Area Related to Circles Test

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Area Related to Circles Test - 31

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Question 1
1 / -0

When the figure above is spun around its vertical axis, the total surface area of the solid formed will be

A
$$144 \pi$$

B
$$108 \pi$$

C
$$72 \pi$$

D
$$36 \pi$$

E
$$9 \pi + 12$$

Solution

The figure formed when the figure is rotated about the vertical axis is a hemisphere, therefore, the total surface area of the figure is the area of the hemisphere that is $$2\pi r^{ 2 }$$ plus the area of circle that is $$\pi r^{ 2 }$$.
It is given that the radius $$r=6$$, the area of the solid is as follows:
$$A=2\pi r^{ 2 }+\pi r^{ 2 }$$
$$=2\pi (6)^{ 2 }+\pi (6)^{ 2 }$$
$$=72\pi +36\pi$$
$$ =108\pi$$
Question 2
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Referring to the figure, if area of square $$1$$ is $$16$$ square units, area of square $$2$$ is $$48$$ square units, then the area of square $$3$$ is

A
$$56$$

B
$$64$$

C
$$78$$

D
$$96$$

Solution

Since the area of square $$1$$ is $$16$$ square units, its side would be $$4$$ units.
Similarly side of square $$2$$ would be $$\sqrt{48}$$ units
The triangle formed in between three squares becomes a right angled triangle since the angle of a square are right angles.
Therefore side of square $$3$$ can be written as $$\sqrt{(4)^2 +(\sqrt{48})^2}= \sqrt{16 + 48} = 8 $$ units
Area of square $$3$$ becomes $$64$$ square units.
Question 3
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The figure is shown in a rectangle topped with a semicircle. The base of the rectangle is one-third its height. If $$h$$ is the height of the rectangle, what is the area of the figure?

A
$$\left (\dfrac {24 + \pi}{72}\right )h^{2}$$

B
$$\left (\dfrac {12 + \pi}{72}\right )h^{2}$$

C
$$\left (\dfrac {\pi}{3}\right )h^{2}$$

D
$$\left (\dfrac {24 + 9\pi}{8}\right )h^{2}$$

E
$$\left (\dfrac {24 + 9\pi}{4}\right )h^{2}$$

Solution

Given that:
The height of the rectangle is equal to $$h.$$
The base of the rectangle $$=\cfrac h3$$

The base of the rectangle $$=$$ Diameter of the semicircle
$$\therefore$$ Radius of the circle $$=\cfrac h6$$

Now,
Area of the figure $$=$$ Area of the rectangular part $$+$$ Area of the semicircular part
$$=h\times \cfrac h3+\cfrac12\times\pi \times \left(\cfrac h6\right)^2$$
$$=\cfrac{h^2}3+\cfrac{\pi h^2}{72}$$
$$=\left(\cfrac{24+\pi}{72}\right)h^2$$

Hence, option $$A$$ is correct.
Question 4
1 / -0

If the radius of the circle with centre O is $$7$$ and the measure of angle AOB is $$100$$, what is the best approximation to the length of arc AB?

A
$$9$$

B
$$10$$

C
$$11$$

D
$$12$$

E
$$13$$
Question 5
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Find the area of shaded portion, where the length of it is $$14$$ units and radius of the upper semicircle is $$7$$ units.

A
$$196$$

B
$$98$$

C
$$200$$

D
None

Solution

Requried area of figure = Area of shaded portion
If we substract area of semi circle from area of rectangle & then add the area of upper semi circle, we get the net area or alternatively, if we calculate the area of rectangle, we get the net area = $$\displaystyle l\times b$$
$$\displaystyle =14\times 14$$
$$\displaystyle =196$$
Question 6
1 / -0

In the figure given above, radius of a greater circle is $$r\ cm$$. Find the area of non-shaded portion

A
$$\dfrac {9\pi r^{2}}{16} sq. cm$$

B
$$\dfrac {3\pi r^{2}}{4} sq. cm$$

C
$$2\pi r^{2} sq. cm$$

D
$$\dfrac {4\pi r^{2}}{3} sq. cm$$

Solution

Radius of big semicircle $$=r$$
Area of big semicircle $$=\cfrac {\pi r^2}{2}$$

Radius of small semicircles $$=r/2$$
Area of smaller semicircle $$= \cfrac 12 \times \cfrac {\pi r^2}{4}$$
Area of 2 smaller semicircles $$= \cfrac {\pi r^2}{4}$$

Area of non-shaded region $$=$$Area of big semicircle $$+ 2 \times $$ Area of the smaller semicircles
Area of non-shaded region $$=\cfrac {\pi r^2}{2} + \cfrac {\pi r^2}{4}= \cfrac {3 \pi r^2}{4}$$
Question 7
1 / -0

A horse is tied to a pole fixed at one corner of a $$50m\times 50$$ square field of grass of means of a $$20m$$ long rope. What is the area of that part of the field which the horse can gaze?

A
$$1256{m}^{2}$$

B
$$942{m}^{2}$$

C
$$628{m}^{2}$$

D
$$314{m}^{2}$$
Question 8
1 / -0

Find the area of sector when its area is $$6cm$$ and radius $$5cm$$

A
$$15$$ sq.cm

B
$$25$$ sq.cm

C
$$18$$ sq.cm

D
$$27$$ sq.cm

E
$$72$$ sq.cm

Solution
Question 9
1 / -0

The radius of a circle is $$14$$ metres. A chord of this circle subtends an angle of $${60}^{o}$$ at the centre. Find the area of the smaller segment cur off by the chord.

A
$$196\left( \cfrac { \pi }{ 6 } +\cfrac { \sqrt { 3 } }{ 4 } \right) $$

B
$$196\left( \cfrac { \sqrt { 3 } }{ 4 } -\cfrac { \pi }{ 6 } \right) $$

C
$$196\left( \cfrac { \pi }{ 6 } -\cfrac { \sqrt { 3 } }{ 4 } \right) $$

D
$$196\left( \cfrac { \pi }{ 2 } -\cfrac { \sqrt { 3 } }{ 4 } \right) $$

E
NONE OF THESE

Solution
Question 10
1 / -0

A rope by which a calf is tied is increased by $$12m$$ to $$23m$$. How much additional grassy ground shall it graze?

A
$$1020{m}^{2}$$

B
$$1120{m}^{2}$$

C
$$1210{m}^{2}$$

D
$$1021{m}^{2}$$

Solution