RD Sharma Solutions Class 10 Areas Related To Circles Exercise 15.3

RD Sharma Class 10 Solutions Chapter 15 Ex 15.3 PDF Free Download

Exercise 15.3

 

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 triangleAOB )

= \(\frac{\theta }{360}\times\prod r^{2} – \frac{\sqrt{3}}{4}\left ( side\right )^{2}\)

= \(\frac{ 60 }{ 360 }\times \prod 4 ^{2} – \frac{\sqrt{ 3 }}{ 4 }\left ( 4\right )^{2}\)

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 = \(\frac{\theta }{360}\times\prod r^{2} – \frac{\sqrt{3}}{4}\left ( side \right )^{2}\)

We have,

\(\angle POQ = 120 and PQ = 12 cm\)

PL = PQ x ( 0.5 )

= 12 x 0.5 = 6 cm

Since, \(\angle POQ\) = 120

\(\angle POL\)= \(\angle QOL\)= 60

In triangle OPQ, we have

\(sin \theta = \frac{ PL }{ OA }\) ,

sin 60° = \(\frac{ 6 }{ OA }\) ,

OA = \(\frac{ 12 }{ \sqrt{ 3 }}\)

Thus ,OA = \(\frac{ 12 }{ \sqrt{ 3 }}\)

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

\(A = 4 \left \{ 4\pi – 3 \sqrt{ 3 } \right\} cm^{2}\).

 

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 )

= \(\frac{\theta }{360}\times\prod r^{2}\) – 0.5 x OA x OB

= \(\frac{90}{360}\times\prod 14^{2}\) – 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

= \(\prod r^{ 2 } – 56 cm^{2}\)

= 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 \(5\sqrt{2} cm\) . Find the area of both segments.

 

Soln:

Given data: Radius of the circle , r =   \(5\sqrt{2} cm\) = OA = OB

Length of the chord AB = 10cm

In triangle OAB , OA2 +OB2 = \((5\sqrt{2})^{2} + (5\sqrt{2})^{2}\) = 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

=     \(\frac{\theta }{360}\times\prod r^{2}\) – 0.5 x OA x OB

= \(\frac{ 90 }{ 360 }\times \prod (5\sqrt{2})^{2}  – 0.5 x  (5\sqrt{2})^{2} x (5\sqrt{2})^{2}\)

= \(\frac{1100}{7} -\frac{100}{7}=\frac{1000}{7} cm^{2}\)

Therefore, Area of segment (minor) = \(\frac{1000}{7} cm^{2}\).

 

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, \(\angle A + \angle B + \angle O = 180\)

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 OAB

= \(\frac{\theta }{360}\times\prod r^{2}-\frac{\sqrt{ 3 }}{ 4 }(- AB)^{2}\)

On solving the above equation we get,

= \(\frac{308 – 147\sqrt{3}}{3} cm^{2}\)

Therefore, area of the segment (shaded region in the figure ) = \(\frac{308 – 147\sqrt{3}}{3} cm^{2}\).

 

Q 6. Ab is the diameter of a circle with centre ‘O’ . C is a point on the circumference such that \(\angle COB\) = θ. The area of the minor segmentcut off by AC is equal to twice the area of sector BOC. Prove that \(\sin \frac{\theta }{2}.\cos \frac{\theta }{2}\) = \(\prod \left ( \frac{1}{2} – \frac{\theta }{120} \right )\).

 

 Soln:

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

Also, \(\angle COB\) = θ = Angle subtended

Area of sector BOC = \(\frac{\theta }{360}\times \prod r^{2}\)

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

\(\angle AOC\) = 180 – θ \(\angle AOC and \angle BOC\) from linear pair ]

Area of sector = \(\frac{\left ( 180 -\theta \right )}{360}\times\pi\times r^{2} = \frac{\pi \times r^{2}}{2} – \frac{\pi \theta r^{2}}{360}\)

In triangle AOC , drop a perpendicular AM , this bisects \(\angle AOC\) and side AC.

Now, In triangle AMO, \(sin\angle AOM=\frac{ AM }{ OA }=sin(\frac{180 -\theta}{2})=\frac{AM}{r}\)

\(AM = r\sin\left ( 90 -\frac{\theta }{2} \right )=r\cos \frac{\theta }{2}\)

\(\cos \angle AOM =\frac{OM}{OA}= \cos \left ( 90-\frac{\theta }{2} \right )= \frac{OM}{r} \Rightarrow OM = r\sin \frac{\theta }{2}\)

Area of segment=  \(\frac{\pi r^{2}}{2}-\frac{\pi \theta r^{2}}{360}-\frac{1}{2}\left ( AC \times OM \right )\left [ AC = 2 AM \right ]\)

= \(\frac{\pi r^{2}}{2}-\frac{\pi \theta r^{2}}{360}-\frac{1}{2}\left ( 2r\cos \frac{\theta }{2} r\sin \frac{\theta }{2}\right ) = r^{2} \left [ \frac{\pi }{2} -\frac{\pi \theta }{360}-\cos \frac{\theta }{2}\sin \frac{\theta }{2}\right ]\)

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

\(r^{2}\left [ \frac{\pi }{2} -\frac{\pi \theta }{360} – \cos \frac{\theta }{2}.\sin \frac{\theta }{2} \right ] = 2r^{2}\left [ \frac{\pi\theta }{360} \right ]\)

On solving the above equation we get,

\(cos\frac{\theta }{2}×\sin\frac{\theta }{2} =\pi ( \frac{1}{2}-\frac{\theta }{120})\)

Hence proved that,  \(\cos \frac{\theta }{2}.\sin\frac{\theta }{2} =\pi \left ( \frac{1}{2} – \frac{\theta }{120} \right )\).

 

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 \(8 \sin \frac{\theta }{2}. \cos \frac{\theta}{2} + \pi = \frac{\pi \theta }{45}\).

 

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 \(\angle AOM\)

\(\angle AOM = \angle MOB = \frac{1}{2} \left ( 0 \right ) = \frac{\theta }{2} , AB = 2 AM\)

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

\(\frac{\theta }{360}\times \pi r^{2} – \frac{1}{2} \times AB \times OM = r^{2} \left [ \frac{\pi \theta }{360} \right ] – \frac{1}{2} . 2 r \sin \frac{\theta}{2} . \cos \frac{\theta}{2}\)

Area of segment = \(\frac{1}{8}\)  ( area of circle )

\(r^{2}\left [ \frac{\pi \theta }{360} – \sin \frac{\theta }{2} . \cos \frac{\theta }{2}\right ] = \frac{1}{8} \pi r^{2}\)

On solving the above equation we get,

\(8 \sin \frac{\theta }{2}. \cos \frac{\theta}{2} + \pi = \frac{\pi \theta }{45}\)

Hence proved, \(8 \sin \frac{\theta }{2}. \cos \frac{\theta}{2} + \pi = \frac{\pi \theta }{45}\)..

 

 

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