NCERT Solutions for Class 10 Maths Exercise 11.1 (NEW SESSION)

Solve the followings Questions.

Unless stated otherwise, use π =22/7.

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.
Answer:
Let the radius of the third circle be R.
Circumference of the circle with radius R = 2πR
Circumference of the circle with radius 19 cm = 2π × 19 = 38π cm
Circumference of the circle with radius 9 cm = 2π × 9 = 18π cm
Sum of the circumference of two circles = 38π + 18π = 56π cm
Circumference of the third circle = 2πR = 56π
⇒ 2πR = 56π cm
⇒ R = 28 cm
The radius of the circle which has circumference equal to the sum of the circumferences of the two circles is 28 cm.

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.
Answer:
Let the radius of the third circle be R.
Area of the circle with radius R = πR2
Area of the circle with radius 8 cm = π × 82 = 64π cm2
Area of the circle with radius 6 cm = π × 62 = 36π cm2
Sum of the area of two circles = 64π cm2 + 36π cm2 = 100π cm2
Area of the third circle = πR2 = 100π cm2
⇒ πR= 100π cm2
⇒ R2 = 100 cm2
⇒ R = 10 cm
Thus, the radius of the new circle is 10 cm.

3. 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.

https://www.pw.live/chapter 12-Area Related to Circles Exercise 12.1/image006.jpg

Answer:

Diameter of Gold region= 21 cm
Radius of gold region= 21/2= 10.5 cm


Area of Gold Region = πr²
= π(10.5)² =( 22/7)×  110.25 = 346.5 cm²
Area of Gold Region= 345.5 cm²

Radius of red region = Radius for gold + red region= 10.5 + 10.5= 21 cm
Area of Red Region = π²(21² – 10.5²)

[Area of a ring= π (R²-r²), where R= radius of outer ring & r= radius of inner ring]

= 22/7 (21² – 10.5²)        [ a²-b²= (a+b)(a-b)]
= 22/7 (21 + 10.5)(21 – 10.5)
= (22/7 )x 31.5 x 10.5 = 1039.5 cm²

Area of Red Region = 1039.5 cm²


Radius of blue region = Radius of blue region = Now radius for gold + red+ blue region= 21+10.5= 31.5 cm

Area of Blue Region = π(31.5² – 21²)
= 22/7 (31.5² – 21²)
= 22/7 (31.5 +21)(31.5 – 21)
= (22/7 )x 52.5 x 10.5 = 1732.5 cm²

Area of Blue Region =1732.5 cm²


Now,
Radius of black region= radius for gold + red+ blue + black region= 31.5+10.5= 42 cm

Area of Black Region = π(42² – 31.5²)

= 22/7 (42²-31.5² )
= 22/7 (42+31.5)(42-31.5)
= (22/7 )x 73.5 x 10.5 = 2425.5 cm²

Area of Black Region =2425.5 cm²

Now
Radius of white region= radius for gold + red+ blue + black+ white region= 42+10.5= 52.5 cm

Area of White Region= π(52.5² – 42²)
= 22/7 (52.5²-42² )
= 22/7 (52.5+42)(52.5-42)
= (22/7 )x 94.5 x 10.5 = 3118.5 cm²

Area of white Region =3118.5 cm²

 

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?

Answer:

The car travels in 10 minutes =66 / 6
= 11km
= 1100000cm
Circumference of the wheel = distance covered by the wheel in one revolution Thus, we have,
Circumference = 2 × 22/7 × 80/2 = 251.43cm
Thus, the number of revolutions covered by the wheel in 1100000cm = 1100000 / 251.43 ≈ 4375

5. 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:

(A) 2 units

(B) π units

(C) 4 units

(D) 7 units

Answer:

(A) Circumference = Area

Correct option is A)
Perimeter of circle  = 2πr
Area of circle  = πr²
According to the Question,
Perimeter of circle = Area of circle
2πr = πr²
or, 2 = r
or, r = 2 units

CHAPTER NAMEOLD NCERTNEW NCERT 
Real NumbersEXERCISE 1.1 
EXERCISE 1.21.1CLICK HERE
EXERCISE 1.31.2CLICK HERE
EXERCISE 1.4
PolynomialsEXERCISE 2.12.1CLICK HERE
EXERCISE 2.22.2CLICK HERE
EXERCISE 2.3
EXERCISE 2.4
Pair of Linear Equations in Two VariablesEXERCISE 3.1
EXERCISE 3.23.1CLICK HERE
EXERCISE 3.33.2CLICK HERE
EXERCISE 3.43.3CLICK HERE
EXERCISE 3.5
EXERCISE 3.6
EXERCISE 3.7
Quadratic EquationsEXERCISE 4.14.1CLICK HERE
EXERCISE 4.24.2CLICK HERE
EXERCISE 4.3
EXERCISE 4.44.3CLICK HERE
Arithmetic ProgressionsEXERCISE 5.15.1CLICK HERE
EXERCISE 5.25.2CLICK HERE
EXERCISE 5.35.3CLICK HERE
EXERCISE 5.45.4 (Optional)CLICK HERE
TrianglesEXERCISE 6.16.1CLICK HERE
EXERCISE 6.26.2CLICK HERE
EXERCISE 6.36.3CLICK HERE
EXERCISE 6.4
EXERCISE 6.5
EXERCISE 6.6
Coordinate GeometryEXERCISE 7.17.1CLICK HERE
EXERCISE 7.27.2CLICK HERE
EXERCISE 7.3
EXERCISE 7.4
Introduction to TrigonometryEXERCISE 8.18.1CLICK HERE
EXERCISE 8.28.2CLICK HERE
EXERCISE 8.3
EXERCISE 8.48.3CLICK HERE
Some Applications of TrigonometryEXERCISE 9.19.1CLICK HERE
CirclesEXERCISE 10.110.1CLICK HERE
EXERCISE 10.210.2CLICK HERE
Construction
Areas Related to CirclesEXERCISE 12.1
EXERCISE 12.211.1CLICK HERE
EXERCISE 12.3
Surface Areas and VolumesEXERCISE 13.112.1CLICK HERE
EXERCISE 13.212.2CLICK HERE
EXERCISE 13.3
EXERCISE 13.4
EXERCISE 13.5
StatisticsEXERCISE 14.113.1CLICK HERE
EXERCISE 14.213.2CLICK HERE
EXERCISE 14.313.3CLICK HERE
EXERCISE 14.4
ProbabilityEXERCISE 15.114.1CLICK HERE
EXERCISE 15.2

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