NCERT Solutions for Class 9 Ganita Manjari Chapter 6

NCERT Solutions for Class 9 Ganita Manjari Chapter 6 Measuring Space: Perimeter and Area समष्टि मापन- परिमाप और क्षेत्रफल

Exercise 6.1

Question 1:
The perimeter of a circle is 44 cm. What is its radius?
Solution
Perimeter of a circlel = 44 cm
\(\quad\) 2\(\pi\)r = 44
\(\quad\) 2\(\frac{22}{7}\)r = 44
\(\quad\quad\quad\) r = 44 \(\times \frac{7}{2 \times 22}\) = 7 cm

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Question 2:
Calculate, correct to 3 significant figures, the circumference of a circle with:
(i) radius 7 cm (ii) radius 10 cm (iii) radius 12 cm.
Soution
(i) Radius (r) = 7 cm
Circumference = $2\pi r$
= 2 × 3.142 × 7
= 43.988 
= 44.0 cm

(ii) Radius (r) = 10 cm
Circumference = $2\pi r$
= 2 × 3.142 × 10
= 62.84 
= 62.8 cm

(iii) Radius (r) = 12 cm
Circumference = $2\pi r$
= 2 × 3.142 × 12
= 75.408 
= 75.4 cm

Question 3:
Calculate the length of the arc of a circle if:
(i) the radius is 3.5 cm, and the angle at the centre is 60°, and
(ii) the radius is 6.3 m, and the angle at the centre is 120°.
Solution
(i) Radius (r) = 3.5 cm
Angle at the centre (θ) = 60°
Length of the arc = $\frac{\theta }{360°}$ × $2\pi r$
\(\quad\quad\quad\) = $\frac{60°}{360°}$ × 2 × $\frac{22}{7}$ × 3.5
\(\quad\quad\quad\) = $\frac{1}{6}$ × 2 × $\frac{22}{7}$ × $\frac{35}{10}$
\(\quad\quad\quad\) = $\frac{11}{3}$ 
\(\quad\quad\quad\) = 3.67 cm

(ii) Radius (r) = 6.3 m
Angle at the centre (θ) = 120°
Length of the arc = $\frac{\theta }{360°}$ × $2\pi r$
\(\quad\quad\quad\) = $\frac{120°}{360°}$ × 2 × $\frac{22}{7}$ × 6.3
\(\quad\quad\quad\) = $\frac{1}{3}$ × 2 × $\frac{22}{7}$ × $\frac{63}{10}$
\(\quad\quad\quad\) = $\frac{66}{5}$ 
\(\quad\quad\quad\) = 13.2 cm

Question 4:
Find the perimeter of a sector (i.e., the curved portion as well as the two straight portions) of a circle of radius 14 cm and sector angle 75°.
Solution
Radius of the circle (r) = 14 cm
Angle of the sector (θ) = 75°
Perimeter of a sector = length of arc + 2r
\(\quad\quad\quad\quad\quad\) = $\frac{\theta }{360°}$ × $2\pi r$ + 2r
\(\quad\quad\quad\quad\quad\) = 2r $(\frac{\theta }{360°}\times \pi +1)$
\(\quad\quad\quad\quad\quad\) = 2 × 14 $(\frac{75°}{360°}\times \frac{22}{7}+1)$
\(\quad\quad\quad\quad\quad\) = 28 $(\frac{15}{72}\times \frac{22}{7}+1)$
\(\quad\quad\quad\quad\quad\) = 28 $(\frac{5}{24}\times \frac{22}{7}+1)$
\(\quad\quad\quad\quad\quad\) = 28 $(\frac{55}{84}+1)$
\(\quad\quad\quad\quad\quad\) = 28 $\left(\frac{55+84}{84}\right)$
\(\quad\quad\quad\quad\quad\) = 28 $\left(\frac{139}{84}\right)$
\(\quad\quad\quad\quad\quad\) = $\frac{139}{3}$ cm = 46$\frac{1}{3}$ cm
Therefore, the perimeter of the sector is 46$\frac{1}{3}$ cm.

Quesiton 5:
Find the perimeters of the following shapes (taking the arcs to be quarter or half or three-quarters of a circle, as appropriate) (Fig. 6.14i to 6.14ix):

Solution
(i) Radii of semicircles, r = 30 cm
Length of lines = 80 + 80 = 160 cm
Perimeter = Length of semicircles + Length of lines
\(\quad\quad\quad\) = $2\pi r$ + 160
\(\quad\quad\quad\) = 2 × $\frac{22}{7}$ × 30 + 160
\(\quad\quad\quad\) = $\frac{1320}{7}$ +160
\(\quad\quad\quad\) = $\frac{1120+1320}{7}$ = $\frac{2440}{7}$ = 348$\frac{4}{7}$ m

(ii) Radius of outer semicircle, r = 6 cm
Radius of inner semicircle = 4 cm
Length of lines = $\frac{12\text{–}8}{2}$ = $\frac{4}{2}$ = 2 + 2 = 4 cm
Perimeter = Length of semicircle + Length of semicircle + Length of lines
\(\quad\quad\quad\) = $\pi \times 6$ + $\pi$ × 4 + 4
\(\quad\quad\quad\) = 6$\pi$ + 4$\pi$ + 4
\(\quad\quad\quad\) = 10$\pi$ + 4
\(\quad\quad\quad\) = 10 × $\frac{22}{7}$ + 4
\(\quad\quad\quad\) = $\frac{220}{7}$ + 4
\(\quad\quad\quad\) = $\frac{220+28}{7}$ = $\frac{248}{7}$ = 35$\frac{3}{7}$ cm

Question 6:
If the diameter of a car tyre is 56 cm, then:
(i) How far does the car need to travel for the tyre to complete one revolution?
(ii) How many revolutions does the tyre make if the car travels 10 km?
Soution
Diameter of the tyre = 56 cm
Radius, r = \(\frac{56}{2}\) = 28 cm
(i) Distance covered in one revolution
\(\quad\quad\quad) = Circumference of the tyre = πd
\(\quad\quad\quad) = $\frac{22}{7}$ × 56
\(\quad\quad\quad) = 176 cm
Therefore, the car needs to travel 176 cm to complete one revolution.

(ii) Distance travelled by the car = 10 km
= 10 × 1000 × 100
= 1000000 cm
Number of revolutions made by the tyre = $\frac{Totaldistance}{Circumferenceofthetyre}$ = $\frac{1000000}{176}$ = 5681.81
Therefore, the tyre made 5681 revolutions to cover 10 km.

Question 7:
Find the total perimeter of all the petals in each of the given flowers.

Question 8:
The ratio of the perimeters of two circles is 5:4. What is the ratio of their radii?
Solution
Let the radii of the two circles be r_1​ and r₂​.
The ratio of the perimeters of the two circles = 5 : 4
Thus,
$\frac{2\pi r1}{2\pi r2}$ = $\frac{5}{4}$
$\frac{r1}{r2}$ = $\frac{5}{4}$
Hence, the ratio of their radii is 5 : 4.