Circumference and area of a Circle
Category : 7th Class
The boundary of the circle is called circumference.
Circumference of a circle \[=2\pi r.\]
Find the circumference of a circle whose diameter is 4.2 cm.
(a) 11.1 cm
(b) 112.2cm
(c) 13.3cm
(d) 13.2cm
(e) None of these
Answer: (d)
Explanation
Diametre of circle = 4.2 cm
\[\therefore \]Radius of circle = 2.1 cm
Circumference \[=2\pi r=2\times \frac{22}{7}\times 2.1=2\times 22\times 0.3=13.2cm\]
Thus the circumference of this circle = 13.2 cm.
Find the radius of circle whose circumference is 88 m.
(a) 14m
(b) 15m
(c) 120m
(d) 25m
(e) None of these
Answer: (a)
Explanation
Circumference (c) = 88 m
\[C=27\pi r=88\Rightarrow 2\times \frac{22}{7}\times r=88\]
Or \[r\frac{7\times 88}{2\times 22}=7\times 2=14\,m\]
Thus the radius of this circle = 14 m
Find the area of a rhombus shaped field, whose each of the sides is 14 cm and the altitude is 1.6 dm.
(a) \[224c{{m}^{2}}\]
(b) \[214c{{m}^{3}}\]
(c) \[224{{m}^{2}}\]
(d) \[224c{{m}^{6}}\]
(e) None of these
Answer: (a)
Explanation
Base of rhombus = 14 cm
Altitude = 1.6 dm = 16 cm
Area of rhombus = Base \[\times \] Altitude
\[=14\times 16=224c{{m}^{2}}\]
How many tiles \[20cm\times 20cm\] each will be required to pave a footpath 1 m wide carried round the outside of a plot \[28\text{ }m\times 18m\]?
(a) 2,400
(b) 5, 2000
(c) 9,900
(d) 5, 250
(e) None of these
Answer: (a)
Explanation
Area of the outer rectangle
\[=(2,800+2\times 100)\times (1,800+\text{2}\times 100)=3,000\times 2,000\text{ }c{{m}^{2}}\]
Area of the inner rectangle \[=2,800\times 1,800\,c{{m}^{2}}\]
Area of the Footpath
\[=3,000\times 2,000-2,800\times 1,800\text{ }9,60,000\text{ }c{{m}^{2}}\]
Required number of tiles
\[=\frac{area\,of\,footpath}{area\,of\,a\,tile}=\frac{9,60,000}{20\times 20}=2,400\]
The base of a parallelogram is twice its height. If the area of parallelogram is \[512\text{ }c{{m}^{2}}\] then find the base.
(a) \[33c{{m}^{2}}\]
(b) \[32c{{m}^{2}}\]
(c) \[40c{{m}^{2}}\]
(d) \[199c{{m}^{2}}\]
(e) None of these
Answer: (b)
A rhombus of side equal to 65 cm has an area of 2,016 cm2. Find its diagonals.
(a) 30
(b) 148
(c) 32
(d) 55
(e) None of these
Answer: (c)
A parallelogram has sides 60 cm and 40 cm and one of its diagonal is 80cm. Find its area.
(a) \[600\sqrt{15}cm\]
(b) \[600\text{ }\sqrt{15}\text{ }c{{m}^{2}}\]
(c) \[650\sqrt{51\,}c{{m}^{2}}\]
(d) \[500\text{ }\sqrt{51\,}c{{m}^{2}}\]
(e) None of these
Answer: (b)
A wire, when bent in the form of a square, encloses an area of \[484\text{ }c{{m}^{2}}.\] If the same wire is bent in the form of a circle then find the area enclosed by it.
(a) \[161c{{m}^{2~}}\]
(b) \[616c{{m}^{2}}\]
(c) \[616\,c{{m}^{2~}}\]
(d) \[916\,c{{m}^{2}}\]
(e) None of these
Answer: (c)
Explanation
Area of square\[~=484\text{ }c{{m}^{2}}\]
Side of square \[\sqrt{484}=22cm\]
Perimeter of square \[=4\times 22\text{ }cm=88\text{ }cm\]
Let r be the radius of the circle.
Same wire is bent in the form of a square and circle.
Therefore, circumference of the circle = perimeter of the square.
or 2\[2\pi r=88\]
\[2\frac{22}{7}\times r=88\Rightarrow r=\frac{88\times 7}{2\times 22}=14\,cm\]
Thus area of circle \[=\pi {{r}^{2}}=\frac{22}{7}\times 14\times 14=616\,c{{m}^{2}}\]
Semicircular lawns are attached to the edges of a rectangular field measuring \[42\text{ }m\text{ }\times 35m.\] Find the area of the total field.
(a) \[3895.6{{m}^{2}}~~\]
(b) \[3818.5{{m}^{2}}\]
(c) \[3735.6{{m}^{2}}\]
(d) \[3899.9{{m}^{2}}\]
(e) None of these
Answer: (b)
Explanation
Area of lawn = Area of rectangle \[ABCD\text{ }+2\times \]Area of semicircle with diameter \[AB+2\times \]Area of semicircle with diametre BC
\[=\text{ }Length\times breadth+2\left( \frac{\pi {{R}^{2}}}{2} \right)+2\left( \frac{\pi {{r}^{2}}}{2} \right)\]
Where R = radius of bigger semicircle
\[=\frac{42}{2}=21\,m.\]
r = radius of smaller semicircle \[=\frac{35}{2}\,m\]
Now area of lawn
\[=42\times 35+\frac{22}{7}\times 21\times 21+\frac{22}{7}\times \frac{35}{2}\times \frac{35}{2}\]
\[=1470+1386++962.5=3818.5\,{{m}^{2}}\]
Diametre of a wheel of a bus is 770 cm. How much distance will it cover in 100 revolutions?
(a) 2.85km
(b) 5.85km
(c) 9.85km
(d) 2.42km
(e) None of these
Answer: (d)
The length of the minute hand of a clock is 10.5 cm long. Find the area swept by the minute hand in 10 minutes.
(a) \[57.75c{{m}^{2}}\]
(b) \[95.75c{{m}^{2}}\]
(c) \[~85.78c{{m}^{2~}}\]
(d) \[99.99c{{m}^{2}}\]
(e) None of these
Answer: (a)
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