Question

A chord of a circle of radius 7cm subtends a right angle at the centre. The area
of minor segment is
a) 28 cm2 b)77 cm2
c)49 cm2
d)14 cm2

Answer 1

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Questions & Answers

CBSE

Mathematics

Grade 9

Area of Segment of a circle

Question

Answers

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A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding:

i) minor segment

ii) major sector

Answer

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Hint: The diagram of the circle can be drawn, either direct formulas can be applied or geometry of subtracting larger area from smaller to get the required area.

Formulas to be used:

Area of sector =

θ

360

×πr2 where, r is the radius of the circle.

Area of triangle =

1

2

×b×h where, b and h are the base and height of the triangle respectively.

Complete step-by-step answer:

i) Minor segment of the circle is formed by XYZ

The area of this segment can be given by the difference of area formed by the sector OXYZ and triangle XOZ.

Area of sector OXYZ =

θ

360

×πr2

r = 10 cm (given)

θ = 90 °

Substituting these values:

⇒

90

360

×3.14×(10)2=

90

360

×

314

(10)2

×(10)2=

1

4

×314

$π$=3.14

= 78.5

Thus, Area of sector OXYZ = 78.5 cm2 _______ (1)

Area of triangle XYZ =

1

2

×b×h

b = h = 10 cm (base and height of triangle are equal to radius of the circle)

Substituting these values:

→

1

2

×10×10

= 5 X 10

= 50

Thus, Area of triangle XYZ = 50 cm2 _______ (2)

Required area of segment = equation (1) – equation (2)

= (78.5 – 50) cm2

Required area of segment = 28.5 cm2

Therefore, the area of the corresponding minor segment is 28.5 cm2

Answer 2

Given:

Radius of circle = 7cm

Angle subtended by chord = 90°

To find:

Area of the minor sector.

Solution:

We have a circle of radius 7cm. A chord AB subtends an angle of 90° at the centre. On joining OA and OB which are the radii of the circle, we obtain two portions: major sector and minor sector. A segment of a circle is the region bounded by chord AB and the corresponding arc lying between the chord's endpoints. The minor sector formed is a quadrant. The minor segment is APB. We need to find the area of this minor segment.

Area of the segment can be determined by subtracting the area of ΔAOB from the area of quadrant OAPB.

ΔAOB is a right triangle. Hence, its base and height are radii of the circle.

Area of ΔAOB = $\frac{1}{2}$ × base × height

$ar(AOB)=\frac{1}{2} *7*7=\frac{49}{2}=24.5cm^{2}$

Area of quadrant OAPB = $\frac{1}{4}$ × Area of circle

$ar(OAPB)=\frac{1}{4}\pi r^{2}$

where $r$ is the radius of the circle

$ar(OAPB)=\frac{1}{4}*\frac{22}{7}* 7*7=\frac{77}{2}=38.5cm^{2}$

Area of minor segment APB = Area of quadrant OAPB - Area of triangle AOB

$ar(APB)=\frac{77}{2}- \frac{49}{2}=\frac{28}{2}=14cm^{2}$

Hence, out of the given options, option (d) is the correct answer.

The area of a minor segment obtained by a chord of a circle of radius 7cm subtending an angle of 90° is 14cm². Option (d) is the correct answer.