Area Related to Circles Test

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Area Related to Circles Test - 19

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Weekly Quiz Competition

Question 1
1 / -0

If the sum of the circumferences of two circles with radii $$R_1$$ and $$ R_2$$ is equal to the circumference of a circle of radius $$R$$, then

A
$$R_{1}+R_{2}=R$$

B
$$R_{1}+R_{2}>R$$

C
$$R_{1}+R_{2} < R$$

D
Nothing definite can be said about the relation among $$R_1, R_2$$ and R.

Solution

The circumference of circle with radius $${ R }_{ 1 }=2\pi { R }_{ 1 }$$.
and the circumference of circle with radius$$ { R }_{ 2 }=2\pi { R }_{ 2 }.$$
$$\therefore$$ The Sum of Circumferences, $$Sum=2\pi \left( { R }_{ 1 }+{ R }_{ 2 } \right)$$ .
Again the circumference of circle with radius $${ R }=2\pi { R }.$$
$$ \therefore$$ By given condition,
$$ 2\pi \left( { R }_{ 1 }+{ R }_{ 2 } \right) =2\pi { R }\Longrightarrow { R }_{ 1 }+{ R }_{ 2 }=R$$.
Question 2
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A horse is placed for grazing inside a square field 12 cm long and is tethered to one corner by a rope 8 cm long. The area it can graze is

A
50.18 sq cm

B
50.28 sq cm

C
50.38 sq cm

D
50.48 sq cm

Solution

Let the grazed portion is shown by shaded quadrant ABC,
Hence, area which horse can graze
$$= \frac{1}{4}\pi (8)^2 = \frac{1}{4}\times \frac{22}{7}\times 64$$
$$= 50.28$$ sq cm
Question 3
1 / -0

A circular disc of radius 10 cm is divided into sectors with angles $$120^{\circ}$$ and $$150^{\circ}$$ then the ratio of the area of two sectors is

A
4 : 5

B
5 : 4

C
2 : 1

D
8 : 7

Solution

Area of sector formed from angle $$\theta$$ = $$\frac{\theta}{260} \pi r^2$$, where r is the radius of the circle
Now, if angle is 120, 150 then the ratio of area of sector will be:
= $$\dfrac{\dfrac{120}{360} \pi r^2}{\dfrac{150}{360} \pi r^2}$$
= $$\dfrac{120}{150} = 4:5$$
Question 4
1 / -0

If circumference of a circle is $$110\ cm$$, then its diameter is

A
$$\cong \ 25 \ cm$$

B
$$\cong 15\ cm$$

C
$$\cong 20 \ cm$$

D
$$35 \ cm$$

Solution

$$\Rightarrow$$ The circumference of a circle $$=110\,cm$$

$$\Rightarrow$$ $$2\pi r=110$$

$$\Rightarrow$$ $$2\times \dfrac{22}{7}\times r=110$$

$$\Rightarrow$$ $$r=110\times \dfrac{7}{22}\times \dfrac{1}{2}$$

$$\Rightarrow$$ $$r=\dfrac{35}{2}$$

$$\Rightarrow$$ $$r=17.5\,cm$$

$$\Rightarrow$$ Diameter of circle $$(d)=2r=2\times 17.5\,cm=35\,cm$$
Question 5
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Semi-circular lawns are attached to the edges of a rectangular field measuring $$42 \ m \times 35\ m$$. The area of the total field is:

A
$$3818.5 m^{2}$$

B
$$8318 m^{2}$$

C
$$5813 m^{2}$$

D
$$1358 m^{2}$$

Solution

Given, length of field $$ = 42 \ m$$
Breadth of field $$ = 35 \ m$$
Radius of semi circles along length $$ = \cfrac{42}2 \ m = 21\ m$$
Radius of semi circles along breadth $$ = \cfrac{35}2\ m = 17.5\ m$$
Now,
Total area of the field $$ = $$ Area of rectangle $$+$$ Area of circle formed along length $$+$$ Area of circle formed along breadth
$$ = l\times b + \pi (21)^2 + \pi( 17.5)^2 \ m^2$$
$$ = (42\times 35 + 1386 + 962.5)\ m^2$$ (take $$\pi = \dfrac {22}{7}$$)
$$ = 3818.5 \ m^2$$

So, Total Area of the field $$ = 3818.5 m^2$$
Question 6
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The radius of a circle is increased by $$1$$ cm, then the ratio of the new circumference to the new diameter is

A
$$\pi +2$$

B
$$\pi +1$$

C
$$\pi $$

D
$$\pi- \dfrac{1}{2}$$

Solution

$$\frac{Circumference}{Diameter}=\frac{2 \pi (r+1)}{2(r+1)}$$
$$\frac{Circumference}{Diameter}=\pi$$
Question 7
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The figure is made up of an isosceles triangle and a semi-circle. Find the perimeter of the figure.

A
$$2(7-2\pi ) cm. $$

B
$$(7+2\pi ) cm. $$

C
$$2(7+2\pi ) cm. $$

D
$$(7-2\pi ) cm. $$

Solution

Perimeter of semicircular arc = $$\pi \times \dfrac d2$$
$$=\pi \times \dfrac82$$
$$=4\pi\ cm$$

Total Perimeter = Perimeter of semi-circular arc $$+$$ Sum of outer sides of triangle $$ABC$$
Total Perimeter $$= l(BOC) + CA+AB$$
$$= 7 + 7 + 4\pi$$
$$=14+4\pi$$
$$=2(7+2\pi)$$ cm
Question 8
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The number of revolutions made by a wheel of diameter $$56 cm$$ in covering a distance of $$1.1 km$$ is $$\left ( Use \pi = \frac{22}{7} \right )$$

A
$$31.25$$

B
$$56.25$$

C
$$625$$

D
$$62.5$$

Solution

Total distance covered by the wheel $$ = 1.1 km = 1.1\times 100000cm = 110000 cm$$
diameter of wheel $$ = 56 cm$$
So, $$r = \dfrac {56}{2}$$
$$ r = 28 cm$$
Now,Circumference of wheel = Distance covered in one round
$$ = 2\pi\times r = 2\pi\times 28$$
So, distance covered in one round $$ = 176 cm $$
Number of revolutions $$= \dfrac {total\ distance\ covered}{Circumference}$$
$$ = \dfrac {110000}{176}$$
So, number of revolutions covered by wheel $$ = 625$$
Question 9
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Find the area enclosed between the circles. Given that each circle is having a radius of $$2\ cm.$$

A
$$4\sqrt{3}\ cm^{2}$$

B
$$(2\sqrt{3}-\pi )\ cm^{2}$$

C
$$2\sqrt{3}-2\pi )\ cm^{2}$$

D
$$2(2\sqrt{3}-\pi )\ cm^{2}$$

Solution

The radii of the circles with center $$A$$ and $$B$$ is equal to $$r=2\ cm.$$ Since they touch each other $$AB$$ will pass through the point of contact of those circles and,
$$AB=2r$$
$$=2\times 2$$
$$=4\ cm$$

Similarly,
$$BC=4\ cm$$
$$AC=4\ cm$$

$$\therefore\ \triangle ABC$$ is an equilateral triangle. So,
$$ar\left( {\triangle ABC} \right) = \dfrac{{\sqrt 3 }}{4}{a^2}$$
$$ = \dfrac{{\sqrt 3 }}{4} \times 4 \times 4$$
$$ = 4\sqrt3 \ cm^2$$

$$ \triangle ABC$$ crops $$3$$ sectors with each of them having radius equal to $$2\ cm$$ and central angle equal to respective vertex angle of the $$\triangle ABC$$ that is equal to $$60^o.$$ So, the three sectors are equal in area.

Sum of area of $$3$$ sectors $$=$$ $$3$$ $$\times$$ Area of $$1$$ sector
$$=3\times \dfrac{\theta}{360^o}\times\pi r^2$$
$$=3\times \dfrac{60^o}{360^o}\times\pi \times2^2$$
$$=2\pi\ {cm}^2$$

Now, area between circles $$=$$ Area of $$\triangle ABC$$ $$-$$ Area of $$3$$ sectors
$$=4\sqrt3-2\pi$$
$$=2\left({2\sqrt3-\pi}\right)\ {cm}^2$$
Question 10
1 / -0

The diameter of a circle is $$10 cm$$. Find the length of the arc, when the corresponding central angle is $$270^o$$. $$(\pi =3.14)$$

A
$$23.55 cm$$

B
$$20.57 cm$$

C
$$13.57 cm$$

D
$$11.57 cm$$

Solution

Radius of the circle $$ = \dfrac {Diameter}{2} = 5 cm $$

Length

of an arc subtending an angle $$ \theta = \dfrac { \theta }{ 360 }

\times 2\pi R $$ where R is the radius of the circle.

So, length of the arc $$ = \dfrac {270} {360} \times 2 \times 3.14\times 5 = 23.55 cm $$

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Re-Attempt Weekly Quiz Competition

Area Related to Circles Test - 19

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Area Related to Circles Test - 19

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Question 1

$$R_{1}+R_{2}=R$$

$$R_{1}+R_{2}>R$$

$$R_{1}+R_{2} < R$$

Nothing definite can be said about the relation among $$R_1, R_2$$ and R.

Solution

Question 2

50.18 sq cm

50.28 sq cm

50.38 sq cm

50.48 sq cm

Solution

Question 3

4 : 5

5 : 4

2 : 1

8 : 7

Solution

Question 4

$$\cong \ 25 \ cm$$

$$\cong 15\ cm$$

$$\cong 20 \ cm$$

$$35 \ cm$$

Solution

Question 5

$$3818.5 m^{2}$$

$$8318 m^{2}$$

$$5813 m^{2}$$

$$1358 m^{2}$$

Solution

Question 6

$$\pi +2$$

$$\pi +1$$

$$\pi $$

$$\pi- \dfrac{1}{2}$$

Solution

Question 7

$$2(7-2\pi ) cm. $$

$$(7+2\pi ) cm. $$

$$2(7+2\pi ) cm. $$

$$(7-2\pi ) cm. $$

Solution

Question 8

$$31.25$$

$$56.25$$

$$625$$

$$62.5$$

Solution

Question 9

$$4\sqrt{3}\ cm^{2}$$

$$(2\sqrt{3}-\pi )\ cm^{2}$$

$$2\sqrt{3}-2\pi )\ cm^{2}$$

$$2(2\sqrt{3}-\pi )\ cm^{2}$$

Solution

Question 10

$$23.55 cm$$

$$20.57 cm$$

$$13.57 cm$$

$$11.57 cm$$

Solution

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