NCERT Solutions Class 10 Maths Exercise 12.1 of Chapter 12-Areas related to circles - Future Study Point

# NCERT Solutions Class 10 Maths Exercise 12.1 -Areas related to circles

NCERT Solutions Class 10 Maths Exercise 12.1 of Chapter 12-Areas related to circles are the solutions of unsolved questions of NCERT maths text book of class 10 prescribed by CBSE. All NCERT solutions of unsolved questions are created by an expert of future study point. You can study here sample papers, solutions of previous year question papers, tips for the preparation of government entrance exams, and carrier in online jobs.

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## Download pdf of NCERT Solutions of class 10 maths chapter 12-Areas related to circles

Pdf-NCERT solutions of class 10 maths chapter 12-Areas related to Circle

## NCERT Solutions for Class 10 Mathsย  ย Chapter 12-Areas related to circles

Exercise 12.1- Area Related to Circle

Exercise 12.2- Areas related to Circle

Exercise 12.3-Areas Related to the Circle

Q1. The radii of the two circles 19 cm and 9 cm respectively . Find the radius of the circles which has a circumference equal to the sum of the circumferences of two circles.

Ans.ย  Circumference of the circle = 2ฯr, where r is the radius of the circle and ฯ= 22/7

In the question, we are given the radii of two circles r1 = 19cm ,r2 = 9 cm

Let the radius of the required circle = R

According to question

Circumference of the required circle = Sum of the circumferences of two circles.

2ฯR = 2ฯr1+ 2ฯr2

2ฯR = 2ฯ(r1 + r2)

R = 19 + 9 = 28

Therefore the radius of the circle is = 28 cm

Q2.The radii of the two circles 8 cm and 6 cm respectively. Find the radius of the circle having an area equal to the sum of the areas of two circles.

Ans.

Area of the circle = ฯrยฒ, where r is the radius of the circle and ฯ= 22/7

In the question, we are given the radii of two circles r1 = 8cm ,r2 = 6cm

Let the radius of the required circle = R

According to question

Area of the required circle = Sum of the areas of two circles.

ฯRยฒ = ฯr1ยฒ+ ฯr2ยฒ

ฯRยฒ = ฯ(r1ยฒ + r2ยฒ)

Rยฒ= 8ยฒ + 6ยฒ= 64+ 36 = 100

R = 10

Therefore the radius of the circle is = 10 cm

Q3. The given figure depicts an archery target with its 5 scoring areas from the centre outwards as Gold, Red, Blue, Black, and White. The diameter representing the 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.(Use ฯ = 22/7).

Ans. We are given the diameter of circle representing gold region = 21 cm

The radius of gold region is = 21/2 = 10.5 cm

Width of each band of colour is = 10.5 cm

Area of gold region =are of 1 st circle = ฯrยฒ = (22/7)ร10.5 ร10.5 = 346.5 sq.cm

Area of red band = Area of 2 nd circleย  – Area of 1 st circle

= ฯ[(radius of 2 nd circle)ยฒ- (radius of 1 st circle)ยฒ]

=22/7[(10.5 +10.5)ยฒ- (10.5)ยฒ]

22/7[ 21ยฒ -(10.5)ยฒ]

22/7[ 441 – 110.25]

22/7(330.75) = 22ร47.25= 1039.5 sq.cm

Area of blue band = Area of 3 rd circleย  – Area of 2 ndย  circle

= ฯ[(radius of 3 rd circle)ยฒ- (radius of 2 nd circle)ยฒ]

=22/7[(21+10.5)ยฒ- (21)ยฒ]

22/7[ 31.5ยฒ -(21)ยฒ]

22/7[ 992.25 – 441]

22/7(551.25) = 22ร78.75= 1732.5 sq.cm

Area of black band = Area of 5 th circleย  – Area of 3 rdย  circle

= ฯ[(radius of 4 th circle)ยฒ- (radius of 3 rd circle)ยฒ]

=22/7[(31.5+10.5)ยฒ- (31.5)ยฒ]

22/7[ 42ยฒ -(31.5)ยฒ]

22/7[ย  1764 – 992.25]

22/7(771.75) = 22ร110.25= 2425.5 sq.cm

Area of white band = Area of 5 th circleย  – Area of 4 th circle

= ฯ[(radius of 5 th circle)ยฒ- (radius of 4 th circle)ยฒ]

=22/7[(42+10.5)ยฒ- (42)ยฒ]

22/7[( 52.5)ยฒ -(42)ยฒ]

22/7[ย  2756.25 – 1764]

22/7(992.25) = 22ร141.75= 3118.5 sq.cm

Q4. The wheels of a car of diameter 80 cm each . How many complete revolutions does each wheel make in 10 minutes when the car is travelling at the speed of 66 km/h.

Ans. We are given the diameter of the wheel = 80 cm

The radius of the wheel,r = 80/2 = 40 cm

One revolution of the wheels corresponds to the distance traveled by the car= Circumference of the wheel = 2ฯr = (2ร22/7 )40 = 1760/7 cm

The speed of car is given = 66 km/h means the car travels 66 km in 1 h

1h = 60 minutes and 66 km = 66000 m = 6600000 cm

In 60 minutes car travels = 6600000 cm

In 1 minute the car travel = 6600000/60 = 110000 cm

In 1 revolution car travels = 1760/7 cm

The number of revolutions of wheels in 1 minute = The distance traveled by the car in 1 minute/ distance traveled in revolutions

= 110000 รท 1760/7

= 110000 ร 7/1760

= 437.5 revolutions

In 10 minutes the number of revolutions ,the wheels have = 437.5 ร10 = 4375

Q5.Tick the correct answer in the following and justify your choice. If the perimeter and area of a circle are numerically equal, then the radius of the circle is.

(i) 2 unitsย  ย (ii) ฯย  unitsย  ย (iii) 4ย  unitsย  ย  (iv) 7 units

Ans.

Let the radius of the circle is = r

Circumference of the circle is = 2ฯr

The area of the circle is = ฯrยฒ

According to question

ฯrยฒ = 2ฯr

r = 2

Hence the radius of the circle is 2 units, so the answer is (i) 2 units

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