Areas Related Circles Exercise 15.3 | Extra Documents, Videos & Tests for Class 10 PDF Download

Q1. AB is a chord of a circle with center O and radius 4 cm. AB is of length 4 cm and divides the circle into two segments. Find the area of the minor segment. 

Soln:

Given data:

Radius of the circle with center‘O’, r = 4 cm = OA = OB

Length of the chord AB = 4 cm

OAB is an equilateral triangle and angle AOB = 60° + θ

Angle subtended at centre θ = 60°

Area of the segment ( shaded region ) = ( area of sector) – ( area of triangle AOB )

On solving the above equation, we get,

= 58.67 – 6.92 = 51.75 cm2

Therefore, the required area of the segment is 51.75 cm2

Q2. A chord PQ of length 12 cm subtends an angle 120 at the center of a circle. Find the area of the minor segment cut off by the chord PQ. 

Soln:

We know that,

Area of the segment

We have, 

PL = PQ x ( 0.5 )

= 12 x 0.5 = 6 cm

In triangle OPQ, we have 

Now using the value of r and angle θ we will find the area of minor segment. 

Q 3. A chord of circle of radius 14 cm makes a right angle at the centre. Find the areas of minor and major segments of the circle. 

Soln:

Given data:

Radius ( r ) = 14 cm

Angle subtended by the chord with the centre of the circle, θ = 90°

Area of minor segment ( ANB ) = ( area of ANB sector ) – ( area of the triangle AOB )

 - 0.5 x OA x OB 

 – 0.5 x 14 x 14 = 154 – 98 = 56 cm2 

Therefore the area of the minor segment ( ANB ) = 56 cm2

Area of the major segment (other than shaded) = area of circle – area of segment ANB

= 3.14 x 14 x 14 – 56 = 616 – 56 = 560 cm2

Therefore, the area of the major segment = 560 cm2.

 Q 4.  A chord 10 cm long is drawn in a circle whose radius is Find the area of both segments.

Soln: 

Given data: Radius of the circle , r  

Length of the chord AB = 10cm 

n triangle OAB , OA2 +OB2 =  = 50 + 50 = 100 =( AB )2 

Hence, pythogoras theorem is satisfied.

Therefore OAB is a right angle triangle.

Angle subtended by the chord with the centre of the circle, θ = 90°

Area of segment (minor) = shaded region = area of sector – area of triangle OAB

Therefore, Area of segment (minor) 

Q5. A chord AB of circle of radius 14 cm makes an angle of 60° at the centre. Find the area of the minor segment of the circle. 

Soln:

Given data: radius of the circle (r) = 14 cm = OA = OB

Angle subtended by the chord with the centre of the circle, θ = 60°

In triangle AOB, angle A = angle B [angle opposite to equal sides OA and OB ] = x

By angle sum property, 

X + X + 60° =180°

2X = 120°, X = 60°

All angles are 60°, triangle OAB is equilateral OA = OB = AB

= area of the segment (shaded region in the figure) = area of sector– area of triangle OAB

On solving the above equation we get, 

Therefore, area of the segment (shaded region in the figure ) 

Q 6. Ab is the diameter of a circle with centre ‘O’ . C is a point on the circumference such that  The area of the minor segmentcut off by AC is equal to twice the area of sector BOC. Prove that  

 Soln:

Given data:  AB is a diameter of circle with centre O,

Also,  Angle subtended 

Area of sector BOC 

Area of segment cut off by AC = (area of sector) – (area of triangle AOC) 

Area of sector  

In triangle AOC , drop a perpendicular AM , this bisects and side AC. 

Now, In triangle AMO, 

Area of segment= 

Area of segment by AC = 2 (Area of sector BOC) 

On solving the above equation we get, 

Hence proved that,

Q 7.  A chord a circle subtends an angle θ at the center of the circle. The area of the minor segment cut off by the chord is one-eighth of the area of the circle. Prove that 

Soln:

Let the area of the given circle be = r

We know that, area of a circle = π r2

AB is a chord, OA and OB are joined. Drop a OM such that it is perpendicular to AB, this OM bisects AB as well as 

Area of segment cut off by AB = (area of sector) – (area of the triangle formed) 

Area of segment = 1/8 ( area of circle ) 

On solving the above equation we get, 

Hence proved,