Perimeter and Area Test

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Perimeter and Area Test - 26

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Weekly Quiz Competition

Question 1
1 / -0

The area ofthe shaded portion in the given figure is

A
7.5 $$\pi$$ sq. units

B
6.5 $$\pi$$ sq. units

C
5.5 $$\pi$$ sq. units

D
4.5 $$\pi$$ sq. units

Solution

Given that:
Radius of bigger semi circle $$R=5$$ units
Area of Bigger semi circle $$A=\dfrac{\pi}{2} R^2\Rightarrow 12.5\ {\pi}$$ unit
Radius of smaller semi circle $$r=5-1\Rightarrow 4$$ units
Area of Bigger semi circle $$a=\dfrac{\pi}{2} r^2\Rightarrow 8\pi$$ unit
Hence,
Area of shaded portion $$=A-a\Rightarrow 12.5\ \pi-8\ \pi\Rightarrow 4.5\ \pi$$ unit$$^2$$
Question 2
1 / -0

A rope by which a cow is tethered is increased from 16 m to 23 m. How much additional ground can the cow graze now?

A
800 $$m^2$$

B
8580 $$m^2$$

C
858 $$m^2$$

D
858 $$cm^2$$

Solution

Given that, a rope by which a cow is tethered is increased from $$16\ m$$ to $$23\ m$$
To find out: The additional area that the cow can graze now
We know that area of a circle of radius $$r =πr^2$$
$$\therefore \ $$ Additional ground area $$= \pi ((23)^2 - (16)^2)$$
$$\Rightarrow \dfrac{22}{7}(529-256)\quad \quad \left[\because \ \pi=\dfrac{22}{7}\right]$$

$$\Rightarrow \dfrac{22}{7}(273)$$

$$= 858 \ m^2$$

Hence, the cow can now graze an additional area of $$858 \ m^2$$.
Question 3
1 / -0

If the side of square is $$21$$ cm, then the circumference of the circle in the figure is:

A
$$33$$ cm

B
$$1386$$ cm

C
$$66$$ cm

D
$$132 $$ cm

Solution

From the figure, we can say that,
Side of square $$=$$ diameter of the circle
$$\therefore$$ radius $$r=\dfrac{21}{2}=10.5cm$$
Now Circumference of the circle $$=2\pi r$$
$$=2\times \dfrac{22}{7}\times \dfrac{21}{2}=66cm$$
Question 4
1 / -0

If the difference between the circumference and radius of a circle is 37 cm, then the area of the circle is

A
$$111 cm^2$$

B
$$148 cm^2$$

C
$$259 cm^2$$

D
$$154 cm^2$$

Solution

Let, the radius be$$ r cm$$
Circumference will be,
$$2πr=2\times 3.14r=6.28r$$
According to question, $$6.28r-1.r=37$$
$$\Rightarrow 5.28r=37$$
$$\Rightarrow r=\dfrac{37}{5.28}$$
$$\Rightarrow r=$$approx. $$7 cm$$
Area$$=\pi {{r}^{2}}=\dfrac{22}{7} \times 7\times 7=154c{{m}^{2}}$$

Hence, this is the answer.
Question 5
1 / -0

If the circumference of a circle is $$88$$ cm, then the area of the circle (in sq.cm) is: $$\displaystyle \left ( \text{take} \ \pi = \frac{22}{7} \right )$$

A
$$\displaystyle 196\pi $$

B
$$\displaystyle 92\pi $$

C
$$\displaystyle 64\pi $$

D
$$\displaystyle 48\pi $$

Solution

$$Circumference =\displaystyle 2\pi r=88\Rightarrow r=\frac{44\times 7}{22}=14\ cm$$

$$\displaystyle \therefore area=\pi r^{2}=\pi \times 14\times 14$$ $$ cm^{2}=196\pi \ cm^{2}$$
Question 6
1 / -0

If the circumference of a circle is reduced by 50%, then the area will be reduced by

A
50%

B
25%

C
75%

D
12.5%

Solution

Let the original radius be $$r$$.
So, the area of circle $$=\pi r^2$$ $$....... (1)$$

Since, the circumference of the circle is reduced by $$50\%$$.
It means that the radius of the circle is also reduced by $$50\%$$.

Then,
The new radius $$=0.5r$$

Therefore, the new area
$$=\pi (0.5r)^2$$
$$=0.25\pi r^2$$

Therefore, the required $$\%$$
$$=\dfrac{\pi r^2-0.25\pi r^2}{\pi r^2}\times 100$$
$$=\dfrac{0.75\pi r^2}{\pi r^2}\times 100$$
$$=75\%$$

Hence, this is the answer.
Question 7
1 / -0

One side of a parallelogram is 8 cm If the corresponding altitude is 6 cm then its area is given by

A
24 $$ \displaystyle cm ^{2} $$

B
36 $$ \displaystyle cm ^{2} $$

C
40 $$ \displaystyle cm ^{2} $$

D
48 $$ \displaystyle cm ^{2} $$

Solution

Given One side of parallelogram is 8 cm and altitude is 6 cm
Here base=8 cm and height(altitude)=6 cm
We know that area of parallelogram=$$base \times hieght$$
Here base=8 cm and height(altitude)=6 cm
Then area of parallelogram=$$8\times 6=48 cm^{2}$$
Question 8
1 / -0

The height of a parallelogram of area $$ \displaystyle 350 cm^{2} $$ and base 25 cm is

A
12 cm

B
13 cm

C
14 cm

D
15 cm

Solution

Given the area of parallelogram is 350 sq cm and base is 25 cm
Then area of parallelogram if height h and base 25
$$\therefore 350=base \times height\Rightarrow 25\times h=350\Rightarrow h=14$$ cm
Question 9
1 / -0

The degree measure of the circumference of the circle is always

A
$$ \displaystyle 0^{\circ} $$

B
$$ \displaystyle 90^{\circ} $$

C
$$ \displaystyle 180^{\circ} $$

D
$$ \displaystyle 360^{\circ} $$

Solution

$$\Rightarrow$$ The degree measure for the circumference of a circle is $$360^o$$.
$$\Rightarrow$$ This is because the sum of total angles in a circle is $$360^o$$.
$$\Rightarrow$$ Whenever we want to get the length of an arc which is part of the circumference of a circle we will always refer to the total angle of a circle which is $$360^o.$$
Question 10
1 / -0

The side of a square is $$2\ cm$$ and semicircles are constructed on each side of the square, then the area of the whole figure is

A
$$(4 + 8\pi)\ cm^2$$

B
$$(4 + 4 \pi)\ cm^2$$

C
$$4 \pi\ cm^2$$

D
$$8 \pi\ cm^2$$

Solution

We know that area of semi circle of radius $$r$$ is $$\dfrac{πr^2}{2} $$ and area of square od side $$a$$ is $$a^2$$
Given that:
Side of square $$=2\ cm$$
Total area $$=$$ Area of square$$+\ 4\times$$Area of semi circle
Total area $$=2^2$$$$+\ 4\times\dfrac{\pi\times 2^2}{2}$$
$$=(4+8\pi)\ cm^2$$

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Perimeter and Area Test - 26

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Perimeter and Area Test - 26

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Question 1

7.5 $$\pi$$ sq. units

6.5 $$\pi$$ sq. units

5.5 $$\pi$$ sq. units

4.5 $$\pi$$ sq. units

Solution

Question 2

800 $$m^2$$

8580 $$m^2$$

858 $$m^2$$

858 $$cm^2$$

Solution

Question 3

$$33$$ cm

$$1386$$ cm

$$66$$ cm

$$132 $$ cm

Solution

Question 4

$$111 cm^2$$

$$148 cm^2$$

$$259 cm^2$$

$$154 cm^2$$

Solution

Question 5

$$\displaystyle 196\pi $$

$$\displaystyle 92\pi $$

$$\displaystyle 64\pi $$

$$\displaystyle 48\pi $$

Solution

Question 6

50%

25%

75%

12.5%

Solution

Question 7

24 $$ \displaystyle cm ^{2} $$

36 $$ \displaystyle cm ^{2} $$

40 $$ \displaystyle cm ^{2} $$

48 $$ \displaystyle cm ^{2} $$

Solution

Question 8

12 cm

13 cm

14 cm

15 cm

Solution

Question 9

$$ \displaystyle 0^{\circ} $$

$$ \displaystyle 90^{\circ} $$

$$ \displaystyle 180^{\circ} $$

$$ \displaystyle 360^{\circ} $$

Solution

Question 10

$$(4 + 8\pi)\ cm^2$$

$$(4 + 4 \pi)\ cm^2$$

$$4 \pi\ cm^2$$

$$8 \pi\ cm^2$$

Solution

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