Perimeter and Area Test

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Perimeter and Area Test - 19

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Weekly Quiz Competition

Question 1
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A floor which measures $$15m\, \times\, 8m$$ is to be laid with tiles measuring $$50cm\, \times\, 25cm$$. Find the number of tiles required.
Further, if a carpet is laid on the floor so that a space of 1 m exists between its edges and the edges of the floor, what fraction of the floor is uncovered.

A
$$960; \displaystyle \frac{7}{20}$$

B
$$860; \displaystyle \frac{20}{7}$$

C
$$760; \displaystyle \frac{7}{13}$$

D
$$660; \displaystyle \frac{7}{15}$$

Solution

We know, $$50 cm = 0.5m $$ and
$$25 cm = 0.25m$$
Tiles along the length of the room=$$\dfrac{15}{0.5}$$ = $$30$$ tiles
Tiles along the length of the room=$$\dfrac{8}{0.25}$$ =$$32$$ tiles
Total number of tiles required =$$30\times 32 $$ =$$960$$ tiles
To leave $$1 m$$ between the carpet and wall on all sides, the carpet needs to be $$2 m$$ shorter in each dimension.
Therefore,
$$(15-2)m(8-2)m$$
$$=(13\times 6)m$$
$$=78 m^2$$
Therefore, the room area is $$15\times 8$$ = $$120 m^2$$
The carpet area is $$78 m^2$$
The fraction of the area uncovered =$$\dfrac{120-78}{120}$$= $$\dfrac{42}{120}$$=$$\dfrac{7}{20}$$
Question 2
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The shaded portion of the figure, given along side,shows two concentric circles.
If the circumference of the two circles are $$396 \:cm$$ and $$374\: cm$$, find the area of the shaded portion.

A
$$1347.5 \: cm^{2}$$

B
$$1047.5 \: cm^{2}$$

C
$$2347.5 \: cm^{2}$$

D
$$1247.5 \: cm^{2}$$

Solution

Given,
$$2\pi r=374cm$$
$$2\pi R=396cm$$
Therefore,
$$2\pi r=374cm$$
$$=>r=\cfrac{374\times7}{2\times 22}$$
$$=59.5$$
$$2\pi R=396cm$$
$$=>R=\cfrac{396\times7}{2\times 22}$$
$$=63$$
Area of shaded portion=$$(\pi R^2-\pi r^2)cm^2$$
=$$\pi [(63)^2-(59.5)^2]cm^2$$
=$$\pi (3969-3540.25)$$
=$$1347.5cm^2$$
Question 3
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The area of a circle is $$394.24$$ $$cm^2$$. What is the radius of the circle ?

A
$$11.2 cm$$

B
$$12.1 m$$

C
$$13.5 m$$

D
$$15.0 m$$

Solution

Area of the circle $$=394.24 \ cm^2$$
$$\implies \pi r^2=394.24$$
$$\implies r^2=\dfrac{394.24\times 7}{22}$$
$$\implies r^2=125.44$$
$$\implies r=11.2\ cm$$
Question 4
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The circumference of a circular field is $$308 m$$, Find its Area.

A
$$7050 m^2$$

B
$$7546 m^2$$

C
$$7946 m^2$$

D
$$8129 m^2$$

Solution

Let the radius of the circle be $$r$$ cm
Circumference of the circle=$$308$$
Therefore,
$$2\pi r=308$$

$$\Rightarrow r=\dfrac{308\times 7}{2\times 22}$$

$$r=49m$$

Therefore,
Area of the circle$$=\pi \times (49)^2$$

$$=\dfrac{22}{7}\times 2401$$

$$=7546m^2$$
Question 5
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The diameter of a circle is $$28 cm$$. Find its circumference

A
$$32 cm$$

B
$$48 cm$$

C
$$88 cm$$

D
$$98 cm$$

Solution

Diameter of a circle $$=28$$ cm
Thus, circumference $$=\pi \times 28$$ cm
$$=\dfrac{22}{7} \times 28$$ cm
$$=22 \times 4$$ cm
$$=88$$ cm
Question 6
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The area enclosed by the circumferences of two concentric circles is $$346.5\, cm^2$$. If the circumference of the inner circle is $$88$$ cm, then the radius of the outer circle is

A
$$88$$ cm

B
$$17.5$$ cm

C
$$35$$ cm

D
$$9.3$$ cm

Solution

$$ Given-\\ O\quad is\quad the\quad centre\quad of\quad two\quad concentric\quad circles\quad with\quad outer$$

$$ radius\quad OA={ r }_{ 2 }\quad \& \quad inner\quad cicumference=C=88cm.$$

$$ ar.ring\quad between\quad the\quad circles=346.5{ cm }^{ 2 }.$$

$$ To\quad find\quad out-$$

$$ { r }_{ 2 }=?$$

$$ Solution-$$

$$ inner\quad cicumference=88cm.$$

$$ Let\quad the\quad radius\quad of\quad the\quad inner\quad circle\quad be\quad { r }_{ 1 }.$$

$$ \therefore \quad { r }_{ 1 }=\dfrac { C }{ 2\pi } =\dfrac { 346.5 }{ 2\times \frac { 22 }{ 7 } } cm=14cm.$$

$$ So\quad ar.ring=ar.outer\quad circle-ar.inner\quad circle=\pi \left\{ { { r }_{ 2 } }^{ 2 }-{ { r }_{ 1 } }^{ 2 } \right\} $$

$$ \Longrightarrow 346.5=\dfrac { 22 }{ 7 } \times \left\{ { { r }_{ 2 } }^{ 2 }-{ { r }_{ 1 } }^{ 2 } \right\} $$

$$ { { \Longrightarrow r }_{ 2 } }^{ 2 }=306.25$$

$$ \Longrightarrow r_2=17.5cm.$$

$$ \therefore \quad The\quad radius\quad of\quad the\quad outer\quad circle=17.5cm.$$

$$ Ans-\quad Option\quad B. $$
Question 7
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The radii of two circles are $$25\ cm$$ and $$18\ cm$$. Find the radius of the circle which has a circumference equal to the sum of the circumference of these two circles.

A
$$43\ cm$$

B
$$12\ cm$$

C
$$34\ cm$$

D
$$40\ cm$$

Solution

The circumference of a circle $$=2\pi r$$
Let $$R$$ be the radius of the circle which has a circumference equal to the sum of the circumferences of the two circles.
$$2\pi R=2\pi \times 25+2\pi \times 18$$
$$\implies R=25+18=43\ cm$$
Question 8
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The diameter of a circle is $$28 cm$$. Find the one-fourth of its area.

A
$$389 cm^{2}$$

B
$$512 cm^{2}$$

C
$$616 cm^{2}$$

D
$$154 cm^{2}$$

Solution

The diameter of a circle is $$28cm$$
Therefore,
Radius of the circle=$$\dfrac{28}{2}=14cm$$
Therefore,
Area of the circle$$=\pi (14)^2$$

$$=\dfrac{22}{7} \times 196$$

$$=616cm^2$$

So, $$\dfrac 14 \times 616 = 154 cm^2$$
Question 9
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Directions For Questions

The cross-section of a piece of metal $$4$$ m in length is shown below. Calculate :

...view full instructions

The area of the cross-section;

A
133.5 $$cm^{3}$$

B
187.5 $$cm^{3}$$

C
138.5 $$cm^{3}$$

D
183.5 $$cm^{3}$$

Solution

The given diagram can be divided in two parts, rectangle ABCD and triangle CEF.
$$AB=12$$cm
$$BE=16$$cm
$$FD=7.5$$cm
$$AD=BC=10$$cm
Now,
$$AB=CD$$ (Opposite sides of a rectangle)
$$=>AB=CF+FD$$
$$=>12=CF+7.5$$
$$=>CF=4.5$$cm
Again,
$$BE=BC+CE$$
$$=>16=10+CE$$
$$=>CE=6$$cm
Area of cross section=Area of rectangle ABCD+$$\triangle$$CEF
$$=AB\times AD+\dfrac{1}{2}\times CE\times EF$$
$$=10\times 12+\dfrac{1}{2}\times 6\times 4.5$$
$$=133.5cm^2$$
Question 10
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Find the area of a ring shaped region enclosed between two concentric circles of radii $$20$$ cm and $$15$$ cm.

A
550 $$cm^{2}$$

B
425 $$cm^{2}$$

C
496 $$cm^{2}$$

D
810 $$cm^{2}$$

Solution

Let us given $$R = 20$$ $$cm $$ & $$r = 15$$ $$cm $$
The area of a ring shaped region$$=\pi(20)^2-\pi(15)^2$$
$$=(400-225)\times \dfrac{22}{7}$$
$$=(175)\times \dfrac{22}{7}$$
$$ =550$$ sq. cm