Perimeter and Area Test

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Perimeter and Area Test - 20

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Weekly Quiz Competition

Question 1
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The circumference of a circular field is $$528\ m$$. Then its radius is

A
$$84\ m$$

B
$$80\ m$$

C
$$76\ m$$

D
$$78\ m$$

Solution

Let $$r$$ be the radius of the circle
Circumference of the circle $$=528$$
$$\Rightarrow 2\pi r=528$$
$$\Rightarrow r=\dfrac{528\times 7}{22\times 2}$$
$$\Rightarrow r=84\ m$$
Question 2
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The following figure shows a square cardboard $$ABCD$$ of side $$28\ cm.$$ Four identical circles of the largest possible size are cut from this card as shown below.
Find the area of the remaining cardboard.

A
$$168$$ $$cm^{2}$$

B
$$158$$ $$cm^{2}$$

C
$$148$$ $$cm^{2}$$

D
$$138$$ $$cm^{2}$$

Solution

Let the radius of each circle be $$r.$$

Then,
$$4r=28\ cm$$
$$r=7\ cm$$

Therefore,
Area of $$4$$ circle $$=4\times \pi (7)^2$$
$$=4\times \dfrac{22}{7}\times 7\times 7$$
$$=616\ cm^2$$

Area of square $$=28^2$$
$$=784\ cm^2$$

Area of remaining cardboard $$=$$ Area of square $$-$$ Area of $$4$$ circles
$$=784-616$$
$$=168\ cm^2$$
Question 3
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The area of a circle is $$\displaystyle 2464\:m^{2}$$, then the diameter is

A
$$56$$ m

B
$$154$$ m

C
$$176$$ m

D
None of the above

Solution

Since, Area of circle $$= 2464m^2$$
$$\therefore \displaystyle \frac{\pi d^{2}}{4}=2464$$

$$\Rightarrow \displaystyle d^{2}=\dfrac{2464\times 4}{\dfrac{22}{7}}=3136$$

$$ \Rightarrow \displaystyle d=\sqrt{3136}=56m$$
Question 4
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If the difference between the circumference and diameter of a circle is $$30 cm$$ then what is the radius of the circle?

A
$$7 cm$$

B
$$15 cm$$

C
$$22 cm$$

D
$$2 cm$$

Solution

Let r be the radius of the circle
Then $$\displaystyle 2\pi r-2r=30\Rightarrow 2r\left ( \pi -1 \right )=30$$
$$\displaystyle \Rightarrow r\left ( \frac{22}{7}-1 \right )=15\Rightarrow r\times \frac{15}{7}=15\Rightarrow r=\frac{15\times 7}{15}=7\:cm$$
Question 5
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The area of a circular plot is $$3850$$ square meters. What is the circumference of the plot ?

A
$$240 \text{ m}$$

B
$$210 \text{ m}$$

C
$$220 \text{ m}$$

D
$$260 \text{ m}$$

Solution

If $$r$$ is the radius of the plot then area is given as $$\pi r^2$$
as per the question,
$$\displaystyle\pi r^{2}=3850$$
$$\Rightarrow r^{2}=\dfrac{3850\times 7}{22}=1225$$
$$\displaystyle \Rightarrow r= 35\text{ m}$$
since circumference $$=2\pi r$$
$$=2\times \dfrac{22}{7}\times 35\text{ m}$$
$$=220\text{ m}$$
Question 6
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In the given figure, the area enclosed between two concentric circles is $$808.5\ cm^2$$. The circumference of the outer circle is $$242\ cm$$. Calculate the radius of the inner circle

A
$$32\ cm$$

B
$$35\ cm$$

C
$$23\ cm$$

D
$$48\ cm$$

Solution

Let the radius of the inner circle be $$r$$ and the 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, $$\pi(R^2-r^2)=808.5$$

$$\Rightarrow {\left(\cfrac{77}{2}\right)}^2-r^2=\cfrac{808.5 \times 7}{22}$$

$$\Rightarrow \cfrac{5929}{4}-r^2=257.25$$

$$\Rightarrow r^2=1225$$

$$\Rightarrow r=35$$

Thus, Radius of the inner circle $$=35\ cm$$
Question 7
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The area enclosed between two concentric circles is $$770$$ $$cm^2$$. If the radius of the outer circle is $$21$$ $$cm$$, calculate the radius of the inner circle.

A
$$7$$ $$cm$$

B
$$14$$ $$cm$$

C
$$2.1$$ $$cm$$

D
$$35$$ $$cm$$

Solution

Let radius of the inner circle be $$r$$ and radius of the outer circle be $$R=21cm$$

Area enclosed between the two concentric circles $$=\pi(R^2-r^2)=770cm^2$$

$$\Rightarrow 770=\pi(R^2-r^2)$$

$$\Rightarrow 770=\dfrac{22}{7}(21^2-r^2)$$

$$\Rightarrow \dfrac{770\times 7}{22}=441-r^2$$

$$\Rightarrow 245=441-r^2$$

$$\Rightarrow r^2=196$$

$$\Rightarrow r=\sqrt{196}$$

$$\Rightarrow r=14$$

Thus, radius of the inner circle $$=14$$ $$cm$$
Question 8
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The inner circumference of a circular track $$14\ m$$ wide is $$440\ m$$. The radius of the outer circle is

A
$$70\ m$$

B
$$56\ m$$

C
$$77\ m$$

D
$$84\ m$$

Solution

Let $$R $$ and $$r $$ be tha radii of the outer and inner circle.
So by given $$R-r = 14m$$
Inner circumference $$= \displaystyle 2\pi r=2\times \frac{22}{7}\times r=\frac{44r}{7}$$
Given $$\displaystyle \frac{44r}{7}=440\Rightarrow r=\frac{440\times 7}{44}=70\ m$$
$$\displaystyle \therefore $$ Radius of outer circle $$= 70\ m + 14\ m = 84\ m$$
Question 9
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Find the the perimeter of the figure given correct to one decimal place.

A
$$56.0m$$

B
$$56.6m$$

C
$$57.7m$$

D
$$57.9m$$

Solution

The perimeter of the given figure $$= AB + CD + EF + GH + 4 \times$$ ( Perimeter of a quadrant of the circle ) $$....(1)$$

Now, $$AB = EF = 20 m - (2 m + 2 m) = 16 m$$

Similarly, $$CD = GH = 10 m - (2 m + 2 m) = 6 m$$

Now, the perimeter of the quadrant having radius $$2 \,\,m\,\,$$

$$ = \dfrac { 2\pi r }{ 4 }$$

$$= \dfrac { \pi r }{ 2 }$$

$$ =\dfrac { \pi \times 2 }{ 2 }$$

$$ =\pi \,\, $$

So, the perimeter of the $$4$$ quadrants having radius $$ 2\,\, m \,\,= \,\,4\pi\,\, $$

Put all the respective values in $$(1)$$, we get

Perimeter of the given figure $$= 16 m + 6 m + 16 m + 6 m + 4$$ $$\pi \,\,\, $$

$$= 44 m + (4 \times 3.14 )$$

$$ = 44 + 12.56 $$

$$= 56.56 m \sim 56.6 m$$

Hence, the perimeter of the given figure correct to one decimal place is 56.6 m.
Question 10
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What is the perimeter of the given figure correct to one decimal place?

A
$$56.6$$ m

B
$$46.5$$

C
$$56.7$$

D
$$23.7$$

Solution

Perimeter(P)=AB + EF + CD + GH + 4 x Arc AH
$$\displaystyle =2\times AB+2\times EF+$$ Circumference of circle with radius 2
$$\displaystyle =2\times (20-4)m+2\times (10-4)m+2\times \frac{22}{7}\times 2$$
= 32 m + 12 m + 12.57 m =56.6m