Area Related to Circles Test

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Area Related to Circles Test - 26

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Weekly Quiz Competition

Question 1
1 / -0

Determine the area of the shaded segment.

A
$$10$$

B
$$11$$

C
$$12$$

D
$$13$$

Solution

$$A_{segment}=A_{sector}-A_{triangle}$$
$$A_{sector}= \dfrac{n}{360}\pi r^2$$
$$= \dfrac{60}{360}\pi 6^2$$
= 18.84 $$cm^2$$
$$A_{triangle}=\dfrac{1}{2}bh$$
$$b=sin60^0*6=$$$$3\sqrt{3}$$
$$h=cos60^0*6=$$$$3$$

$$A_{triangle}=\dfrac{1}{2}bh$$
= $$\dfrac{1}{2}3\sqrt{3}\times 3$$
= $$7.79$$
$$A_{segment}=18.84-7.79 \approx 11$$
Question 2
1 / -0

Calculate the area of a segment of a circle with a central angle of $$165$$ degrees and a radius of $$4$$. Express answer to nearest integer.

A
$$10$$

B
$$20$$

C
$$30$$

D
$$40$$

Solution

$$A_{segment}=A_{sector}-A_{triangle}$$
$$A_{sector}= \dfrac{n}{360}\pi r^2$$
$$= \dfrac{165}{360}\pi 4^2$$
= 23.02 $$cm^2$$
$$A_{triangle}=\dfrac{1}{2}bh$$
b = $$2\sqrt{3}$$
h = 2
$$A_{triangle}=\dfrac{1}{2}bh$$
= $$\dfrac{1}{2}2\sqrt{3}\times 2$$
= $$3.46$$
$$A_{segment}=23.02-3.46 \approx 20$$
Question 3
1 / -0

What is the area of the sector?

A
170 $$cm^2$$

B
100 $$cm^2$$

C
110 $$cm^2$$

D
130 $$cm^2$$

Solution

$$A_{sector}= \dfrac{n}{360}\pi r^2$$
$$= \dfrac{135}{360}\pi 12^2$$
$$169.714$$
$$\approx 170 cm^2$$
Question 4
1 / -0

Determine the length of the arc $$ADC$$.

A
$$65.2 ft$$

B
$$75.4 ft$$

C
$$86.1 ft$$

D
$$90.2 ft$$

Solution

Radius, $$r=16\ ft$$
Central Angle, $$\theta = 270^o$$

Arc length $$=\dfrac{\theta}{360^o}\times 2 \pi r$$
$$=$$ $$\dfrac{270^o}{360^o}\times 2 \times \pi \times 16$$
$$=$$ $$75.4\ ft$$

Therefore, the length of the given arc is $$75.4\ ft.$$
Question 5
1 / -0

Find the arc length of the sector.

A
$$3\pi$$

B
$$2\pi$$

C
$$\pi$$

D
$$\frac{\pi}{2}$$

Solution

Arc length = $$\frac{\theta}{360}2 \pi r$$
= $$\frac{90}{360}2 \pi 2$$
= $$\pi$$
Question 6
1 / -0

Calculate the arc length for the given diagram.

A
$$1.23\text{ in}$$

B
$$2.23\text{ in}$$

C
$$4.23\text{ in}$$

D
$$5.23\text{ in}$$

Solution

Arc length of a sector having radius $$r$$ and central angle $$\theta$$ is given by,
$$l = \dfrac{\theta }{{360}} \times 2\pi r$$

So, by using the above formula,
$$\begin{aligned}{}l &= \frac{\theta }{{360}} \times 2\pi r\\& = \frac{{50}}{{360}} \times 2 \times \pi \times 6\\& = 5.23\text{ in}\end{aligned}$$
Question 7
1 / -0

In the figure, $$ABCD$$ is a rectangular and $$\angle F = 90^{\circ}$$. If $$\overline{FD}=\overline{DC}, \overline{AB}=6$$, and the perimeter of rectangular $$ABCD$$ is $$30$$, Calculate $$\overline{FA}$$

A
$$6$$

B
$$3\sqrt{5}$$

C
$$8$$

D
$$5\sqrt{3}$$

Solution

Rectangle $$ABCD$$ has its side $$AB$$ as $$6$$ and its perimeter as $$30$$.
So, side $$AB = CD = 6$$ and $$BC = DA = 9$$
Since $$FD = DC, FD$$ also becomes equal to $$6$$.
Now in $$\Delta AFD, \angle F = 90^o$$
So, $$(AF)^2 + (FD)^2 = (AD)^2$$
$$\Rightarrow (AF)^2 + 36 = 81$$
$$\Rightarrow (FA)^2 = 45$$
$$\Rightarrow FA = 3\sqrt{5}$$
Question 8
1 / -0

Points $$A$$ and $$B$$ lie on circle with centre $$O$$ (not shown). $$AO=3$$ and $$\angle AOB ={120}^{\circ}$$. Find the area of sector $$AOB.$$

A
$$\dfrac{\pi}{3}$$

B
$$\pi$$

C
$$3 \pi$$

D
$$9 \pi$$

Solution

Area of a sector is given by $$\cfrac{\theta}{360} \times \pi \times r^2$$ where $$\theta$$ is the angle made by the sector, $$r$$ is the radius of the circle.

Here, $$\theta = 120^\circ$$ and $$r = 3$$

$$\therefore$$ Area of sector $$= \cfrac{120}{360} \times \pi \times 9 $$

$$= 3\pi$$
Question 9
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The length of the arc $$OP$$ is

A
$$16.28 cm$$

B
$$12.28 cm$$

C
$$15.28 cm$$

D
$$19.28 cm$$

Solution

Arc length = $$\dfrac{\theta}{360}2 \pi r$$
= $$\dfrac{65}{360}\times 2 \times \pi \times 17$$
= $$19.28 cm$$
Question 10
1 / -0

The trapezoid is divided into a rectangle and two triangles as shown in the above figure. Find the combined area of the two shaded triangles. ( All lengths are in inches)

A
$$4$$

B
$$6$$

C
$$9$$

D
$$12$$

E
$$14$$

Solution

The base of the triangles is $$=\dfrac{1}{2}\times (12-6)=3$$.
Hence the two triangles are right angles isosceles as base and height are both equal to $$3$$ units.
Area of one triangle $$=\dfrac{1}{2}\times \mathrm{Base}\times \mathrm{Height} =\dfrac{1}{2}\times 3\times 3=\dfrac{9}{2} \ \mathrm{sq. units}$$
Therefore, the required area is $$2\times ($$ Area of the triangle $$)$$$$=2\times \dfrac{9}{2}\ \text{sq. units}$$
$$=9\ \mathrm{sq. units}$$