Area of a Circle
Trending Questions
- 1320 sq.m
- 1300 sq.m
- 1400 sq.m
- 1420 sq.m
Radius of a circle is doubled. What will be the ratio of areas of the new circle to the area of the given circle?
2:3
4:8
4:1
5:3
- ₹ 878
- ₹ 887
- ₹ 918
- ₹ 981
Two circles touch each other internally and their centres are O and O′ as shown. The sum of their areas is 180 π sq. cm. and the distance between their centres is 6 cm. What is the diameter of the larger circle?
16 cm
12 cm
18 cm
24 cm
If the area to be covered for TV telecast is doubled, then the height of the transmitting antenna (TV tower) will have to be
doubled
halved
quadrupled
kept unchanged
The circumference of a circle is directly proportional to the length of its radius, while the area it encloses is directly proportional to the square of the length of its radius. If the area of a circle is multiplied by , then its circumference is multiplied by ______.
- 2 πr2
- 2 πr3
- πr2
- πr3
Write True or False and justify your answer:
The cost of leveling the ground in the form of a triangle having the sides 51m, 37m and 20m at the rate of Rs.3 per m2 is Rs.918.
Find the area of the shaded region in Fig, if radii of the two concentric circles with centre O are 7 cm and 14 cm respectively and ∠AOC=40∘.
50 cm2
48 cm2
51 cm2
36 cm2
In the given figure, a circle with cenre O has diameter 112 cm. Two equal circles whose diameter is half the diameter of the bigger circle are drawn inside the circle with centre O. If the shaded portion has been cut-out from the two smaller circles as shown in the figure, then the area of the remaining part of the circle is equal to
2 times area of shaded figure
None of these
area of shaded figure
3 times area of shaded figure
If the perimeter of a circle is equal to that of a square, then the ratio of area of circle to the square is ______.
14 : 11
7 : 22
11 : 14
22 : 07
A circle with centre O and radius 7 cm is circumscribing a square ABCD of side 8 cm.
Find the area of the shaded region. [Take π=227]
100 cm2
64 cm2
90 cm2
154 cm2
- True
- False
A farmer has to plough the field for five different types of crops naming A, B, C, D and E in a circular field. If the diameter of the field, where crops A, B and C are grown is 126 units and every other band is 21 units wide. Then, choose the correct option.
i) Area of region, where crop A is grown =1386 sq. units
ii) Area of region, where crop B is grown =22176 sq. units
iii) Area of region, where crops A, B, C and D are grown =5544 sq. units
iv) Area of region, where crop E is grown =34650 sq. units
i) only
ii) and iii) only
ii), iii) and iv)
i), ii) and iii)
Find the area of the figure ABCDE given below:
64 sq. cm
70 sq. cm
30 sq. cm
72 sq. cm
- 135 m2
- 125 m2
- 125√3 m2
- 135√3 m2
(a) 14 cm2
(b) 14.1 cm2
(c) 14.11 cm2
(d) 14.1124 cm2
If the larger circle of radius 14 cm has four smaller circles of radius 3.5 cm, then the area of the shaded portion is
400 cm2
426 cm2
462 cm2
482 cm2
The adjacent figure shows two arcs P and Q. Arc P is part of the circle with centre O and radius OA=10 cm. Arc Q is a part of the circle with centre M and radius AM=5 cm, where M is the mid-point of AB. Area enclosed by the arc Q and chord AB is
25π cm2
252π cm2
20π cm2
202π cm2
How would you go about finding the area of the shaded region?
(Area of I + Area of III) + (Area of II + Area of IV)
(Area of I + Area of IV) + (Area of II + Area of III)
(Area of I + Area of II) × 2
(Area of I + Area of II) + (Area of III + Area of IV)
- 21 cm
- 10 cm
- 35 cm
- 28 cm
Write True or False and justify your answer:
In a triangle, the sides are given as 11cm, 12cm and 13cm. the length of the altitude is 10.25 cm corresponding to the side having 12 cm.
Find the diameter of a circle in cm whose area is equal to the sum of the area of the two circles of radii 24 cm and 7 cm.
25
62
50
31
- The radius of the quarter circles in figure 1 is equal to the radius of the circle in figure 2
- The radii of the quarter circles are equal.
If the sum of the areas of two circles with radii R1 and R2 is equal to the area of a circle of radius R, then
R1 + R2 = R
R1 + R2 < R
R12 + R22 < R2
R12 + R22 = R2