RD Sharma Solutions Class 10 Areas Related To Circles Exercise 15.3

RD Sharma Solutions Class 10 Chapter 15 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, \(8 \sin \frac{\theta }{2}. \cos \frac{\theta}{2} + \pi = \frac{\pi \theta }{45}\)..

 

 


Practise This Question

In a marathon of 5 km, Manish could run for 2 km.If  Kushagra ran for 12 of the total length of the marathon, who covered more distance?