S.chand books class 8 maths solution chapter 26 Circumference and Area of a Circle exercise 26

 EXERCISE 26


Q1 | Ex-26 | Class 8 | S.Chand | Composite maths | Circumference and Area of a Circle |myhelper

Question 1

Find the circumference and area of each of the following circles given:
(i) radius=14cm
(ii) radius=4.2cm
(iii) diameter=56cm
(iv) diameter=16.1cm

Sol :













Q2 | Ex-26 | Class 8 | S.Chand | Composite maths | Circumference and Area of a Circle |myhelper


Question 2

Find the diameter of the circle whose
(i) Circumference is 308cm
(ii) Area=962.5cm2

Sol :














Q3 | Ex-26 | Class 8 | S.Chand | Composite maths | Circumference and Area of a Circle |myhelper

Question 3

Two wire circles of diameters 28cm and 35cm are cut and then joined to form a big circle of radius k cm. Find the value of k cm .Find the value of k .

Sol :












Q4 | Ex-26 | Class 8 | S.Chand | Composite maths | Circumference and Area of a Circle |myhelper


Question 4

The circumference of circular pond is 396m .What is its area ?

Sol :













Q5 | Ex-26 | Class 8 | S.Chand | Composite maths | Circumference and Area of a Circle |myhelper

Question 5

The diameter of a wheel is 98cm . In how many revolutions will it cover a distance of 1540m ?

Sol :











Q6 | Ex-26 | Class 8 | S.Chand | Composite maths | Circumference and Area of a Circle |myhelper

Question 6

If the difference between the circumference and diameter of a circle is 30cm , then what is the radius of the circle ?

Sol :











Q7 | Ex-26 | Class 8 | S.Chand | Composite maths | Circumference and Area of a Circle |myhelper

Question 7

Find the perimeter and area of each of the following shapes represented by bold lines .




Sol :

Perimeter











AREA















Q8 | Ex-26 | Class 8 | S.Chand | Composite maths | Circumference and Area of a Circle |myhelper


Question 8

A gear 12cm in diameter is turning a gear 16cm in diameter.When the smaller gear has 42 revolutions. How many revolutions has the larger one made ?

Sol :

Given, a gear 12cm in diameter is turning a gear 18cm in diameter
The smaller gear has 42 revolutions.
We need to find how many revolutions has the larger one made.
Distance covered by smaller gear is calculated by multiplying circumference and number of revolutions
Now, Distance covered by small gear

=2πr×42

$=2\pi \times \frac{12}{2}\times 42$

=504π


Now, number of revolution by larger gear

$=\frac{504 \pi}{2\pi \times 9}$

=28

Hence, the larger gear made 28 revolutions.



Q9 | Ex-26 | Class 8 | S.Chand | Composite maths | Circumference and Area of a Circle |myhelper

 

Question 9

A circular road runs round a circular ground. If the differences between the circumferences of the outer circle and inner circle is 66 meter. Find the width of the road.

Sol :

Circumferences of outer circle-Circumferences of inner circle=66

2πR-2πr=66

2π(R-r)=66

$(R-r)=\frac{66}{2}\times \frac{7}{22}$

$(R-r)=\frac{33\times 7}{22}$

(R-r)=10.5m

Width of the road=10.5m




Q10 | Ex-26 | Class 8 | S.Chand | Composite maths | Circumference and Area of a Circle |myhelper


Question 10

A circular water fountain 6.6m in diameter is surrounded by a path of width 1.5m .What is the area of the path ? (Write the answer correct up to 2 places of decimal)

Sol :

Area of circular fountain=$\frac{22}{7} \times \left(\frac{6.6}{2}\right)^2$

$=\frac{22}{7}\times (3.3)^2$

$=\frac{22}{7}\times 10.89$

$=\frac{239.58}{7}$

=34.22m2


Area of outer circle (fountain +width)$=\frac{22}{7}\times \left(\frac{6.6}{2}+1.5\right)^2$

$=\frac{22}{7}\times \left(3.3+1.5\right)^2$

$=\frac{22}{7}\times (4.8)^2$

$=\frac{22\times 23.04}{7}$

$=\frac{506.88}{7}$

=72.41m2


Area of path=(72.41-34.22)m2

=38.19m2




Q11 | Ex-26 | Class 8 | S.Chand | Composite maths | Circumference and Area of a Circle |myhelper


Question 11

The area enclosed between the circumference of two concentric circles is 16π cm2 and their radii are in the ratio 5 : 3 . What is the area of the outer circle ?

Sol :

Area of outer circle-Area of inner circle=16π

π(5x)2-π(3x)2=16π

π[(5x)2-(3x)2]=16π

25x2-9x2=16

x2(25-9)=16

$x^2=\frac{16}{16}$

x=1


Area of outer circle=π(5×1)2

=π(5)2

=25π

$=\frac{22\times 25}{7}$

$=\frac{550}{7}$

=78.57cm2




Q12 | Ex-26 | Class 8 | S.Chand | Composite maths | Circumference and Area of a Circle |myhelper


Question 12

What is the area of the largest circle that can be drawn inside a square of side 28cm ?

Sol :

Area of largest circle inside square=πr2

$=\frac{22}{7}\times \left(\frac{28}{2}\right)^2$

$=\frac{22}{7}\times (14)^2$

=22×2×14

=616cm2




Q13 | Ex-26 | Class 8 | S.Chand | Composite maths | Circumference and Area of a Circle |myhelper


Question 13

Calculate the shaded area in each of the following figures






Sol :

(i)

 Total area= area of triangle + area of 1/4 of circle

=1/2 b×h+1/4 πr2

$=\frac{1}{2}\times 7\times 7+\frac{1}{4}\times \frac{22}{7}\times \left(\frac{14}{2}\right)^2$

$=\frac{49}{2}+\frac{1}{2}\times \frac{11}{7}\times \left(7\right)^2$

$=\frac{49}{2}+ \frac{539}{14}$

$=\frac{49\times 7+539\times 1}{14}$

$=\frac{49\times 7+539\times 1}{14}$

$=\frac{343+539}{14}=\frac{882}{14}$

=63cm2


(ii) 

Diameter of semi circle=Hypotenuse of triangle

H2=B2+A2

H2=122+162

H2=144+256

H2=400

H=20=D

$r=\frac{D}{2}=\frac{20}{2}=10$


Total Area=Area of semi circle-Area of triangle

$=\frac{1}{2}\pi r^2-\frac{1}{2}\times 12\times 16$

$=\frac{1}{2}\times \frac{22}{7}\times  (10)^2-\frac{1}{2}\times 12\times 16$

$= \frac{11}{7}\times 100- 6\times 16$

$= \frac{1100}{7}-96=\frac{1100-96\times 7}{7}$

$=\frac{1100-96\times 7}{7}$

$=\frac{1100-672}{7}=\frac{428}{7}$

$=61\frac{1}{7}$cm2


(iii)

Total area =Area of triangle(at top)+Area of rectangle(middle)+Area of semi circle(bottom)

 Area of triangle=1/2 base×height

$=\frac{1}{2}\times 7\times (15-10)$

$=\frac{7\times 5}{2}$cm2

$=\frac{35}{2}$cm2

=17.5cm2


Area of rectangle=length×breadth

=7×10

=70cm2


Area of semi circle=1/2 πr2

$=\frac{1}{2}\times \frac{22}{7}\times \left(\frac{7}{2}\right)^2$

$=\frac{1}{2}\times \frac{22}{7}\times \frac{7}{2}\times \frac{7}{2}$

$=\frac{1}{2}\times 11\times \frac{7}{2}=\frac{77}{4}$

=19.25cm2


Total area=17.5cm2+70cm2+19.25cm2

$=106\frac{3}{4}$cm2


(iv)

7=OA=OC=radius of bigger circle

Area of bigger circle=1/2 πR2

$=\frac{1}{2}\times \frac{22}{7}\times 7^2$

=11×7

=77cm2

Radius for smaller circle $=\frac{7}{2}$cm

Area of smaller circle=πr2

$=\frac{22}{7}\times \left(\frac{7}{2}\right)^2$

$=\frac{22}{7}\times \frac{7}{2}\times \frac{7}{2}$

$=11\times \frac{7}{2}=\frac{77}{2}$cm2

=38.5cm2


Area of shaded region=Area of bigger circle-Area of smaller circle

=(77-38.5)cm2

=38.5cm2




Q14 | Ex-26 | Class 8 | S.Chand | Composite maths | Circumference and Area of a Circle |myhelper

Question 14

Four equal sized maximum circular plates are cut off from a square paper sheet of area 784cm2. What is the circumference of each plate ?

Sol :

Area of square sheet=784cm2

a2=784

a=28cm

Diameter of single circular plate$=\frac{1}{2}28$

2r=14cm

$r=\frac{14}{2}$cm

r=7cm

Circumference of single circular plate=2πr

$=2\times \frac{22}{7}\times 7$

=2×22

=44cm



Q15 | Ex-26 | Class 8 | S.Chand | Composite maths | Circumference and Area of a Circle |myhelper

Question 15

A circular path is enclosed by two concentric circles. If the outer circumference of the path is 88cm and the uniform width of the path is 3.5cm, find the area of the path.

Sol :

2πR=88

$R=\frac{88}{2}\times \frac{7}{22}$

$=44\times \frac{7}{22}=4\times \frac{7}{2}$

R=2×7cm

R=14cm

Area of bigger circle =πR2

$=\frac{22}{7}\times (14)^2$

=22×2×14cm2

=616cm2


Radius of inner circle=14-3.5

=10.5cm

Area of inner circle (small circle)=πr2

$=\frac{22}{7}\times (10.5)^2$

=22×1.5×10.5

=346.5cm2


Area of path=Area of bigger circle-Area of inner circle

=616-346.5

=269.5cm2

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