In this chapter, we provide RD Sharma Class 10 Ex 15.3 Solutions Chapter 15 Areas Related To Circles for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 10 Ex 15.3 Solutions Chapter 15 Areas Related To Circles pdf, Now you will get step by step solution to each question.

Textbook | NCERT |

Class | Class 10 |

Subject | Maths |

Chapter | Chapter 15 |

Chapter Name | Areas Related To Circles |

Exercise | 15.3 |

Category | RD Sharma Solutions |

**RD Sharma Solutions for Class 10 Chapter** **15**** Areas Related To Circles** Ex 15.3 Download PDF

**Areas Related To Circles**Ex 15.3 Download PDF

**Chapter 15: Areas Related To Circles Exercise – 15.3**

**Question: 1**

**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.**

**Solution:**

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 cm^{2}

Therefore, the required area of the segment is 51.75 cm^{2}

**Question: 2**

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.

**Solution:**

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.

**Question: 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.

**Solution:**

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 × πr^{2 }– 0.5 × OA × OB

= 90/360 × π14^{2} – 0.5 × 14 × 14

= 154 – 98 = 56 cm^{2}

Therefore the area of the minor segment (ANB) = 56 cm^{2}

Area of the major segment (other than shaded) = area of circle – area of segment ANB = πr^{2 }– 56 cm^{2}

= 3.14 × 14 × 14 – 56

= 616 – 56

= 560 cm^{2}

Therefore, the area of the major segment = 560 cm^{2}.

**Question: 4**

**A chord 10 cm long is drawn in a circle whose radius is **5√2 cm**. Find the area of both segments.**

**Solution:**

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.

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 = θ/360 × πr^{2} – 0.5 × OA × OB

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

**Question: 5**

**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.**

**Solution:**

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)

**Question: 6**

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

**Solution:**

Given data: AB is a diameter of circle with centre O, Also, ∠COB = θ

= Angle subtended Area of sector BOC = θ/360 × πr^{2}

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)

On solving the above equation we get,

Hence proved that,

**Question: 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

**Solution:**

Let the area of the given circle be = r

We know that, area of a circle = πr^{2}

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)

On solving the above equation we get,

Hence proved,

**All Chapter RD Sharma Solutions For Class10 Maths**

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