Perimeter and Area Test

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Perimeter and Area Test - 9

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Weekly Quiz Competition

Question 1
1 / -0

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

A
$$24\: cm^2$$

B
$$36\: cm^2$$

C
$$40\: cm^2$$

D
$$48\: cm^2$$

Solution

Area of parallelogram $$=$$ length $$\times$$ height
Here altitude is nothing but the height
$$\therefore $$ Area of parallelogram $$= 8 \times 6$$
$$= 48\: cm^2$$
Question 2
1 / -0

Circumference of a circle is .....................

A
$$\pi r$$

B
$$2\pi r$$

C
$$3\pi r$$

D
$$4\pi r$$

Solution

Circumference of a circle whose radius is $$r$$ is given by $$2\pi r.$$
Question 3
1 / -0

The area of a circle is the measurement of the region enclosed by its

A
Radius

B
Centre

C
Circumference

D
Area

Solution

As shown in the above figure, area is the region enclosed inside the boundary/circumference of the circle.
Question 4
1 / -0

Area of a circle is ...................

A
$$\pi r^2$$

B
$$2\pi r$$

C
$$2\pi r^2$$

D
$$4\pi r$$

Solution

Area covered within the boundary of circle = $$\pi \times r^2$$, where $$r$$ is a radius of a circle.
Question 5
1 / -0

The sides of a triangle are (3p 4) cm, (2p 5) cm and (2p + 5) cm. Which of the following expressions gives its perimeter?

A
(7p 4) cm

B
(5p + 5) cm

C
5p 7 cm

D
7p + 4cm
Question 6
1 / -0

What is the area of the circular ring included between two concentric circles of radius $$14$$ cm and $$10.5$$ cm ?

A
$$255\ cm^2$$

B
$$148\ cm^2$$

C
$$324\ cm^2$$

D
$$269\ cm^2$$

Solution

Let $$R$$ be the radius of the larger circle
Hence $$R=14\ cm$$

Let $$r$$ be the radius of the smaller circle
Hence $$r=10.5\ cm$$

Area of the circular ring $$=$$ Area of the larger circle - Area of the smaller circle

$$=\pi R^2 - \pi r^2$$

$$=\dfrac { 22 }{ 7 } \times \left( { R }^{ 2 } - { r }^{ 2 } \right)$$

$$= \dfrac { 22 }{ 7 } \times \left( { 14 }^{ 2 } - { 10.5 }^{ 2 } \right)$$

$$=269.5 \ { cm }^{ 2 }$$

$$\approx 269\ { cm }^{ 2 } $$
Question 7
1 / -0

A path of width 8 m runs around a circular park whose radius is $$38\ m$$. Find the area of the path.

A
$$2112\ m^2$$

B
$$2834\ m^2$$

C
$$3212\ m^2$$

D
$$1578\ m^2$$

Solution

Inner radius (Radius of the park) $$=38\ m$$
Outer radius (Radius of the park $$+$$ path width of the park) $$=38+8=46\ m$$
Area of the path $$=\pi(46^2-38^2)$$
$$=\cfrac{22}{7}(2116-1444)$$
$$=\cfrac{22}{7}(672)$$
$$=2112\ m^2$$
Question 8
1 / -0

In the given figure, the area enclosed between two concentric circles is $$808.5\ cm^2$$. The circumference of the outer circle is $$242\ cm$$. Find the width of the ring.

A
$$1.2\ cm$$

B
$$2.4\ cm$$

C
$$3.5\ cm $$

D
$$4.2\ cm$$

Solution

Let radius of the inner circle be $$r$$ and 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, $$\pi(R^2-r^2)=808.5$$
$$\Rightarrow {\left(\cfrac{77}{2}\right)}^2-r^2=\cfrac{808.5 \times 7}{22}$$
$$\Rightarrow \cfrac{5929}{4}-r^2=257.25$$
$$\Rightarrow r^2=1225$$
$$\Rightarrow r=35$$
Radius of the inner circle $$=35\ cm$$
$$\therefore$$ width of the ring $$=(R-r)$$
$$=(38.5-35)\ cm$$
$$=3.5\ cm$$
Question 9
1 / -0

The radius of a circle is $$5\: m$$. Find the circumference of the circle whose area is $$49$$ times the area of the given circle.

A
$$220 \: m$$

B
$$120 \: m$$

C
$$320 \: m$$

D
$$420 \: m$$

Solution

Radius of given circle $$=5\ m$$
Therefore,
Area of given circle $$=\pi R^2=\pi (5)^2$$
$$=25\pi\ m^2$$
Let radius of required circle $$ =r$$
Therefore, According to the given condition,
$$\pi r^2=49\times 25\pi$$
$$\Rightarrow r^2=(7\times 5)^2$$
$$\Rightarrow r=35m$$
Therefore,
Circumference $$=2\pi r$$
$$=2\times \dfrac { 22 }{ 7 }\times35$$
$$=220\ m$$
Question 10
1 / -0

The circumference of a circular field is $$528\ m$$. Then its area is

A
22176 $$m^2$$

B
26400 $$m^2$$

C
10080 $$m^2$$

D
84000 $$m^2$$

Solution

$$ Given-\\ the\quad circumference=C\quad of\quad a\quad circular\quad field=528m.\\ To\quad find\quad out-\\ the\quad ar.circle=?\\ Solution-\\ Let\quad the\quad radius\quad of\quad the\quad circle\quad be\quad r.\\ Then\quad r=\cfrac { C }{ 2\pi } \quad when\quad C=circumference\quad of\quad the\quad circle.\\ \Longrightarrow r=\cfrac { 528 }{ 2\times \cfrac { 22 }{ 7 } } cm=84m.\\ \therefore \quad ar.circle=\pi { r }^{ 2 }=\cfrac { 22 }{ 7 } \times { 84 }^{ 2 }{ m }^{ 2 }=22176{ m }^{ 2 }.\\ Ans-\quad Option\quad A. $$

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Re-Attempt Weekly Quiz Competition

Perimeter and Area Test - 9

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Perimeter and Area Test - 9

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SHARING IS CARING

Question 1

$$24\: cm^2$$

$$36\: cm^2$$

$$40\: cm^2$$

$$48\: cm^2$$

Solution

Question 2

$$\pi r$$

$$2\pi r$$

$$3\pi r$$

$$4\pi r$$

Solution

Question 3

Radius

Centre

Circumference

Area

Solution

Question 4

$$\pi r^2$$

$$2\pi r$$

$$2\pi r^2$$

$$4\pi r$$

Solution

Question 5

(7p 4) cm

(5p + 5) cm

5p 7 cm

7p + 4cm

Question 6

$$255\ cm^2$$

$$148\ cm^2$$

$$324\ cm^2$$

$$269\ cm^2$$

Solution

Question 7

$$2112\ m^2$$

$$2834\ m^2$$

$$3212\ m^2$$

$$1578\ m^2$$

Solution

Question 8

$$1.2\ cm$$

$$2.4\ cm$$

$$3.5\ cm $$

$$4.2\ cm$$

Solution

Question 9

$$220 \: m$$

$$120 \: m$$

$$320 \: m$$

$$420 \: m$$

Solution

Question 10

22176 $$m^2$$

26400 $$m^2$$

10080 $$m^2$$

84000 $$m^2$$

Solution

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