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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.
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
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.
We know that, Area of the segment
We have, ∠POQ = 120 and PQ = 12cm
PL = PQ × (0.5) = 12 × 0.5 = 6 cm
Since, ∠POQ = 120
∠POL = ∠QOL = 60
In triangle OPQ, we have
Now using the value of r and angle θ. We will find the area of minor segment.
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.
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)
= θ/360 × πr2 - 0.5 × OA × OB
= 90/360 × π142 - 0.5 × 14 × 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 = πr2 - 56 cm2
= 3.14 × 14 × 14 - 56
= 616 - 56
= 560 cm2
Therefore, the area of the major segment = 560 cm2.
A chord 10 cm long is drawn in a circle whose radius is 5√2 cm. Find the area of both segments.
Given data: Radius of the circle, r = 5√2 cm = OA = OB
Length of the chord AB = 10 cm
In triangle OAB,
Hence, Pythagoras theorem is satisfied.
Therefore OAB is a right angle triangle.
Area of segment (minor) = shaded region = area of sector - area of triangle
OAB = θ/360 × πr2 - 0.5 × OA × OB
Therefore, Area of segment (minor) = 1000/7 cm2.
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.
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, ∠A + ∠B + ∠O = 180
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
On solving the above equation we get,
Therefore, area of the segment (shaded region in the figure)
Ab is the diameter of a circle with centre 'O'. C is a point on the circumference such that ∠COB = θ. The area of the minor segment cut off by AC is equal to twice the area of sector BOC. Prove that
Given data: AB is a diameter of circle with centre O, Also, ∠COB = θ
= Angle subtended Area of sector BOC = θ/360 × πr2
Area of segment cut off by AC = (Area of sector) - (Area of triangle AOC)
∠AOC = 180 - θ ∠AOC and ∠BOC from linear pair] Area of sector
In triangle AOC, drop a perpendicular AM, this bisects ∠AOC and side AC.
Now, In triangle AMO,
Area of segment
Area of segment by AC = 2 (Area of sector BOC)
Hence proved that,
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
Let the area of the given circle be = r
We know that, area of a circle = πr2
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
∠AOM
Area of segment cut off by AB = (area of sector) - (area of the triangle formed)
Hence proved,
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Chapter 15: Areas Related To Circles Exercise...