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CBSE Questions for Class 7 Maths Perimeter And Area Quiz 4 - MCQExams.com
CBSE
Class 7 Maths
Perimeter And Area
Quiz 4
A circular grassy plot of land, 42 m in diameter, has a path 3.5 m wide running around it on the outside. Find the cost graveling the path at Rs 4 per square meter.
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$$Rs 2002$$
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$$Rs 2003$$
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$$Rs 2004$$
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$$Rs 2000$$
Explanation
Given : Radius ($$r$$) $$=21m$$
Radius ($$R$$) $$=24.5m$$
Required area $$=\pi (R^{2}-r^{2})$$
Required area $$=\pi (24.5^{2}-21^{2})=500.5$$
Cost $$=500.5\times 4$$
$$=2002Rs$$
The sides of a triangle are (3p 4) cm, (2p 5) cm and (2p + 5) cm. Which of the following expressions gives its perimeter?
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(7p 4) cm
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(5p + 5) cm
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5p 7 cm
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7p + 4cm
In fig. 3, the area of the shaded region is :
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$$\pi \left ( r_1+r_2 \right )$$
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$$\pi \left ( r_1^{2}+r_2^{2} \right )$$
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$$\pi \left ( r_1-r_2 \right )$$
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$$\pi \left ( r_2^{2}-r_1^{2} \right )$$
Explanation
The area of the given shaded region will be :
The area of the smaller circle subtracted from the larger circle
Area of larger circle is $$\pi r_{2}^{2}$$ and
Area of smaller circle is $$\pi r_{1}^{2}$$
Then, the ares of shaded region is $$\pi r_{ 2 }^{ 2 } -\pi r_{ 1 }^{ 2 }=\pi(r_{2}^{2}-r_{1}^{2})$$
If the base of a parallelogram is $$8\ cm$$ and its altitude is $$5\ cm$$, then its area is equal to
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$$15\ cm^{2}$$
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$$20\ cm^{2}$$
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$$40\ cm^{2}$$
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$$10\ cm^{2}$$
Explanation
Area of parallelogram $$=Base \times height $$
$$=8\times 5\\ =40\ { cm }^{ 2 }$$
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Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
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Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
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Assertion is correct but Reason is incorrect
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Assertion is incorrect but Reason is correct
Explanation
Let the radius of the circle be $$r.$$
Circumference $$ = 2 \times \pi \times r$$
$$\Rightarrow 176=2\times\dfrac{22}7 \times r$$
$$\Rightarrow r=\dfrac{176\times 7}{2\times 22}$$
$$\Rightarrow r=28\ cm$$
We used the statement given in the reason to prove that the assertion is correct hence, option $$A$$ is correct.
If a circular grass lawn of $$35\ m$$ in radius has a path $$7\ m$$ wide running around it on the outside, then the area of the path is
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$$1450\ m^2$$
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$$1576\ m^2$$
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$$1694\ m^2$$
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$$3368\ m^2$$
Explanation
Radius of bigger circle(with the path) = $$35 + 7 = 42\ m.$$
So, $$R = 42$$ $$cm$$ & $$r = 35 $$ $$cm$$
Thus area of the path $$=$$ Area of bigger circle $$-$$ Area of smaller circle
$$\therefore$$ Required area $$= \pi (42)^2 - \pi (35)^2 = \dfrac{22}{7} \times (42 + 35)(42 - 35) = 22 \times 77 = 1694\ m^2$$
A pond that is $$100$$ m in diameter is surrounded by a circular grass walk that is $$2$$ m wide. How many square meters of grass is there for the walk?
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$$98 \pi $$
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$$100 \pi $$
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$$102 \pi $$
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$$204 \pi $$
Explanation
Area of grass on walk $$ = $$ Area of outer circle $$-$$ Area of the pond
$$ = \pi \times 52^2 - \pi\times 50^2$$
$$ = \pi (2704 - 2500)$$
$$ = \pi\times 204$$
So, Area of grass for the walk $$ = 204\pi m^2$$
Find the area of a circular ring formed by two concentric circles whose radii are $$5.7\ cm$$ and $$4.3\ cm$$ respectively (Take $$\pi =3.1416$$)
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$$43.98$$ sq. cm
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$$53.67$$ sq. cm
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$$47.24$$ sq. cm
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$$38.54$$ sq. cm
Explanation
Let the radii of the outer and inner circles be $$r_2$$ and $$r_1$$ respectively; we have
$$Area =\pi r_2^2 -\pi r_1^2 =\pi (r_2^2 - r_1^2)$$
$$=\pi (r_2 -r_1)(r_2+r_1)$$
$$=\pi (5.7 -4.3)(5.7 + 4.3)=\pi \times 1.4 \times 10$$ sq. cm
$$=3.1416 \times 14 sq. cm. =43.98$$ sq. cm
If the radius of a circle is $$\displaystyle \frac{7}{\sqrt{\pi}}$$ cm, then the area of the circle is equal to
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$$\displaystyle \frac{49}{\pi} cm^2$$
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$$\displaystyle \frac{154}{\pi} cm^2$$
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$$154 cm^2$$
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$$49 cm^2$$
Explanation
Area of the circle $$ = \pi \left (\cfrac{7}{\sqrt{\pi}}\right )^2 =\cfrac{\pi (49)}{\pi}=49 cm^2$$
Now,$$ \cfrac{154}{\pi}=\cfrac{154 \times 7}{22} =49cm^2$$
A track is in the form of a ring whose inner circumference is $$352\ m$$ and the outer circumference is $$396\ m$$. The width of the track is:
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$$44\ m$$
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$$14\ m$$
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$$22\ m$$
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$$7\ m$$
Explanation
Let the radii of inner and outer circles be $$r$$ and $$R$$, respectively.
Circumference of inner circle $$ = 352\ m$$
$$\therefore 2\pi r = 352$$
$$2\times \dfrac{22}{7}\times r=352$$
$$\Rightarrow r = 56\ m$$
Similarly, Circumference of outer circle $$ = 396 m$$
$$\therefore 2\pi R = 396$$
$$2\times \dfrac{22}{7}\times R=396$$
$$\Rightarrow R = 63\ m$$
Now, Width of the track $$ = R - r$$
$$ = 63 - 56$$
$$ = 7\ m$$
So, width of the track $$ = 7\ m$$
If area of circular field is $$6.16 \text{ m}^2$$, then its diameter will be
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$$6.8 \ \text{m}$$
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$$2.8 \ \text{m}$$
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$$12 \ \text{m}$$
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$$6.84 \ \text{m}$$
Explanation
Area of circular field $$=6.16\,\text{m}^2$$
Area of a circle $$=\pi r^2$$
$$\Rightarrow$$ $$6.16=\dfrac{22}{7}\times r^2$$
$$\Rightarrow$$ $$6.16\times \dfrac{7}{22}=r^2$$
$$\Rightarrow$$ $$r^2=\dfrac{43.12}{22}$$
$$\Rightarrow$$ $$r^2=1.96$$
$$\Rightarrow$$ $$r=1.4\,\text{m}$$
$$\Rightarrow$$ $$d=2r=1.4\times 2=2.8\text{ m}$$
$$\therefore$$ The diameter of a circular field is $$2.8\,\text{m}.$$
A circular grassy plot of land, $$42\ m$$ in diameter, has a path $$3.5\ m$$ wide running round it on the outside. Find the cost of gravel ling the path at Rs. $$4$$ per square metre.
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Rs. $$2002$$
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Rs. $$2003$$
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Rs. $$2004$$
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Rs. $$2000$$
Explanation
We have, Radius of the plot $$ = 21\ m$$
Radius of plot including the path $$ = 21 + 3.5 = 24.5\ m$$
Area of the path $$ = \pi (24.5^2) - \pi (21^2)$$
$$ = \pi(24.5^2 - 21^2)$$
$$ = \pi (24.5 - 21) (24.5 + 21)$$
$$ = \cfrac {22}{7} \times 3.5\times 45.5$$
$$ = 500.5\ m^2$$
Total cost of gravelling the path $$ = 500.5\times 4$$
$$ =$$ Rs. $$2002$$
A certain right angled triangle has its area numerically equal to its perimeter. The length of its each side is an even integer. What is the perimeter?
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$$24\ units$$
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$$36\ units$$
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$$32\ units$$
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$$30\ units$$
Explanation
Given that, the measure of each side of the right triangle is an even integer unit.
Also, its perimeter and area are numerically equal.
Let us start from even integers: $$ 2,4,6,8,10,12,14,$$ etc.
Out of them, only $$\left( 6,8,10 \right)$$ make Pythagorean triplet
$$ \because { 8 }^{ 2 }+{ 6 }^{ 2 }={ 10 }^{ 2 }$$
Also, $$8+6+10=24 \left( Perimeter \right) $$
And, $$\cfrac { 1 }{ 2 } \times 8\times 6=24 \left( Area \right) $$
$$\therefore$$ This group of even integers comply with all the given conditions.
$$ \therefore$$ Perimeter $$= 24$$ units.
The adjacent sides of a parallelogram are 10 cm and 12 cm. The diagonal joining the ends of these sides is 14 cm. Its area is
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$$48 \sqrt{6}$$ $$cm^2$$
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$$48 \sqrt{5}$$ $$cm^2$$
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$$48 \sqrt{3}$$ $$cm^2$$
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none of these
Explanation
Area of the parallelogram $$ABCD=2\times \text{ area } \triangle BCD$$
Now, in $$\triangle BCD$$
$$S=\dfrac { 12+10+14 }{ 2 } =18\text{ cm}$$
Area of $$\triangle BCD=\sqrt { s\left( s-a \right) \left( s-b \right) \left( s-c \right) }$$
$$=\sqrt { 18\left( 18-12 \right) \left( 18-10 \right) \left( 18-14 \right) } $$
$$=\sqrt { 18\times 6\times 8\times 4 } $$
$$=\sqrt { 3\times 6\times 6\times 2\times 4\times 4 } $$
$$=4\times 6\sqrt { 6 } $$
$$=24\sqrt { 6 } \text{ cm}^{ 2 }$$
$$\therefore $$ Area of the parallelogram $$ABCD=2\times 24\sqrt { 6 } =48\sqrt { 6 } \text{ cm}^{ 2 }$$
The circumference of a circular ground is $$88$$ metres. A strip of land, $$3$$ metres wide, inside and along circumference of the ground is to be levelled. What is the budgeted expenditure if the levelling costs Rs. $$7$$ per square metre?
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$$1050$$
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$$1125$$
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$$1325$$
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$$1650$$
Explanation
Circumference of ground $$ = 88\ m$$
$$\Rightarrow 2\pi r = 88$$
$$\Rightarrow r = 14\ m$$ ....take $$\pi = \dfrac{22}{7}$$
Area of strip to be levelled $$ = $$ Area of ground $$-$$ Area of inner circle
$$ = \pi 14^2 - \pi 11^2$$
$$ = 235.72\ m^2$$ ....take $$\pi = \dfrac{22}{7}$$
Total cost of levelling the strip (if the cost of levelling $$1\ m^2$$ is Rs. $$7$$) $$ = 235.72\times 7$$
$$ =$$ Rs. $$1650$$
A flooring tile has the shape of a parallelogram whose base is $$24\: cm$$ and the corresponding height is $$10\: cm$$. How many such tiles are required to cover a floor of area $$1080$$ $$m^2$$?
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$$20000$$
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$$35000$$
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$$45000$$
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$$65000$$
Explanation
Area of the parallelogram $$=$$ Base$$\times $$Height
So area of each tile $$=24\times 10=240 \:cm^2$$
Area of the floor $$=1080 \: m^2=(1080\times 100\times 100) \: cm^2$$
$$\therefore$$ Required number of tiles $$=\dfrac{\text{Area of the floor}}{\text{Area of each tile}}=\dfrac{10800000}{240}=45000$$
The cost of fencing a circular field at the rate of $$Rs\:.240\: per\: metre$$ is $$Rs. \: 52,800$$ . The field is to be ploughed at the rate of $$Rs. 12.50 \: per \: m^{2}$$. Find the cost of ploughing the field.
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$$Rs. \:48,125$$
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$$Rs. \:38,125$$
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$$Rs. \:10,125$$
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$$Rs. \:11,125$$
Explanation
Cost of fencing at the rate of Rs.$$240$$ per meter=$$Rs.52800$$
Therefore,
Circumference=$$\dfrac{52800}{240}$$ = $$220$$
Let r be the radius of the field
Therefore,
$$2\pi r=220$$
$$\Rightarrow r=\dfrac{220\times 7}{2\times 22}$$
$$\Rightarrow r=35$$
Therefore,
Area of the field=$$\pi r^2$$
=$$\dfrac{22}{7}\times 35\times 35$$
=$$3850 m^2$$
Cost of fencing the field = $$12.5\times 3850$$
=$$Rs.48125$$
The circumference of a circular field is $$308\:m$$ . Find its
area.
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$$7546 \: m^{2}$$
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$$6546 \: m^{2}$$
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$$7046 \: m^{2}$$
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$$7846 \: m^{2}$$
Explanation
Let the radius of the circle be $$r$$ cm
Circumference of the circle=$$308$$
Therefore,
$$2\pi r=308$$
$$=>r=\dfrac{308\times 7}{2\times 22}$$
$$=>r=49m$$
Therefore,
Area of the circle=$$\pi \times (49)^2$$
=$$\frac{22}{7}\times 2401$$
=$$7546m^2$$
The diameter of a circle is $$28 \:cm$$. Find its
area.
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$$616 \: cm^{2}$$
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$$516 \: cm^{2}$$
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$$116 \: cm^{2}$$
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$$216 \: cm^{2}$$
Explanation
The diameter of a circle is $$28cm$$
Therefore,
Radius of the circle=$$\dfrac{28}{2}=14cm$$
Therefore,
Area of the circle=$$\pi (14)^2$$
=$$\dfrac{22}{7}\times196$$
=$$616cm^2$$
The diagram, given below, shows two paths drawn inside a rectangular field 80 m long and 45 m wide. The widths of the two oaths are 8 m and 15 m as shown. Find the area of the shaded portion.
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$$1195\, m^{2}$$
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$$8895\, m^{2}$$
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$$1155\, m^{2}$$
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$$1895\, m^{2}$$
Explanation
Given, rectangular field is 80 m long and 45 m wide.
Two paths are drawn inside the rectangular field. The widths of the two oaths are 8 m and 15 m.
The two paths intersect. The length and breath of the intersection is $$8$$ m and $$15$$ m.
Area of the shaded region $$=(8\times 80) + (15\times 45) - (15\times 8)$$
$$=640+675-120$$ sq. m
$$=1195$$ sq. m
Find the length and the breadth of the lawn.
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26 m and 10 m
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35 m and 12 m
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41 m and 16 m
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49 m and 18 m
Explanation
Since, in the given figure
Outer length$$=30m$$ and Outer breadth$$=12m$$
As per given, On the three sides of the lawn there are flower-beds having a uniform width of 2 m.
$$\therefore$$ length of lawn$$=30-2-2=26m$$
and
breadth of lawn$$=12-2=10m$$
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.
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$$1347.5 \: cm^{2}$$
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$$1047.5 \: cm^{2}$$
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$$2347.5 \: cm^{2}$$
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$$1247.5 \: cm^{2}$$
Explanation
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=$$(\pi R^2-\pi r^2)cm^2$$
=$$\pi [(63)^2-(59.5)^2]cm^2$$
=$$\pi (3969-3540.25)$$
=$$1347.5cm^2$$
The area of a circle is $$394.24$$ $$cm^2$$. What is t
he radius of the circle ?
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$$11.2 cm$$
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$$12.1 m$$
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$$13.5 m$$
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$$15.0 m$$
Explanation
Area of the circle $$=394.24 \ cm^2$$
$$\implies \pi r^2=394.24$$
$$\implies r^2=\dfrac{394.24\times 7}{22}$$
$$\implies r^2=125.44$$
$$\implies r=11.2\ cm$$
The circumference of a circular field is $$308 m$$, Find its
Area.
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$$7050 m^2$$
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$$7546 m^2$$
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$$7946 m^2$$
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$$8129 m^2$$
Explanation
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$$=\pi \times (49)^2$$
$$=\dfrac{22}{7}\times 2401$$
$$=7546m^2$$
The diameter of a circle is $$28 cm$$. Find its c
ircumference
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$$32 cm$$
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$$48 cm$$
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$$88 cm$$
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$$98 cm$$
Explanation
Diameter of a circle $$=28$$ cm
Thus, circumference $$=\pi \times 28$$ cm
$$=\dfrac{22}{7} \times 28$$ cm
$$=22 \times 4$$ cm
$$=88$$ cm
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.
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$$960; \displaystyle \frac{7}{20}$$
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$$860; \displaystyle \frac{20}{7}$$
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$$760; \displaystyle \frac{7}{13}$$
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$$660; \displaystyle \frac{7}{15}$$
Explanation
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
=$$\dfrac{120-78}{120}$$
= $$\dfrac{42}{120}$$
=$$\dfrac{7}{20}$$
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.
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$$43\ cm$$
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$$12\ cm$$
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$$34\ cm$$
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$$40\ cm$$
Explanation
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.
$$2\pi R=2\pi \times 25+2\pi \times 18$$
$$\implies R=25+18=43\ cm$$
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
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$$88$$ cm
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$$17.5$$ cm
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$$35$$ cm
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$$9.3$$ cm
Explanation
$$ Given-\\ O\quad is\quad the\quad centre\quad of\quad two\quad concentric\quad circles\quad with\quad outer$$
$$ radius\quad OA={ r }_{ 2 }\quad \& \quad inner\quad cicumference=C=88cm.$$
$$ ar.ring\quad between\quad the\quad circles=346.5{ cm }^{ 2 }.$$
$$ To\quad find\quad out-$$
$$ { r }_{ 2 }=?$$
$$ Solution-$$
$$ inner\quad cicumference=88cm.$$
$$ Let\quad the\quad radius\quad of\quad the\quad inner\quad circle\quad be\quad { r }_{ 1 }.$$
$$ \therefore \quad { r }_{ 1 }=\dfrac { C }{ 2\pi } =\dfrac { 346.5 }{ 2\times \frac { 22 }{ 7 } } cm=14cm.$$
$$ So\quad ar.ring=ar.outer\quad circle-ar.inner\quad circle=\pi \left\{ { { r }_{ 2 } }^{ 2 }-{ { r }_{ 1 } }^{ 2 } \right\} $$
$$ \Longrightarrow 346.5=\dfrac { 22 }{ 7 } \times \left\{ { { r }_{ 2 } }^{ 2 }-{ { r }_{ 1 } }^{ 2 } \right\} $$
$$ { { \Longrightarrow r }_{ 2 } }^{ 2 }=306.25$$
$$ \Longrightarrow r_2=17.5cm.$$
$$ \therefore \quad The\quad radius\quad of\quad the\quad outer\quad circle=17.5cm.$$
$$ Ans-\quad Option\quad B. $$
The diameter of a circle is $$28 cm$$. Find the o
ne-fourth of its area.
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$$389 cm^{2}$$
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$$512 cm^{2}$$
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$$616 cm^{2}$$
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$$154 cm^{2}$$
Explanation
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, $$\dfrac 14 \times 616 = 154 cm^2$$
Area of the shaded region is
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$$96\ m^2$$
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$$15\ m^2$$
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$$81\ m^2$$
0%
$$111\ m^2$$
Explanation
We know that area of rectangle of sides $$l$$ & $$b$$ = $$ l \times b $$.
Area of the shaded region
$$=$$ Area of bigger rectangle $$-$$ area of smaller rectangle
$$=\left( 12\times 8 \right) -\left( 5\times 3 \right) $$
$$=96-15=81\ m^2$$
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