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A chord of a circle of radius 10 cm subtends a right angle at the centre of the circle. Find the area of the corresponding minor segment. [Use π = 3.14] - Mathematics

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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 minor segment. [Use π = 3.14.]

A chord of a circle of radius 10 cm subtends a right angle at the centre of the circle. Find the area of the corresponding minor segment. [Use π = 3.14]

Sum

Solution 1

Let AB be the chord of a circle subtending an angle of 90° at the centre O of the circle.

Area of ​​sector = `theta/(360^circ) x pi^2`

`= (90^circ)/(360^circ) xx 314 /100 xx 10xx10` cm2

= `1/4 xx 314` cm2

`= 157/2` cm2

= 78.5 cm

Corresponding minor segment = ΔAOB

= 78.5 cm2 − `[1/2 xx10 xx10]`

= 78.5 cm2 − 50 cm2

= 28.5 cm

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Solution 2

Area of minor segment = Area of sector AOBC − Area of right triangle AOB

`=theta/360^circpi("OA")^2 - 1/2xx` OA× OB `

`= 90^circ/360^circxx3.14(10)^2 - 1/2xx10xx10`

= 78.5 − 50

= 28.5 cm2

Hence, the area of minor segment is 28.5 cm2

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Notes

Students should refer to the answer according to their question and preferred marks.

  Is there an error in this question or solution?
Chapter 18: Area of Circle, Sector and Segment - Exercise 18A [Page 835]

APPEARS IN

RS Aggarwal Mathematics [English] Class 10
Chapter 18 Area of Circle, Sector and Segment
Exercise 18A | Q 50 | Page 835
NCERT Mathematics [English] Class 10
Chapter 12 Areas Related to Circles
Exercise 12.2 | Q 4. (i)

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