Class 9 - Chapter 6: Measuring Space: Perimeter and Area

Question 1.

Find the area of a sector of a circle with radius 7 cm if the angle of the sector is 60°.

Solution:

Given:

Radius, r = 7 cm

Angle of the sector, θ = 60°

We know that:

Area of a sector = (θ/360°) × πr²

Taking π = 22/7,

Area of the sector = (60°/360°) × (22/7) × 7²

= (1/6) × (22/7) × 49

= (1/6) × 22 × 7

= 154/6

= 77/3 cm²

≈ 25.67 cm²

Answer: The area of the sector is 77/3 cm², or approximately 25.67 cm².

Question 2.

Find the area of a quadrant of a circle whose circumference is 44 cm.

Solution:

Given:

Circumference of the circle = 44 cm

We know that:

Circumference = 2πr

Therefore,

2πr = 44

Taking π = 22/7,

2 × (22/7) × r = 44

44r/7 = 44

r = 7 cm

Now,

Area of the circle = πr²

= (22/7) × 7²

= 154 cm²

A quadrant is one-fourth of a circle.

Area of a quadrant = (1/4) × πr²

= (1/4) × 154

= 38.5 cm²

Answer: The area of the quadrant is 38.5 cm².

Question 3.

The length of the minute hand of a clock is 7 cm. Find the area swept by the minute hand in 10 minutes.

Solution:

Given:

Length of the minute hand = Radius of the circle = 7 cm

Time taken for one complete revolution = 60 minutes

Angle swept in 60 minutes = 360°

Therefore, the angle swept in 10 minutes is:

θ = (10/60) × 360°

= 60°

We know that:

Area swept = Area of the sector

= (θ/360°) × πr²

Taking π = 22/7,

Area swept = (60°/360°) × (22/7) × 7²

= (1/6) × 22 × 7

= 154/6

= 77/3 cm²

≈ 25.67 cm²

Answer: The area swept by the minute hand is 77/3 cm², or approximately 25.67 cm².

Question 4.

A chord of a circle of radius 10 cm subtends 90° at the centre. Find the areas of the corresponding:

(i) Minor sector

(ii) Major sector

Use π = 3.14.

Solution:

Given:

Radius, r = 10 cm

Angle of the minor sector = 90°

Angle of the major sector = 360° − 90° = 270°

(i) Area of the minor sector

Area of a sector = (θ/360°) × πr²

Area of the minor sector = (90°/360°) × 3.14 × 10²

= (1/4) × 3.14 × 100

= 314/4

= 78.5 cm²

Answer: The area of the minor sector is 78.5 cm².

(ii) Area of the major sector

Area of the major sector = (270°/360°) × πr²

= (3/4) × 3.14 × 10²

= (3/4) × 314

= 235.5 cm²

Answer: The area of the major sector is 235.5 cm².

Verification:

Area of the minor sector + Area of the major sector = Area of the circle

78.5 + 235.5 = 314 cm²

Question 5.

A chord of a circle of radius 15 cm subtends an angle of 60° at the centre of the circle. Find the areas of the corresponding minor and major segments of the circle.

Use π ≈ 3.14 and √3 ≈ 1.73.

Solution:

Given:

Radius, r = 15 cm

Angle of the minor sector = 60°

Step 1: Find the area of the circle.

Area of the circle = πr²

= 3.14 × 15²

= 3.14 × 225

= 706.5 cm²

Step 2: Find the area of the minor sector.

Area of the minor sector = (θ/360°) × πr²

= (60°/360°) × 706.5

= (1/6) × 706.5

= 117.75 cm²

Step 3: Find the area of triangle AOB.

Let O be the centre and AB be the chord.

OA = OB = 15 cm

Angle AOB = 60°

Since OA = OB and the angle between them is 60°, triangle AOB is an equilateral triangle.

Area of an equilateral triangle = (√3/4) × side²

Area of triangle AOB = (√3/4) × 15²

Using √3 ≈ 1.73,

= (1.73/4) × 225

= 97.3125 cm²

Step 4: Find the area of the minor segment.

Area of the minor segment = Area of the minor sector − Area of triangle AOB

= 117.75 − 97.3125

= 20.4375 cm²

≈ 20.44 cm²

Answer: The area of the minor segment is approximately 20.44 cm².

Step 5: Find the area of the major segment.

Area of the major segment = Area of the circle − Area of the minor segment

= 706.5 − 20.4375

= 686.0625 cm²

≈ 686.06 cm²

Answer: The area of the major segment is approximately 686.06 cm².

Question 6.

A sector of a circle has radius 15 cm and area 150 cm². Find the length of the corresponding arc.

Solution:

Given:

Radius, r = 15 cm

Area of the sector = 150 cm²

Let the angle of the sector be θ.

We know that:

Area of the sector = (θ/360°) × πr²

Therefore,

150 = (θ/360°) × π × 15²

150 = (θ/360°) × 225π

θ/360° = 150/(225π)

= 2/(3π)

Now, the length of the arc is:

Arc length = (θ/360°) × 2πr

= (2/3π) × 2π × 15

= 60π/(3π)

= 20 cm

Answer: The length of the corresponding arc is 20 cm.

Question 7.

A chord of a circle of radius r subtends an angle of 60° at the centre of the circle. Show that the area of the corresponding minor segment of the circle is equal to:

πr²(1/6 − √3/(4π))

Solution:

Let O be the centre of the circle and AB be the chord.

Given:

OA = OB = r

Angle AOB = 60°

The area of the minor segment is obtained by subtracting the area of triangle AOB from the area of the minor sector.

Area of the minor segment = Area of the minor sector − Area of triangle AOB

Step 1: Find the area of the minor sector.

Area of the minor sector = (θ/360°) × πr²

= (60°/360°) × πr²

= πr²/6

Step 2: Find the area of triangle AOB.

Since OA = OB = r and angle AOB = 60°, triangle AOB is an equilateral triangle.

Area of an equilateral triangle = (√3/4) × side²

Therefore,

Area of triangle AOB = (√3/4)r²

Step 3: Find the area of the minor segment.

Area of the minor segment = Area of the minor sector − Area of triangle AOB

= πr²/6 − (√3/4)r²

Taking r² as a common factor,

= r²(π/6 − √3/4)

Taking π outside the bracket,

= πr²(1/6 − √3/(4π))

Hence,

Area of the minor segment = πr²(1/6 − √3/(4π)).

Hence proved.

Question 8.

An equilateral triangle is inscribed in a circle of radius r. Show that the ratio of the area of the triangle to the area of the circle is equal to 3√3/(4π) ≈ 0.413.

Solution:

Let the side of the equilateral triangle be a and the radius of the circle be r.

For an equilateral triangle, the circumradius is:

r = a/√3

Therefore,

a = √3r

Step 1: Find the area of the triangle.

Area of an equilateral triangle = (√3/4)a²

Substituting a = √3r,

Area of the triangle = (√3/4)(√3r)²

= (√3/4) × 3r²

= 3√3r²/4

Step 2: Find the area of the circle.

Area of the circle = πr²

Step 3: Find the required ratio.

Ratio = Area of the triangle / Area of the circle

= (3√3r²/4)/(πr²)

Cancelling r²,

= 3√3/(4π)

Using √3 ≈ 1.732 and π ≈ 3.14,

= (3 × 1.732)/(4 × 3.14)

= 5.196/12.56

≈ 0.414

Using more precise values, the ratio is approximately 0.4135.

Answer: The required ratio is 3√3/(4π), approximately 0.4135.

Hence proved.

Question 9.

A square is inscribed in a circle of radius r. Show that the ratio of the area of the square to the area of the circle is equal to 2/π ≈ 0.637.

Solution:

Let the side of the square be a and the radius of the circle be r.

The diagonal of the inscribed square is equal to the diameter of the circle.

Diagonal of the square = 2r

We know that the diagonal of a square is a√2.

Therefore,

a√2 = 2r

a = 2r/√2

a = √2r

Step 1: Find the area of the square.

Area of the square = a²

= (√2r)²

= 2r²

Step 2: Find the area of the circle.

Area of the circle = πr²

Step 3: Find the required ratio.

Ratio = Area of the square / Area of the circle

= 2r²/(πr²)

Cancelling r²,

= 2/π

Taking π ≈ 3.14,

Ratio = 2/3.14

≈ 0.637

Answer: The required ratio is 2/π, approximately 0.637.

Hence proved.

Question 10.

A regular hexagon is inscribed in a circle of radius r. Show that the ratio of the area of the hexagon to the area of the circle is equal to 3√3/(2π) ≈ 0.827.

Can you see why the answer is exactly twice the answer to Question 8?

Solution:

Let the radius of the circle be r.

A regular hexagon inscribed in a circle can be divided into six equal equilateral triangles by joining the centre to its vertices.

Each side of the regular hexagon is equal to the radius of the circle.

Therefore, side of each equilateral triangle = r.

Step 1: Find the area of one equilateral triangle.

Area of an equilateral triangle = (√3/4) × side²

= (√3/4)r²

Step 2: Find the area of the hexagon.

Area of the hexagon = 6 × Area of one equilateral triangle

= 6 × (√3/4)r²

= 6√3r²/4

= 3√3r²/2

Step 3: Find the area of the circle.

Area of the circle = πr²

Step 4: Find the required ratio.

Ratio = Area of the hexagon / Area of the circle

= (3√3r²/2)/(πr²)

Cancelling r²,

= 3√3/(2π)

Using √3 ≈ 1.732 and π ≈ 3.14,

Ratio = (3 × 1.732)/(2 × 3.14)

= 5.196/6.28

≈ 0.827

Answer: The required ratio is 3√3/(2π), approximately 0.827.

Why is this ratio exactly twice the answer to Question 8?

The area of the inscribed equilateral triangle in Question 8 is:

3√3r²/4

The area of the regular hexagon is:

3√3r²/2

Therefore,

Area of the hexagon = 2 × Area of the equilateral triangle

Hence, the ratio for the hexagon is twice the ratio for the triangle:

3√3/(2π) = 2 × 3√3/(4π)

Hence proved.