Area Related To Circle – Chapter 12.1 (NCERT Solutions) Mathematics (Class 10th)

By | 5th February 2019

SUBJECT – MATHS
CLASS  X (10th)
QUESTION FROM – Area Related To Circle – Chapter 12 (12.1)
ALL EXTRA & IMPORTANT / EXERCISE QUESTION ANSWER WITH VIDEOS


Hi Friends, Welcome To – Study24Hours.Com, Here We Are Giving You Some Very Important Question & Answers From The Chapter Area Related To Circle – Chapter 12. Area Related To Circle – Is The 12th Chapter Of Your Mathematics NCERT Book Of Class X. So Check Out All Of Them & Please Revise All Of These Questions Before Your Exam. So Please Remember To Visit – www.Study24Hours.Com Daily For All Study Materials And More.



Exercise 12.1 – Area Related To Circle – Chapter 12 (NCERT Solutions) Mathematics (Class 10th)

Unless stated otherwise, use π =22/7


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

SolutionAccording to the question, We got here

Radius of circle R1 = 19 cm
Radius of circle-II R2 = 9 cm

Circumference of circle-I = 2πR1 = 2π * 19 cm
Circumference of circle-II = 2πR2 = 2π * 9 cm

Therefore, the Sum of the circumferences of circle-I and circle-II = 2π * 19 + 2π * 9

= 2π(19 + 9)
= (2π * 28) cm

Let R be the radius of the circle-III.
and, Circumference of circle-III is = 2πR

According to the condition, 2πR = 2π * 28

=> R = (2π * 28)/2π
=> R = 28

Thus, the radius of the new circle = 28 cm





Q2. 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 – According to the question, We got here

Radius of circle-I, r1 = 8 cm
So, Area of circle-I = πr12 = π(8)2 cm2
Area of circle-II = πr22 = π(6) 2 cm2

So, Let the radius of the circle-III be R
∴ Area of circle-III = πR2

Now, according to the condition,
πr12 + πr12 = πR2

=> π(8)2 + π(6)2 = πR2
=> π(82 + 62) = πR2
=> 64 + 36 = R2
=> 100 = R2
=> 102 = R2
=> R = 10

Thus, the radius of the new circle = 10 cm.



Q3. Fig. 12.3 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 – According to the question, We got here

Diameter of the innermost region = 21 cm
Radius of the innermost (Gold Scoring) region = 21/2 = 10.5 cm
So, area of Gold region = π(10.5)2



= (22/7) * 10.5 * 10.5
= 22 * 1.5 * 10.5
= 346.50 cm2

Area of Red region = π(10.5 + 10.5)2 – π(10.5)2

= π(21)2 – π(10.5)2
= (22/7) * 21 * 21 – 346.50
= 22 * 21 * 3 – 346.50
= 1386 – 346.50
= 1039.50 cm2

Area of Blue region = π(21 + 10.5)2 – π(21)2

= π(31.5)2 – π(21)2
= π{(31.5)2 – (21)2}
= π(31.5 – 21)(31.5 + 21)
= (22/7) * 10.5 * 52.5
= 22 * 1.5 * 52.5
= 1732.50 cm2

Area of Black region = π(31.5 + 10.5)2 – π(31.5)2

= π(42)2 – π(31.5)2
= π{(42)2 – (31.5)2}
= π(42 – 31.5)(42 + 31.5)
= (22/7) * 10.5 * 73.5
= 22 * 1.5 * 73.5
= 2425.50 cm2

Area of White region = π(42 + 10.5)2 – π(42)2

= π(52.5)2 – π(42)2
= π{(52.5)2 – (42)2}
= π(52.5 – 42)(52.5 + 42)
= (22/7) * 10.5 * 94.5
= 22 * 1.5 * 94.5
= 3118.50 cm2



Q4. 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?

Answer – According to the question, We got here

Given diameter of the wheel = 80
So, the radius of the wheel r = 80/2 = 40
Now circumference of the wheel = 2πr

=2π * 40
= 80π

Now Speed of the car = 66 km/hour = (66*1000 *100)/60

(1 km = 1000m and 1 m = 100 cm, 1 hour = 60 minutes)

= (66*1000 *10)/6 (100 and 60 is divided by 10)
= 11*10000 (66 and 6 is divided by 6)
= 110000 cm/minute

Distance travelled by car in 10 minutes = 110000*10 = 1100000 cm
Let the number of revolution of the wheel is n.
So, n * distance travelled in one minute = Distance travelled in 10 minutes

=> n * circumference of the wheel = Distance travelled in 10 minutes
=> n * 80π = 1100000
=> n = 1100000/80π
=> n = (110000)/8π (1100000 and 80 is divided by 10)
=> n = (110000*7)/(22*8) (π = 22/7)
=> n = (10000*7)/(2*8) (110000 and 22 is divided by 11)
=> n = (5000*7)/8 (10000 and 2 is divided by 2)
=> n = 625*7 (5000 and 8 is divided by 8)
=> n = 4375




So, the number of revolution in 10 minutes at speed of 66 km/hour is 4375



Q5. 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) π units (C) 4 units (D) 7 units
Answer – Numerical area of the circle = Numerical circumference of the circle

=> πr2 = 2πr
=> πr2 – 2πr = 0
=> π(r2 – 2r) = 0
=> r2 – 2r = 0 [Since π ≠ 0]
=> r(r – 2) = 0
=> r = 0 or r = 2

But r cannot be zero. So, r = 2 units.
Thus, the option (A) is the correct answer.



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