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CBSE Questions for Class 7 Maths Perimeter And Area Quiz 1 - MCQExams.com
CBSE
Class 7 Maths
Perimeter And Area
Quiz 1
The area of a circle is the measurement of the region enclosed by its
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Radius
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Centre
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Circumference
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Area
Explanation
As shown in the above figure, area is the region enclosed inside the boundary/circumference of the circle.
Circumference of a circle is .....................
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$$\pi r$$
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$$2\pi r$$
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$$3\pi r$$
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$$4\pi r$$
Explanation
Circumference of a circle whose radius is $$r$$ is given by $$2\pi r.$$
Calculate the radius of a circle whose circumference is $$44\ cm$$?
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$$7\ cm$$.
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$$14\ cm$$.
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$$21\ cm$$.
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None of these
Explanation
Given: Circumference of circle $$=44\space\mathrm{cm}$$
$$\Rightarrow 2\pi r=44$$
$$\Rightarrow 2\times\dfrac{22}{7}\times r=44$$
$$\Rightarrow r=44\times\dfrac{1}{2}\times\cfrac{7}{22}$$
$$\Rightarrow r=7\space\mathrm{cm}$$
$$\therefore$$ Radius of circle $$r=7\space\mathrm{cm}$$
Hence, $$\text{A}$$ is the correct option.
Consider the railway platform which is in square shape having side length $$2\ km$$. Then area of the platform is $$4$$ ____.
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$$m$$
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$$km$$
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$$km^{2}$$
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$$m^{2}$$
Explanation
Area of square platform$$=2\times 2=4 km^2$$
Circumference of a circle is always
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more than three times of its diameter
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three times of its diameter
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less than three times of its diameter
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three times of its radius.
Explanation
Option (a) is correct.
As we know [bat Circumference of a circle $$ 2 \times 3.14 \times r$$
$$\Rightarrow circumference = 3.14 \times d$$
Therefore, Circumference of a circle is more than three of its diameter.
State true or false.
The area of a circle whose radius is $$2.1$$ m is $$13.86\,\, m^2$$
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True
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False
Explanation
Area of a circle with radius $$r$$ is $$ \pi {r}^{2} $$
Area of circle of radius $$ 2.1 \text{ m}$$
$$ = \pi { r }^{ 2 } $$
$$= \dfrac { 22 }{ 7 } \times 2.1\times 2.1 $$
$$=13.86\text{ m}^{ 2 } $$
Area of a circle is ...................
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$$\pi r^2$$
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$$2\pi r$$
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$$2\pi r^2$$
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$$4\pi r$$
Explanation
Area covered within the boundary of circle = $$\pi \times r^2$$, where $$r$$ is a radius of a circle.
What is the area of the circular ring included between two concentric circles of radius $$14$$ cm and $$10.5$$ cm ?
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$$255\ cm^2$$
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$$148\ cm^2$$
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$$324\ cm^2$$
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$$269\ cm^2$$
Explanation
Let $$R$$ be the radius of the larger circle
Hence $$R=14\ cm$$
Let $$r$$ be the radius of the smaller circle
Hence $$r=10.5\ cm$$
Area of the circular ring $$=$$ Area of the larger circle - Area of the smaller circle
$$=\pi R^2 - \pi r^2$$
$$=\dfrac { 22 }{ 7 } \times \left( { R }^{ 2 } - { r }^{ 2 } \right)$$
$$= \dfrac { 22 }{ 7 } \times \left( { 14 }^{ 2 } - { 10.5 }^{ 2 } \right)$$
$$=269.5 \ { cm }^{ 2 }$$
$$\approx 269\ { cm }^{ 2 } $$
In Fig. 9.4, the area of parallelogram ABCD is :
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AD x BM
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BC x BN
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DC x DL
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AD x DL
Explanation
$$ Given-\\ ABCD\quad is\quad a\quad parallelogram.\quad \\ DL\bot AB\quad \& \quad BM\bot AD.\quad \\ To\quad find\quad out-\quad arABCD.\\ Solution-\\ DL\bot AB\quad \therefore \quad DL\quad is\quad the\quad height\quad of\quad ABCD\quad when\quad its\quad base\quad is\quad AB.\\ \therefore \quad arABCD=DL\times AB..\\ Similarly\quad BM\bot AD.\\ \therefore \quad BM\quad is\quad the\quad height\quad of\quad ABCD\quad when\quad its\quad base\quad is\quad AD.\\ \therefore \quad arABCD=BM\times AD\\ Ans-\quad Option\quad A\quad \& \quad C $$
One side of a parallelogram is $$8$$ cm. If the corresponding altitude is $$6$$ cm, then its area is given by
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$$24\: cm^2$$
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$$36\: cm^2$$
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$$40\: cm^2$$
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$$48\: cm^2$$
Explanation
Area of parallelogram $$=$$ length $$\times$$ height
Here altitude is nothing but the height
$$\therefore $$ Area of parallelogram $$= 8 \times 6$$
$$= 48\: cm^2$$
A path of width 8 m runs around a circular park whose radius is $$38\ m$$. Find the area of the path.
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$$2112\ m^2$$
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$$2834\ m^2$$
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$$3212\ m^2$$
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$$1578\ m^2$$
Explanation
Inner radius (Radius of the park) $$=38\ m$$
Outer radius (Radius of the park $$+$$ path width of the park) $$=38+8=46\ m$$
Area of the path $$=\pi(46^2-38^2)$$
$$=\cfrac{22}{7}(2116-1444)$$
$$=\cfrac{22}{7}(672)$$
$$=2112\ m^2$$
In the given figure, the area enclosed between two concentric circles is $$808.5\ cm^2$$. The circumference of the outer circle is $$242\ cm$$. Find t
he width of the ring.
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$$1.2\ cm$$
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$$2.4\ cm$$
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$$3.5\ cm $$
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$$4.2\ cm$$
Explanation
Let radius of the inner circle be $$r$$ and radius of the outer circle be $$R$$
Area enclosed between the two concentric circles $$=\pi(R^2-r^2)$$
Given $$2\pi R=242$$
$$\Rightarrow R=\cfrac{242 \times 7}{22\times 2}$$
$$\Rightarrow R=\cfrac{77}{2}$$
Now, $$\pi(R^2-r^2)=808.5$$
$$\Rightarrow {\left(\cfrac{77}{2}\right)}^2-r^2=\cfrac{808.5 \times 7}{22}$$
$$\Rightarrow \cfrac{5929}{4}-r^2=257.25$$
$$\Rightarrow r^2=1225$$
$$\Rightarrow r=35$$
Radius of the inner circle $$=35\ cm$$
$$\therefore$$ width of the ring $$=(R-r)$$
$$=(38.5-35)\ cm$$
$$=3.5\ cm$$
The circumference of a circular field is $$528\ m$$. Then
its area is
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22176 $$m^2$$
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26400 $$m^2$$
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10080 $$m^2$$
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84000 $$m^2$$
Explanation
$$ Given-\\ the\quad circumference=C\quad of\quad a\quad circular\quad field=528m.\\ To\quad find\quad out-\\ the\quad ar.circle=?\\ Solution-\\ Let\quad the\quad radius\quad of\quad the\quad circle\quad be\quad r.\\ Then\quad r=\cfrac { C }{ 2\pi } \quad when\quad C=circumference\quad of\quad the\quad circle.\\ \Longrightarrow r=\cfrac { 528 }{ 2\times \cfrac { 22 }{ 7 } } cm=84m.\\ \therefore \quad ar.circle=\pi { r }^{ 2 }=\cfrac { 22 }{ 7 } \times { 84 }^{ 2 }{ m }^{ 2 }=22176{ m }^{ 2 }.\\ Ans-\quad Option\quad A. $$
The radius of a circle is $$5\: m$$. Find the circumference of the circle whose area is $$49$$ times the area of the given circle.
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$$220 \: m$$
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$$120 \: m$$
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$$320 \: m$$
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$$420 \: m$$
Explanation
Radius of given circle $$=5\ m$$
Therefore,
Area of given circle $$=\pi R^2=\pi (5)^2$$
$$=25\pi\ m^2$$
Let radius of required circle $$ =r$$
Therefore, According to the given condition,
$$\pi r^2=49\times 25\pi$$
$$\Rightarrow r^2=(7\times 5)^2$$
$$\Rightarrow r=35m$$
Therefore,
Circumference $$=2\pi r$$
$$=2\times \dfrac { 22 }{ 7 }\times35$$
$$=220\ m$$
Find the area of a ring whose outer and inner radii are $$19\ cm$$ and $$16\ cm$$ respectively.
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$$330\ cm^2$$
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$$310\ cm^2$$
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$$320\ cm^2$$
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$$350\ cm^2$$
Explanation
Let us given $$R = 19 $$ $$cm $$ & $$r = 16$$ $$cm $$
Area of outer circle $$=\pi{(19)}^2$$
$$=\cfrac{22}{7}\times 361$$
Area of inner circle $$=\pi{(16)}^2$$
$$=\cfrac{22}{7}\times 256$$
$$\therefore$$ area of the ring $$=\dfrac{22}{7}\times 361-\dfrac{22}{7}\times 256$$
$$=\cfrac{22}{7}(361-256)$$
$$=\cfrac{22}{7}(105)$$
$$=330\ cm^2$$
The circumference of a circular field is $$308 m$$. Find its
radius in metres.
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$$49$$m
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$$59$$m
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$$91$$m
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$$94$$m
Explanation
The circumference of a circular field is $$308m$$
Therefore, $$2\pi r=308$$
$$\Rightarrow r=\cfrac{308\times 7}{2\times 22}$$
$$\Rightarrow r=49m$$
The total boundary length of circle is called
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area
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volume
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circumference
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diameter
Explanation
Boundary length of circle is called circumference
A circle is inscribed in a square as shown below. If the radius of the circle is $$4$$ cm, then the perimeter of the square is ______
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$$28$$ cm
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$$24$$ cm
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$$32$$ cm
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$$36$$ cm
Explanation
From the fig it is clear that,
Diameter of the circle $$=$$ side of the square
So length of side of square $$=2\times$$ $$4$$ cm
$$=8$$ cm
$$\therefore$$ perimeter of the square $$=8 + 8 + 8 + 8 $$
$$= 32$$ cm
Find the missing length of the following figure.
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$$6\text{ cm}$$
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$$4\text{ cm}$$
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$$2\text{ cm}$$
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$$5\text{ cm}$$
Explanation
Let unknown side of the figure is $$x$$
Perimeter $$=$$ sum of all sides of figure
$$23=4+5+8+x$$
$$x=23-17=6\text{ cm}$$.
Find the missing length of the figure given below.
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$$4\text{ cm}$$
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$$2\text{ cm}$$
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$$6\text{ cm}$$
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$$8\text{ cm}$$
Explanation
Let unknown side is $$x$$
Perimeter $$=$$ sum of length.
$$20=6+4+8+x$$
$$x=20-18=2\text{ cm}$$
Perimeter of a circle is called as__________
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Area
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Circumference
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Volume
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None of these
Explanation
Perimeter of a circle is equal to its circumference i.e., $$P=2\pi r$$
Perimeter of a circle is called its
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radius
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area
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diameter
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none of these
Explanation
Perimeter of a circle is called its circumference.
Perimeter of a circle is called as
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area
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circumference
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volume
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none of these
Explanation
Perimeter of a circle is called as circumference.
The formula of circumference is $$2\pi r$$
Option B is correct.
Perimeter of a triangle is the sum of the lengths of all the ............ sides.
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$$4$$
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$$2$$
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$$3$$
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$$6$$
Explanation
Perimeter of a triangle is the sum of the lengths of all the three sides.
The radius of a wheel is $$0.25 m$$. How many rounds will it take to complete the distance of $$11 km$$?
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$$7000$$
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$$8000$$
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$$9000$$
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$$6000$$
Explanation
Distance covered in one round $$=$$ Circumference of wheel
$$\Rightarrow 2\pi r= 2 \times \cfrac{22}{7} \times 0.25 =\cfrac{11}{7}m$$
$$ \displaystyle \therefore $$ Number of rounds to cover $$11 km =\cfrac{\text {Total distance covered}}{\text {Distance covered in one round}}=\cfrac{11\times\:1000}{\frac{11}{7}}=7000$$
What is the area of the given figure? $$ABCD$$ is a rectangle and $$BDE$$ is an isosceles right triangle.
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$$ab$$
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$$\displaystyle ab^{2}$$
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$$cab$$
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$$b\left(a + \dfrac b2\right)$$
If the radii of two concentric circles are $$15\ \text{cm}$$ and $$13\ \text{cm}$$, respectively, then the area of the circulating ring in sq. cm will be:
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$$176$$
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$$178$$
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$$180$$
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$$200$$
Explanation
$$R = 15\ \text{cm}, r = 13\ \text{cm}. $$
Area of the circulating ring $$= \pi (R^2-r^2)$$
$$= \cfrac {22}{7} ({15}^2-{13}^2)\ \text{cm}^2$$
$$= \cfrac {22}{7} \times 28 \times 2\ \text{cm}^2$$
$$=176\ \text{cm}^2$$
The diameter of each circle shown in the given figure is 'd' then the area of the square is given by
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$$9\displaystyle d^{2}$$
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$$9 d$$
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$$3\displaystyle d^{2}$$
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$$\displaystyle \frac{3}{4} d^{2}$$
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$$\displaystyle \frac{4}{3} d^{2}$$
Explanation
There are 3 circles of diameter 'd
So side of the square is 3d and
Hence area is $$side^2=\displaystyle 9d^{2}$$
The area of a circle is 301.84 $$\displaystyle cm^{2} $$ Then its radius is
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$$9.2$$ cm
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$$9.3$$ cm
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$$9.8$$ cm
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9.6
9.6
cm
Explanation
Area of the circle = $$\displaystyle \pi r^{2}$$
According to the question,
$$\displaystyle \pi r^{2}=301.84$$
$$\displaystyle \Rightarrow \frac{22}{7}r^{2}=301.84$$
$$\displaystyle r^{2}=\frac{7\times 301.84}{22}=96.04$$
$$\displaystyle \therefore r=\sqrt{96.04}=9.8\ cm$$
A circular park has a path of uniform width around it. The difference between the outer and inner circumferences of the circular park is $$132$$ m. Its width is
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$$20$$ m
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$$21$$ m
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$$22$$ m
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$$24$$ m
Explanation
Let $$r_1$$ and $$C_1$$ be the radius and the circumference of the inner circle.
Let $$r_2$$ and $$C_2$$ be the radius and the circumference of the outer circle.
The width of the circular path is $$(r_2-r_1)$$ m.
Given,
$$C_2-C_1=132$$ m
$$\Rightarrow 2\pi r_2 - 2\pi r_1=132$$
$$\Rightarrow 2\pi (r_2-r_1)=132$$
$$\Rightarrow r_2-r_1=\cfrac{66}{\dfrac{22}{7}}$$
$$\Rightarrow r_2-r_1=21$$
Thus, width of the circular path is $$21$$ m.
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