Note:
(i) The interior of a circle along with its boundary is called the circular region of the circle.
(ii) By the area of a circle, we mean the area of the circular region.
➢ Area of Sector and Segment of a Circle
Note:
(i) ∠AOB is called the ‘angle of sector’.
(ii) OAPB is the ‘minor sector’ and OAQB is the ‘major sector’.
(iii) APB is the ‘minor segment’ and AQB is the ‘major segment’.
(iv) When we write ‘sector’ and ‘segment’ we will mean the ‘minorsector’ and the ‘minorsegment’ respectively.
➢ Let us remember that
[Unless stated otherwise, use π = 22/7]
Q.1. The radii of the 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.
Sol. We have, r_{1} = 19 cm
► r_{2} = 9 cm
► Circumference of circleI = 2πr_{1} = 2π (19) cm
► Circumference of circleII = 2πr_{2} = 2π (9) cm
► Sum of the circumferences of circleI and circleII = 2π (19) + 2π (9)
= 2π (19 + 9) cm = 2π (28) cm
Let R be the radius of the circleIII.
∴ Circumference of circleIII = 2πR
According to the condition:
► 2πR = 2π (28)
Thus, the radius of the new circle = 28 cm
Q.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.
Sol. We have, Radius of circleI, r_{1} = 8 cm
► Radius of circleII, r_{2} = 6 cm
► Area of circleI = πr_{1}^{2} = π (8)^{2} cm^{2}
► Area of circleII = πr_{2}^{2} = π (6)^{2} cm^{2}
Let the area of the circleIII be R
∴ Area of circleIII = πR^{2}
Now, according to the condition:
π r_{1}^{2} + π r_{2}^{2} = πR^{2}
i.e. π (8)^{2} + π (6)^{2} = πR^{2}
⇒ π (8^{2} + 6^{2})= πR^{2}
⇒ 8^{2} + 6^{2} = R^{2}
⇒ 64 + 36 = R^{2}
⇒ 100 = R^{2}
⇒ 102 = R^{2} ⇒ R = 10
Thus, the radius of the new circle = 10 cm.
Q.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.
Sol. Diameter of the innermost region = 21 cm
Radius of the innermost (Gold Scoring) region = 21/2 = 10.5 cm
► Area of Gold region = π (10.5)^{2} cm^{2}
► Area of the Red region
► Area of Blue region
► Area of Black region
► Area of White region
= π [(42 + 10.5)2] = (42)^{2} cm
Q.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?
Sol. The diameter of a wheel = 80 cm
∴ Radius of the wheel = 80/2 = 40 cm
∴ Circumference of the wheel
⇒ Distance covered by a wheel in one revolution
Distance travelled by the car in 1hr = 66 km = 66 × 1000 × 100 cm
∴ Distance travelled in 10 minutes
Now,
Number of revolutions
Thus, the required number of revolutions = 4375
Q.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) π units
(c) 4 units
(d) 7 units
Ans. (a)
Sol. We have
[Numerical area of the circle] = [Numerical circumference of the circle]
⇒ π r^{2} = 2πr
⇒π r^{2} − 2πr = 0
⇒ r^{2} − 2r = 0
⇒ r (r − 2) = 0r = 0 or r = 2
But r cannot be zero
∴ r = 2 units.
Thus, option (a) is correct.
276 docs149 tests

1. What are some important areas related to circles? 
2. How do you find the circumference of a circle? 
3. How do you find the area of a circle? 
4. What is the relationship between the diameter and radius of a circle? 
5. How are circles and their properties used in reallife applications? 

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