : 3 : 5\).

Final Answer: \(\boxed{1 : 3 : 5}\)

Step 1: The length of the minute hand is given as 5 cm. This is the radius of the circular path made by the tip of the hand. Radius \(r = 5\,\text{cm} = 0.05\,\text{m}\) (converted to SI unit, metres).

Step 2: The time interval is from 6:05 a.m. to 6:40 a.m. That is a total of \(40 - 5 = 35\) minutes.

Step 3: The minute hand completes one full circle (360°) in 60 minutes. So, in 35 minutes it covers: \(\dfrac{35}{60} \times 360^{\circ} = 210^{\circ}\).

Step 4: The area swept by the hand is the area of a sector of a circle. Formula for sector area: \[ A = \frac{\theta}{360^{\circ}} \times \pi r^2 \]

Step 5: Substituting values: \[ A = \frac{210}{360} \times \pi (0.05)^2 \, \text{m}^2 \]

Step 6: Simplify: \[ A = \frac{7}{12} \times \pi (0.0025) \, \text{m}^2 = \frac{0.0175}{12} \pi \, \text{m}^2 \]

Converting back to cm² (since 1 m² = 10,000 cm²): \[ A = \frac{175}{12} \pi \, \text{cm}^2 \approx 45.8 \, \text{cm}^2 \]

Final Answer: The area swept by the minute hand is about \(45.8 \, \text{cm}^2\).

Step 1: Recall the formula for the area of a sector:

\[ A = \frac{\theta}{360^\circ} \times \pi r^2 \]

where:

• \(A\) = area of the sector
• \(\theta\) = central angle in degrees
• \(r\) = radius of the circle

Step 2: Substitute the given values:

Area \(A = 770\,\text{cm}^2, \; \theta = 200^\circ, \; \pi = \dfrac{22}{7}\).

So,

\[ 770 = \frac{200}{360} \times \pi r^2 \]

Step 3: Simplify the fraction:

\[ \frac{200}{360} = \frac{5}{9} \]

So,

\[ 770 = \frac{5}{9} \pi r^2 \]

Step 4: Solve for \(r^2\):

\[ r^2 = \frac{770 \times 9}{5 \pi} \]

Substitute \(\pi = \dfrac{22}{7}\):

\[ r^2 = \frac{6930}{5 \times \tfrac{22}{7}} = 441 \]

So, \(r = \sqrt{441} = 21\,\text{cm}.\)

Step 5: Recall the formula for arc length:

\[ L = \frac{\theta}{360^\circ} \times 2 \pi r \]

Step 6: Substitute the values:

\[ L = \frac{200}{360} \times 2 \pi \times 21 \]

\[ L = \frac{5}{9} \times 42 \pi \]

\[ L = \frac{70}{3} \pi \, \text{cm} \]

Step 7: Approximate using \(\pi \approx 3.1416\):

\[ L \approx 73.3\,\text{cm} \]

Final Answer: The length of the arc is \(\dfrac{70}{3}\pi\,\text{cm} \;\approx\; 73.3\,\text{cm}.\)

Step 1: Recall the formulas

• Area of a sector = \(\dfrac{\theta}{360^\circ} \times \pi r^2\)
• Arc length of a sector = \(\dfrac{\theta}{360^\circ} \times 2\pi r\)
• Here, \(r\) = radius of the circle, \(\theta\) = central angle of the sector.

Step 2: For the first circle (radius = 7 cm, angle = 120°)

Area = \(\dfrac{120}{360} \times \pi \times 7^2\)

= \(\dfrac{1}{3} \times \pi \times 49\)

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

Arc length = \(\dfrac{120}{360} \times 2\pi \times 7\)

= \(\dfrac{1}{3} \times 14\pi\)

= \(\dfrac{14\pi}{3}\,\text{cm}\)

Step 3: For the second circle (radius = 21 cm, angle = 40°)

Area = \(\dfrac{40}{360} \times \pi \times 21^2\)

= \(\dfrac{1}{9} \times \pi \times 441\)

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

Arc length = \(\dfrac{40}{360} \times 2\pi \times 21\)

= \(\dfrac{1}{9} \times 42\pi\)

= \(\dfrac{14\pi}{3}\,\text{cm}\)

Step 4: Observation

The arc lengths are the same (\(\dfrac{14\pi}{3}\,\text{cm}\)), even though the radii and angles are different. However, the areas are different: one is \(\dfrac{49\pi}{3}\,\text{cm}^2\), the other is \(49\pi\,\text{cm}^2\).

Step 1: The outer shape is a square of side \(14\,\text{cm}\).

Area of square = side × side = \(14 \times 14 = 196\,\text{cm}^2\).

Step 2: Inside the square, there is a flower-like figure made by 4 semicircles.

Each semicircle has radius \(3\,\text{cm}\).

Step 3: Area of one semicircle = \(\tfrac{1}{2} \pi r^2\).

= \(\tfrac{1}{2} \pi (3^2) = \tfrac{1}{2} \pi \times 9 = 4.5\pi\,\text{cm}^2\).

Step 4: There are 4 such semicircles.

Total area of 4 semicircles = \(4 \times 4.5\pi = 18\pi\,\text{cm}^2\).

Step 5: The shaded region is the square area minus the 4 semicircles area.

Shaded area = \(196 - 18\pi\,\text{cm}^2\).

Step 6: Approximating \(\pi \approx 3.14\):

\(18 \pi = 18 \times 3.14 = 56.52\,\text{cm}^2\).

Shaded area = \(196 - 56.52 = 139.48 \approx 139.5\,\text{cm}^2\).

Final Answer: \(\mathbf{139.5\,\text{cm}^2}\)

Step 1: Recall the formula for the area of a circle.

The area of a circle is given by \(A = \pi r^2\), where \(r\) is the radius.

Step 2: Substitute the given area.

We are told that the area of the wheel is \(1.54\,\text{m}^2\).

So, \(\pi r^2 = 1.54\).

Step 3: Solve for \(r^2\).

\(r^2 = \dfrac{1.54}{\pi}\).

Taking \(\pi \approx 3.14\):

\(r^2 = \dfrac{1.54}{3.14} \approx 0.49\).

Step 4: Find the radius.

\(r = \sqrt{0.49} = 0.7\,\text{m}\).

Step 5: Find the circumference of the wheel.

The circumference (distance covered in one revolution) is \(C = 2\pi r\).

\(C = 2 \times 3.14 \times 0.7 \approx 4.4\,\text{m}\).

Step 6: Calculate the number of revolutions.

Total distance travelled = \(176\,\text{m}\).

Number of revolutions = \(\dfrac{\text{Total distance}}{\text{Circumference}}\).

\(= \dfrac{176}{4.4} = 40\).

Final Answer: The wheel makes 40 revolutions.

Step 1: Understand the problem.

A circle has a chord of length 5 cm. This chord makes an angle of \(90^\circ\) at the centre. The chord divides the circle into two parts called segments (minor and major). We need to find the difference between the areas of these two segments.

Step 2: Relation between chord length and radius.

Formula: For a chord of length \(c\), subtending angle \(\theta\) at the centre: \[ c = 2r \sin\left(\dfrac{\theta}{2}\right) \]

Here, \(c = 5\,\text{cm}\), \(\theta = 90^\circ\). So, \(c = 2r \sin 45^\circ = 2r \cdot \dfrac{1}{\sqrt{2}} = r\sqrt{2}\). Therefore, \[ r = \dfrac{5}{\sqrt{2}}\,\text{cm} \]

Step 3: Area of the sector (angle = 90°).

Formula: \(\text{Sector area} = \dfrac{\theta}{360^\circ} \times \pi r^2\). Substituting \(\theta = 90^\circ\): \[ \text{Sector area} = \dfrac{90}{360} \pi r^2 = \dfrac{1}{4}\pi r^2 \]

With \(r^2 = (\tfrac{5}{\sqrt{2}})^2 = \tfrac{25}{2}\): \[ \text{Sector area} = \dfrac{1}{4}\pi \times \dfrac{25}{2} = \dfrac{25\pi}{8}\,\text{cm}^2 \]

Step 4: Area of the triangle (isosceles, two sides = radius).

Formula: \(\text{Triangle area} = \tfrac{1}{2}r^2 \sin \theta\). Here, \(\theta = 90^\circ\), so \(\sin 90^\circ = 1\). \[ \text{Triangle area} = \tfrac{1}{2} \times \tfrac{25}{2} = \dfrac{25}{4}\,\text{cm}^2 \]

Step 5: Area of the minor segment.

Minor segment = Sector area – Triangle area \[ = \dfrac{25\pi}{8} - \dfrac{25}{4} = \dfrac{25}{8}(\pi - 2)\,\text{cm}^2 \]

Step 6: Area of the major segment.

Total circle area = \(\pi r^2 = \pi \times \tfrac{25}{2} = \dfrac{25\pi}{2}\,\text{cm}^2\). Major segment = Total circle area – Minor segment \[ = \dfrac{25\pi}{2} - \dfrac{25}{8}(\pi - 2) = \dfrac{25}{8}(3\pi + 2)\,\text{cm}^2 \]

Step 7: Difference of areas.

Difference = Major segment – Minor segment \[ = \dfrac{25}{8} \big[(3\pi + 2) - (\pi - 2)\big] \] Simplify inside the bracket: \(3\pi + 2 - \pi + 2 = 2\pi + 4\). So, \[ \text{Difference} = \dfrac{25}{8} \times (2\pi + 4) = \dfrac{25}{4}(\pi + 2)\,\text{cm}^2 \]

Step 8: Approximate value.

Taking \(\pi \approx 3.14\): \[ \text{Difference} = \dfrac{25}{4}(3.14 + 2) = \dfrac{25}{4} \times 5.14 \approx 32.14\,\text{cm}^2 \]

Final Answer: Difference between the areas of the two segments = \(\dfrac{25}{4}(\pi+2)\,\text{cm}^2 \;\approx\; 32.14\,\text{cm}^2\).

Step 1: Recall the formula for the area of a sector of a circle:

\[ A = \frac{\theta}{360^\circ} \times \pi r^2 \]

where:

• \(\theta\) = angle of the sector (in degrees)
• \(r\) = radius of the circle
• Area is in square units (here, cm²).

Step 2: In this problem, the radius \(r = 21\,\text{cm}\).

Step 3: The given minor sector has angle \(120^\circ\).

So its corresponding major sector will have angle:

\[ 360^\circ - 120^\circ = 240^\circ \]

Step 4: Write the difference of the two areas:

\[ \text{Difference} = A_{240^\circ} - A_{120^\circ} \]

Step 5: Apply the formula:

\[ A_{240^\circ} - A_{120^\circ} = \frac{240}{360}\pi r^2 - \frac{120}{360}\pi r^2 \]

Step 6: Simplify the fractions:

\[ = \frac{240 - 120}{360}\pi r^2 = \frac{120}{360}\pi r^2 \]

Step 7: Substitute \(r = 21\,\text{cm}\):

\[ = \frac{120}{360} \times \pi \times 21^2 \]

Step 8: Simplify step by step:

• \(21^2 = 441\)
• \(\tfrac{120}{360} = \tfrac{1}{3}\)

So,

\[ = \tfrac{1}{3} \times \pi \times 441 = 147\pi\,\text{cm}^2 \]

Step 9: Approximate using \(\pi \approx 3.1416\):

\[ 147 \times 3.1416 \approx 461.8\,\text{cm}^2 \]

Final Answer: The difference of the areas is \(147\pi\,\text{cm}^2 \;\approx\; 461.8\,\text{cm}^2\).