Mathematics solution NCERT

Class 9 - Chapter 6: Measuring Space: Perimeter and Area

NCERTChapter 6Solution- Exercise Set 6.1

Question 1.

The perimeter of a circle is 44 cm. What is its radius?

Solution:

Given,

Circumference of the circle,

C=44\text{ cm}

We know that

C=2\pi r

Substitute the given values.

\begin{aligned} 44 &=2\times\frac{22}{7}\times r \end{aligned} \begin{aligned} 44 &=\frac{44r}{7} \end{aligned}

Multiply both sides by 7.

\begin{aligned} 44\times7 &=44r \end{aligned}

Divide both sides by 44.

\begin{aligned} r &=\frac{44\times7}{44}\\[4pt] &=7\text{ cm} \end{aligned}

Answer:

{r=7\text{ cm}}

Question 2.

Calculate, correct to 3 significant figures, the circumference of a circle with:

(i) Radius 7 cm

Solution:

Given,

r=7\text{ cm}

Using the formula,

C=2\pi r

Substitute the values.

\begin{aligned} C &=2\times\frac{22}{7}\times7\\[4pt] &=44\text{ cm} \end{aligned}

Correct to 3 significant figures,

{C=44.0\text{ cm}}

(ii) Radius 10 cm

Solution:

Given,

r=10\text{ cm}

Using the formula,

C=2\pi r

Substitute the values.

\begin{aligned} C &=2\times\frac{22}{7}\times10\\[4pt] &=\frac{440}{7}\\[4pt] &=62.857142\ldots\text{ cm} \end{aligned}

Correct to 3 significant figures,

{C=62.9\text{ cm}}

(iii) Radius 12 cm

Solution:

Given,

r=12\text{ cm}

Using the formula,

C=2\pi r

Substitute the values.

\begin{aligned} C &=2\times\frac{22}{7}\times12\\[4pt] &=\frac{528}{7}\\[4pt] &=75.428571\ldots\text{ cm} \end{aligned}

Correct to 3 significant figures,

{C=75.4\text{ 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^\circ.

Solution:

Given,

r=3.5\text{ cm}, \qquad \theta=60^\circ

We know that

\text{Arc Length}=\frac{\theta}{360^\circ}\times2\pi r

Substitute the given values.

\begin{aligned} l &=\frac{60}{360}\times2\times\frac{22}{7}\times3.5\\[4pt] &=\frac16\times\frac{44}{7}\times\frac72\\[4pt] &=\frac16\times22\\[4pt] &=\frac{11}{3}\text{ cm}\\[4pt] &\approx3.67\text{ cm} \end{aligned}

Answer:

{l=\frac{11}{3}\text{ cm}\approx3.67\text{ cm}}

(ii) the radius is 6.3 m and the angle at the centre is 120^\circ.

Solution:

Given,

r=6.3\text{ m}, \qquad \theta=120^\circ

Using the formula,

\text{Arc Length}=\frac{\theta}{360^\circ}\times2\pi r

Substitute the values.

\begin{aligned} l &=\frac{120}{360}\times2\times\frac{22}{7}\times6.3\\[4pt] &=\frac13\times\frac{44}{7}\times6.3\\[4pt] &=\frac13\times39.6\\[4pt] &=13.2\text{ m} \end{aligned}

Answer:

{l=13.2\text{ m}}

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^\circ.

Solution:

Given,

r=14\text{ cm},\qquad \theta=75^\circ

The perimeter of a sector is given by

\text{Perimeter}=\text{Arc Length}+2r

First, find the arc length.

\begin{aligned} l &=\frac{75}{360}\times2\times\frac{22}{7}\times14\\[4pt] &=\frac{5}{24}\times88\\[4pt] &=\frac{55}{3}\text{ cm}\\[4pt] &\approx18.33\text{ cm} \end{aligned}

Now, calculate the perimeter.

\begin{aligned} \text{Perimeter} &=l+2r\\[4pt] &=\frac{55}{3}+2(14)\\[4pt] &=\frac{55}{3}+28\\[4pt] &=\frac{139}{3}\text{ cm}\\[4pt] &\approx46.33\text{ cm} \end{aligned}

Answer:

{\text{Perimeter}=\frac{139}{3}\text{ cm}\approx46.33\text{ cm}}

Question 5.

Find the perimeters of the following shapes.

(i)

Solution:

Given,

Length of the straight portion =80 m
Diameter of each semicircle =60 m

Therefore,

r=\frac{60}{2}=30\text{ m}

The figure consists of

Two straight sides of length 80 m each.
Two semicircles, which together form one complete circle.

Length of the curved portion

\begin{aligned} &=2\pi r\\ &=2\times\frac{22}{7}\times30\\ &=\frac{1320}{7}\text{ m} \end{aligned}

Total perimeter

\begin{aligned} P &=80+80+\frac{1320}{7}\\[2mm] &=160+\frac{1320}{7}\\[2mm] &=\frac{2440}{7}\\[2mm] &\approx348.57\text{ m} \end{aligned}

Answer:

{P\approx348.57\text{ m}}

(ii)

Solution:

Outer diameter

12\text{ cm}

Inner diameter

8\text{ cm}

Outer radius

R=\frac{12}{2}=6\text{ cm}

Inner radius

r=\frac{8}{2}=4\text{ cm}

The perimeter consists of

Outer semicircle
Inner semicircle
Two straight portions

Outer semicircular arc

\begin{aligned} &=\pi R\\ &=\frac{22}{7}\times6\\ &=\frac{132}{7}\text{ cm} \end{aligned}

Inner semicircular arc

\begin{aligned} &=\pi r\\ &=\frac{22}{7}\times4\\ &=\frac{88}{7}\text{ cm} \end{aligned}

Length of one straight portion

\frac{12-8}{2}=2\text{ cm}

Total length of both straight portions

2+2=4\text{ cm}

Total perimeter

\begin{aligned} P &=\frac{132}{7}+\frac{88}{7}+4\\[2mm] &=\frac{220}{7}+4\\[2mm] &=\frac{248}{7}\\[2mm] &\approx35.43\text{ cm} \end{aligned}

Answer:

{P\approx35.43\text{ cm}}

(iii)

Solution:

Each curved part is a semicircle.

Diameter of each semicircle

10\text{ cm}

Radius

r=\frac{10}{2}=5\text{ cm}

The figure consists of

Four semicircles.

Four semicircles make

2\text{ complete circles.}

Hence,

\begin{aligned} P &=2\times2\pi r\\ &=4\pi r\\ &=4\times\frac{22}{7}\times5\\[2mm] &=\frac{440}{7}\\[2mm] &\approx62.86\text{ cm} \end{aligned}

Answer:

{P\approx62.86\text{ cm}}

(iv)

Solution:

Each side of the equilateral triangle is

12\text{ cm}

Each side is the diameter of a semicircle.

Therefore,

r=\frac{12}{2}=6\text{ cm}

The figure consists of three semicircular arcs.

Length of one semicircular arc

\begin{aligned} &=\pi r\\ &=\frac{22}{7}\times6\\ &=\frac{132}{7}\text{ cm} \end{aligned}

Total perimeter

\begin{aligned} P &=3\times\frac{132}{7}\\[2mm] &=\frac{396}{7}\\[2mm] &\approx56.57\text{ cm} \end{aligned}

Answer:

{P\approx56.57\text{ cm}}

(v)

Solution:

The diameter of each semicircle is

14\text{ cm}

Hence,

r=\frac{14}{2}=7\text{ cm}

The figure consists of four semicircular arcs.

Four semicircles together form

2\text{ complete circles.}

Therefore,

\begin{aligned} P &=2\times2\pi r\\ &=4\pi r\\ &=4\times\frac{22}{7}\times7\\[2mm] &=88\text{ cm} \end{aligned}

Answer:

{P=88\text{ cm}}

(vi)

Solution:

The total diameter of the large semicircle is

28\text{ cm}

Hence,

R=\frac{28}{2}=14\text{ cm}

The base is divided into four equal parts.

Diameter of each small semicircle

\frac{28}{4}=7\text{ cm}

Therefore,

r=\frac{7}{2}=3.5\text{ cm}

The perimeter consists of

One large semicircular arc.
Four small semicircular arcs.

Length of the large semicircular arc

\begin{aligned} &=\pi R\\ &=\frac{22}{7}\times14\\ &=44\text{ cm} \end{aligned}

Length of one small semicircular arc

\begin{aligned} &=\pi r\\ &=\frac{22}{7}\times3.5\\ &=11\text{ cm} \end{aligned}

Length of four small semicircular arcs

4\times11=44\text{ cm}

Total perimeter

\begin{aligned} P &=44+44\\ &=88\text{ cm} \end{aligned}

Answer:

{P=88\text{ cm}}

(vii)

Solution:

The figure consists of:

One semicircle of diameter 8 cm.
One semicircle of diameter 6 cm.
One semicircle whose diameter is the diagonal of the right triangle.

Step 1: Find the diameter of the third semicircle.

Using Pythagoras' Theorem,

\begin{aligned} d &=\sqrt{8^2+6^2}\\ &=\sqrt{64+36}\\ &=\sqrt{100}\\ &=10\text{ cm} \end{aligned}

Step 2: Find the radii.

\begin{aligned} r_1&=\frac{8}{2}=4\text{ cm},\\[2mm] r_2&=\frac{6}{2}=3\text{ cm},\\[2mm] r_3&=\frac{10}{2}=5\text{ cm}. \end{aligned}

Step 3: Find the perimeter.

The perimeter consists of three semicircular arcs.

\begin{aligned} P &=\pi(r_1+r_2+r_3)\\[2mm] &=\frac{22}{7}(4+3+5)\\[2mm] &=\frac{22}{7}\times12\\[2mm] &=\frac{264}{7}\\[2mm] &\approx37.71\text{ cm} \end{aligned}

Answer:

{P\approx37.71\text{ cm}}

(viii)

Solution:

The diameter of the large semicircle is

4+4+4=12\text{ cm}

Hence,

R=\frac{12}{2}=6\text{ cm}

Each small semicircle has diameter

4\text{ cm}

Therefore,

r=\frac{4}{2}=2\text{ cm}

The perimeter consists of

One large semicircular arc.
Three small semicircular arcs.

Length of the large semicircular arc

\begin{aligned} &=\pi R\\ &=\frac{22}{7}\times6\\ &=\frac{132}{7}\text{ cm} \end{aligned}

Length of one small semicircular arc

\begin{aligned} &=\pi r\\ &=\frac{22}{7}\times2\\ &=\frac{44}{7}\text{ cm} \end{aligned}

Length of three small semicircular arcs

3\times\frac{44}{7} = \frac{132}{7}\text{ cm}

Total perimeter

\begin{aligned} P &=\frac{132}{7}+\frac{132}{7}\\[2mm] &=\frac{264}{7}\\[2mm] &\approx37.71\text{ cm} \end{aligned}

Answer:

{P\approx37.71\text{ cm}}

(ix)

Solution:

The figure consists of

One large semicircle of diameter 20 cm.
One small semicircle of diameter 10 cm.

Step 1: Find the radii.

\begin{aligned} R&=\frac{20}{2}=10\text{ cm},\\[2mm] r&=\frac{10}{2}=5\text{ cm}. \end{aligned}

Step 2: Find the lengths of the arcs.

Large semicircular arc

\begin{aligned} &=\pi R\\ &=\frac{22}{7}\times10\\ &=\frac{220}{7}\text{ cm} \end{aligned}

Small semicircular arc

\begin{aligned} &=\pi r\\ &=\frac{22}{7}\times5\\ &=\frac{110}{7}\text{ cm} \end{aligned}

Step 3: Total perimeter.

\begin{aligned} P &=\frac{220}{7}+\frac{110}{7}\\[2mm] &=\frac{330}{7}\\[2mm] &\approx47.14\text{ cm} \end{aligned}

Answer:

{P\approx47.14\text{ 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?

Solution:

Given,

\text{Diameter of the tyre }(d)=56\text{ cm}

We know that one complete revolution of the tyre covers a distance equal to its circumference.

\text{Circumference}=\pi d

Using

\pi=\frac{22}{7},

we get

\begin{aligned} \text{Circumference} &=\frac{22}{7}\times56\\[4pt] &=22\times8\\[4pt] &=176\text{ cm} \end{aligned}

(i) Distance travelled in one revolution

Therefore, the tyre travels

{176\text{ cm}}

in one complete revolution.

(ii) Number of revolutions in 10 km

First convert 10 km into centimetres.

\begin{aligned} 10\text{ km} &=10\times1000\text{ m}\\[4pt] &=10000\text{ m}\\[4pt] &=10000\times100\text{ cm}\\[4pt] &=1000000\text{ cm} \end{aligned}

Number of revolutions

= \frac{\text{Total distance}}{\text{Distance in one revolution}} \begin{aligned} \text{Number of revolutions} &=\frac{1000000}{176}\\[4pt] &=\frac{62500}{11}\\[4pt] &\approx5681.82 \end{aligned}

Hence, the tyre makes approximately

{5682\text{ revolutions}}

while travelling 10 km.

Answers:

(i)

{176\text{ cm}}

(ii)

{5682\text{ revolutions (approximately)}}

Question 7 (i).

Find the total perimeter of all the petals in the given flower.

Solution:

The given figure is a square of side

14\text{ cm}.

The centres of the arcs are the midpoints of the sides of the square.

Each petal is formed by two quarter-circular arcs.

Step 1: Find the radius of each arc.

The radius is equal to half the side of the square.

\begin{aligned} r &=\frac{14}{2}\\[2mm] &=7\text{ cm} \end{aligned}

Step 2: Find the length of one quarter-circle arc.

We know that

\text{Arc Length} = \frac{\theta}{360^\circ}\times2\pi r

Here,

\theta=90^\circ.

Therefore,

\begin{aligned} l &=\frac{90}{360}\times2\times\frac{22}{7}\times7\\[2mm] &=\frac14\times44\\[2mm] &=11\text{ cm} \end{aligned}

Step 3: Find the perimeter of one petal.

Each petal consists of two quarter-circle arcs.

\begin{aligned} \text{Perimeter of one petal} &=2\times11\\[2mm] &=22\text{ cm} \end{aligned}

Step 4: Find the total perimeter of all petals.

There are four petals in the figure.

\begin{aligned} \text{Total perimeter} &=4\times22\\[2mm] &=88\text{ cm} \end{aligned}

Answer:

{88\text{ cm}}

Question 7 (ii).

Find the total perimeter of all the petals in the given flower.

Solution:

The given figure is a regular hexagon of side

42\text{ cm}.

The centres of the arcs are the vertices of the hexagon.

Each petal is formed by two arcs of a circle.

Step 1: Find the radius of each arc.

The radius of each arc is equal to the side of the hexagon.

r=42\text{ cm}

Step 2: Find the angle subtended by each arc.

Each interior angle of a regular hexagon is

120^\circ.

Hence, each arc is a

120^\circ

arc of a circle.

Step 3: Find the length of one arc.

Using the formula,

\text{Arc Length} = \frac{\theta}{360^\circ}\times2\pi r

Substitute the values.

\begin{aligned} l &=\frac{120}{360}\times2\times\frac{22}{7}\times42\\[4pt] &=\frac13\times2\times22\times6\\[4pt] &=88\text{ cm} \end{aligned}

Step 4: Find the perimeter of one petal.

Each petal consists of two such arcs.

\begin{aligned} \text{Perimeter of one petal} &=2\times88\\[4pt] &=176\text{ cm} \end{aligned}

Step 5: Find the total perimeter of all petals.

There are six petals in the figure.

\begin{aligned} \text{Total perimeter} &=6\times176\\[4pt] &=1056\text{ cm} \end{aligned}

Answer:

{1056\text{ cm}}

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 \quad \text{and} \quad r_2.

The circumference (perimeter) of a circle is given by

C=2\pi r.

According to the question,

\frac{C_1}{C_2}=\frac{5}{4}.

Substitute the formula for circumference.

\begin{aligned} \frac{2\pi r_1}{2\pi r_2} &=\frac{5}{4}. \end{aligned}

Cancel the common factor 2\pi from the numerator and denominator.

\begin{aligned} \frac{r_1}{r_2} &=\frac{5}{4}. \end{aligned}

Therefore, the ratio of the radii is

{5:4}

Answer:

{\text{Ratio of the radii}=5:4}