NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.1 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.1.

NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.1

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 10 |

Subject |
Maths |

Chapter |
Chapter 12 |

Chapter Name |
Areas Related to Circles |

Exercise |
Ex 12.1 |

Number of Questions Solved |
5 |

Category |
NCERT Solutions |

## NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.1

**Ex 12.1 Class 10 Question 1.**

**The radii of two circles are 19 cm and 9 cm respectively. Find the radius of the circle which has circumference equal to the sum of the circumferences of the two circles.**

**Solution:
**

**Given:**radius of 1

^{st}circle (R

_{1}) = 19 cm

∴ Circumference of 1

^{st}circle = 2πR

_{1}= 2π(19) cm

Radius of 2

^{nd}circle (R

_{2}) = 9 cm

∴ Circumference of 2

^{nd}circle = 2πR

_{2}= 2π(9) cm

Let radius of 3

^{rd}circle be R

_{3 }Circumference of 3

^{rd}circle = 2πR

_{3 }According to question,

2πR

_{1}+ 2πR

_{2}= 2πR

_{3}

⇒ 2π(R

_{1}+ R

_{2}) = 2πR

_{3}

⇒ R

_{1}+ R

_{2}= R

_{3 }⇒ 19 + 9 = R

_{3}

⇒ R

_{3}= 28 cm

**Class 10 Maths Chapter 12.1 Question 2.**

**The radii of two circles are 8 cm and 6 cm respectively. Find the radius of the circle having area equal to the sum of the areas of the two circles.**

**Solution:
**

**Given:**radius of 1

^{st}circle (R

_{1}) = 8 cm

Area of 1

^{st}circle = πR

_{1}

^{2}= π(8)

^{2}cm

^{2}

Radius of 2

^{nd}circle (R

_{2}) = 6 cm

Area of 2

^{nd}circle = πR

_{2}

^{2}= π(6)

^{2}cm

^{2 }Let radius of 3

^{rd}circle be R

_{3 }Area of 3

^{rd}circle = πR

_{3}

^{2 }According to question,

πR,

^{2}+ πR

_{2}

^{2}– πR

_{3}

^{2 }⇒ R

_{1}

^{2}+ R

_{2}

^{2}= R

_{3}

^{2 }⇒ (8)

^{2}+ i6)

^{2}– R

_{3}

^{2 }⇒ 64 + 36 = R

_{3}

^{2}⇒ R

_{3}= \( \sqrt{100} \) = 10 cm

**Areas Related To Circles Exercise 12.1 Question 3.**

**The figure depicts an archery target marked with its five scoring regions from the centre outwards as Gold, Red, Blue,**

**Black and White. The diameter of the region representing Gold score is 21 cm and each of the other bands is 10.5 cm wide. Find the area of each of the five scoring regions.**

**Solution:
**Diameter of the region representing gold score is 21 cm

⇒ Radius of the region representing gold region = \(\frac { 21 }{ 2 }\) cm

**Exercise 12.1 Class 10 Question 4.**

**The wheels of a car are of diameter 80 cm each. How many complete revolutions does each wheel make in 10 minutes when the car is travelling at a speed of 66 km per hour?**

**Solution:
**

**Given:**diameter of the wheels of the car = 80 cm

⇒ Radius of the wheel of the car = \(\frac { 80 }{ 2 }\) = 40 cm

Circumference of the wheel = 2πr = 2 x \(\frac { 22 }{ 2 }\) x 40 cm

Speed of the car = 66 km/h

Distance covered in 10 minutes =\(\frac { 66 x 10 }{ 60 }\) = 11 km

= 11 x 1000 x 100 cm = 11,00,000 cm

**Ex 12.1 Class 10 Maths Solution Question 5.**

**Tick the correct answer in the following and justify your choice: If the perimeter and the area of a circle are numerically equal, then the radius of the circle is**

(a) 2 units

(b) n units

(c) 4 units

(d) 7 units

**Solution:**

Let radius of the circle = r units

Perimeter of the circle = 2πr

Area of the circle = πr^{2
}According to question,

Perimeter of the circle = Area of the circle

⇒ 2πr = πr^{2
}⇒ r = 2 units

Hence, option (a) is correct.

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