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CBSE Questions for Class 7 Maths Perimeter And Area Quiz 10 - MCQExams.com
CBSE
Class 7 Maths
Perimeter And Area
Quiz 10
If the perimeter of a circle is $$132$$ cm, find its area.
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$$1356$$ sq. cm
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$$1386$$ sq. cm
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$$1340$$ sq. cm
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$$1436$$ sq. cm
Explanation
Let the radius of the circle is $$r$$ cm
Perimeter of a circle $$=2\pi r=132$$
$$ \Rightarrow 2\times \cfrac{22}{7}\times r=132$$
$$\Rightarrow r=\cfrac{132\times 7}{22\times 2}$$
$$\Rightarrow r=21$$ cm
Area of the circle $$=\pi r^{2}$$
$$ \Rightarrow =\cfrac{22}{7}\times 21\times 21$$
$$ \Rightarrow =1386$$
$$cm^{2}$$
Find the area of a circle whose radius is $$6$$ $$cm$$.
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$$26$$ $$ c{m^2}$$
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$$46$$ $$ c{m^2}$$
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$$56$$ $$ c{m^2}$$
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$$36$$ $$ c{m^2}$$
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none
Explanation
The radius of circle is $$6$$ $$cm$$
The area of circle is given as $$πr^2$$
$$\implies π(6)^2$$ $$cm^2$$
$$\implies 36π$$ $$cm^2$$
One diagonal of a parallelogram is $$7cm$$ and the perpendicular distance of this diagonal from either of the outlying vertices is $$27cm$$. The area of the parallelogram (in sq.cm) is:
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$$180$$
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$$183$$
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$$189$$
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$$198$$
Explanation
Area of parallelogram $$=2 \times $$area of $$\triangle ACD$$
$$=2\times \dfrac{1}{2}\times 27 \times 7$$
$$189\ cm^2$$
A race track is in the form of ring whose inner and outer circumstances are $$352\ meters$$ and $$396\ meters$$ respectively. Find the width of the track.
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$$7\ meters$$
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$$14\ meters $$
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$$14\pi\ meters $$
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$$7\pi\ meters $$
Explanation
Let, $$R, r $$ be the outer and inner radius of the track then,
$$2\pi R = 396 \ m$$
$$2\pi r = 352 \ m $$
Now , $$2\pi R - 2\pi r = 396 - 352$$
$$\implies 2\pi (R - r) = 44 $$
$$\implies 2\times \dfrac{22}{7} \times (R - r) = 44 $$
$$\implies R - r = 44 \times \dfrac{7}{44}$$
$$\therefore R - r = 7 \ m $$
i.e., The width of the track $$ = 7 \ m $$
Option A is correct.
The radius of a circular wheel is $$1.75\ m$$. The number of revolutions that it will make in covering $$11\ kms$$ is:
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$$1000$$
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$$100000$$
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$$100$$
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$$10$$
Explanation
Let It take 'n' revolution
Then $$ n\times 2\pi r = 11,000$$
$$\displaystyle \Rightarrow n\times2\times \frac{22}{7}\times 1.75 = 11000$$
$$\displaystyle \therefore n = 1000$$
If the area of a parallelogram is $$144 \operatorname { cm } ^ { 2 }$$ and its base is $$9 cm$$. then its height is
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$$8 cm$$
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$$12 cm$$
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$$24 cm$$
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$$16 cm$$
Explanation
Area of parallelogram $$= $$ base$$\times$$ height
Given : base $$=9\ cm,$$
area $$=144\ cm^2$$
Let height be $$h$$
$$\Rightarrow$$ $$144\ {cm}^{2}=9cm\times h$$
$$\Rightarrow$$ $$h$$$$=\cfrac{144\ {cm}^{2}}{9\ cm}$$
$$\therefore$$ $$h=16\ cm$$
If the circumference of a circle is 88 cm, then the area of a circle (in sq. cm) is:
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$$196\pi $$
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$$92\pi $$
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$$64\pi $$
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$$48\pi $$
Explanation
Given,
Circumference of the circle,
$$2\pi r=88$$
$$\therefore r=\dfrac{44}{\pi}=14$$
Area of circle,
$$A=\pi r^2$$
$$=\pi \times 14^2$$
$$=196 \pi$$
The area of a circle whose diameter is 1.4 cm is
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$$1.54\ cm^2 $$
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$$0.77\ cm^2 $$
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$$6.16\ cm^2 $$
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None of these
Explanation
Area of cirlce
$$=\pi r^2$$
$$=\pi \left ( \dfrac{d}{2} \right )^2$$
$$=\dfrac{22}{7} \left ( \dfrac{1.4}{2} \right )^2$$
$$=1.54m^2$$
Find the circumference of the circles with the radius 14cm :$$(Take \ \pi = \frac{22}{7})$$
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88 cm
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28 mm
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21 cm
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90 cm
Explanation
Circumference $$ = 2\pi r $$
$$ r = 14 \,cm $$
$$ = 2\pi r = 2\times \frac{22}{7}\times 14 $$
$$ = 2\times 22\times 2 $$
$$ = 4\times 22 $$
$$ = 88 \,cm $$
An ant is moving around a few food pieces of different shapes scattered on the floor. For which food-piece would the ant have to take a longer round? Remember, circumference of a circle can be obtained by using the expression $$c =2\pi r$$, where r is the radius of the circle.
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0%
0%
0%
$$ None\ of\ these$$
Explanation
$$(a)$$Diameter$$=2.8$$cm$$\Rightarrow$$Radius$$=\dfrac{2.8}{2}=1.4$$cm
Perimeter of a circle$$=2\pi r\Rightarrow $$ Perimeter of a semicircle$$=\dfrac{2\pi r}{2}=\pi r$$
$$\therefore$$ Perimeter of the figure$$=\pi r+$$ diameter
$$=\dfrac{22}{7}\times\dfrac{14}{10}+2.8$$cm
$$=4.4+2.8=7.2$$cm
$$(b)$$Perimeter of the semi-circular part$$=\pi r$$
$$=\dfrac{22}{7}\times 1.4$$cm
$$=4.4$$cm
Perimeter of the remaining part$$=1.5+2.8+1.5=5.8$$cm
Total perimeter$$=4.4+5.8=10.2$$
$$(c)$$Perimeter of the semi-circular part$$=\dfrac{\pi d}{2}=\dfrac{22}{7}\times \dfrac{2.8}{2}=4.4$$cm
$$\therefore$$ Perimeter of the figure$$=4.4+2+2=8.4$$cm
Since,$$7.2$$cm$$<8.4$$cm$$<10.2$$cm
$$\therefore$$ perimeter of figure $$b$$ has the longest round.
Find the area of circle whose circumference is $$220cm$$
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3850
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3500
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3700
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3900
Explanation
Circumference of circle $$=220\,cm$$ [ Given ]
Circumference of circle $$=2\pi r$$
$$\Rightarrow$$ $$220=2\times\dfrac{22}{7}\times r$$
$$\Rightarrow$$ $$r=220\times \dfrac{7}{22}\times\dfrac{1}{2}$$
$$\Rightarrow$$ $$r=5\times 7$$
$$\therefore$$ $$r=35\,cm^2$$
$$\Rightarrow$$ Area of circle $$=\pi r^2$$
$$=\dfrac{22}{7}\times 35\times 35$$
$$=22\times 5\times 35$$
$$=3850\,cm^2$$
The area of the concentric circles are $$962.5\ cm^{2}$$ and $$1368\ cm^{2}$$. Find the width of the ring formed by them.
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$$2.1\ cm$$
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$$1 cm$$
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$$5.5\ cm$$
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$$3.5\ cm$$
Explanation
Let the radii of the bigger and smaller circles be $$R\space\mathrm{cm}$$ and $$r \space\mathrm{cm}$$, respectively.
Now,
Area of the bigger circle $$=\pi R^2$$
$$\Rightarrow 1386=\frac{22}{7}\times R^2$$
$$\Rightarrow R^2=\frac{1386\times7}{22}$$
$$\Rightarrow R^2=441$$
$$\Rightarrow R=21$$
$$\therefore$$ Radius of bigger circle $$(R)=21\space\mathrm{cm}$$
Similarly,
Area of the smaller circle $$=\pi r^2$$
$$\Rightarrow 962.5=\frac{22}{7}\times r^2$$
$$\Rightarrow r^2=\frac{962.5\times7}{22}$$
$$\Rightarrow r^2=306.25$$
$$\Rightarrow r=17.5$$
$$\therefore$$ Radius of smaller circle $$(r)=17.5\space\mathrm{cm}$$
Hence, Width of Ring $$= R-r=21-17.5=3.5\space\mathrm{cm}$$
Thus, $$\text{D}$$ is the correct option.
The area of parallelogram if the base is $$36cm$$ and height is $$45cm$$
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1620
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1800
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1250
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1640
Explanation
The base of parallelogram is $$36cm$$
The height is $$45cm$$
The area of parallelogram is $$36\times 45=1620cm^2$$
The area of triangle with base $$4\ cm$$ and height is $$3\ cm$$
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$$6\ cm^2$$
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$$12\ cm^2$$
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$$24\ cm^2$$
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$$3\ cm^2$$
Explanation
The area of triangle is
$$=\dfrac 12\times height \times base \\=\dfrac 12\times 3\times 4=3\times 2=6\ cm^2$$
If the area of a circle is $$154{ cm }^{ 2 }$$, then the perimeter is
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11 cm
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22 cm
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44 cm
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55 cm
If The side and altitude of a parallelogram are 9 cm and 6 cm respectively, then its area is
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$$27 cm ^{2}$$
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$$236cm ^{2}$$
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$$54 cm ^{2}$$
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$$45 cm ^{2}$$
Explanation
the area of a parallelogram is the product of a side and height or altitude.
here Side =$$9 cm$$, Altitude=$$6cm$$
so, area= $$9\times6$$sq. cm=$$54 $$sq. cm
The radius of two concentric circles is $$4\ cm$$ and $$3\ cm.$$ The area of the bounded region will be:-
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$$22\ cm^{2}$$
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$$12\ cm^{2}$$
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$$32\ cm^{2}$$
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$$18\ cm^{2}$$
Explanation
Given:-
$$R=4\ cm$$
$$r=3\ cm$$
Hence, the
area of the bounded region will be,
$$A=\pi (R^2-r^2)$$
$$=\dfrac {22}7(16-9)$$
$$=\dfrac {22}7\times 7\ cm^2$$
By converting the 80.2km into the hectare, the answer will be
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0.08020ha
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8020ha
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802ha
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0.802ha
By converting the 4.8mm into the cm, the answer will be
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0.048cm
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0.48cm
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48cm
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480cm
Ram has a plate in circular form which has design in $$7$$ meter radius .Find the area of design in $$cm^2$$
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$$154cm^2$$
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$$15400cm^2$$
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$$1540000cm^2$$
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$$.154cm^2$$
Explanation
Radius of circular plate $$(r)=7\space\mathrm{m}$$
$$=7\times100\space\mathrm{cm}\quad[\because 1\space\mathrm{m}=100\space\mathrm{cm}]$$
$$=700\space\mathrm{cm}$$
Now, Area of design $$=\pi r^2$$
$$=\dfrac{22}{7}\times700^2$$
$$=1540000\space\mathrm{cm^2}$$
So, $$\text{C}$$ is the correct option.
By converting the $$0.0287m^2$$ into the $$cm^2$$, the answer will be
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$$28.7 \, cm^2$$
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$$2870 \, cm^2$$
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$$287 \, cm^2$$
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None of these
Area of triangle PQR is $$100\, cm^2$$ (Fig ). If altitude QT is $$10\, cm$$, then its base PR is.
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$$ 20 \,cm$$
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$$ 15 \,cm$$
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$$ 10 \,cm$$
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$$5 \,cm$$
Explanation
Option (a) is correct.
Give, area of triangle $$PQR =100\, cm^2$$.
We know drat Area of triangle $$PQR \times \dfrac {1}{2} PR \times QT$$
$$ \Rightarrow 100 = \dfrac {1}{2} \times PR \times 10$$
$$ \Rightarrow PR =\dfrac {100\times 2}{10}$$
$$ \Rightarrow PR = 20\, cm$$
Area of parallelogram ABCD is not equal to.
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$$DE \times DC $$
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$$BE \times AD$$
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$$BF \times DC $$
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$$ BE \times DC$$
Explanation
Option (a) is correct.
Area of parallelogram $$=$$ Base $$\times$$ corresponding height
Hence, area of parallelogram $$ABCD $$
$$AD \times BE =BC \times BE $$
$$\because [AD//BC] $$
Also,
$$BF\times CD =BF\times AB$$
$$\because[AB //CD]$$
Ratio of areas of $$\Delta MNO, \Delta MOP $$ and $$ \Delta MPQ$$ in Fig. is
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$$2: 1: 3 $$
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$$1: 3:2 $$
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$$2: 3: 1 $$
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$$ 1:2:3$$
Explanation
Option (a) is correct
Area of triangle $$MNO = \frac {1}{2} \times NO \times MO$$
$$=10\, cm^2$$
Area of triangle $$MOP= \frac{1}{2} \times PQ \times MQ $$
$$\frac {1}{2} \times 2 \times 5$$
$$= 5\, cm^2$$
Therefore, $$=105$$
Area of triangle $$MOP= \frac{1}{2} \times PQ \times MQ $$
$$\frac {1}{2} \times 6 \times 5$$
$$= 15\, cm^2$$
Therefore, required ratio $$ =10:5:15 =2:1:3$$
In Fig, if $$PR = 12\, cm, QR = 6\, cm$$ and $$PL= 8\, cm$$, then QM is
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$$ 6\, cm$$
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$$ 9\, cm$$
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$$4\, cm$$
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$$ 5\, cm$$
Explanation
Given that $$PL =8\, cm$$
$$PR =12\,cm$$
$$ QR = 6\, cm$$
Now, in right-angled triangle PLR
Using Pythagoras theorem,
$$ \Rightarrow PR^2 = PL^2 + LR^2$$
$$ \Rightarrow LR^2 = 144 - 64$$
$$ =\sqrt {80} = 4\sqrt {5}\ cm$$
$$LR = LQ +QR$$
$$ \Rightarrow LQ = LR - QR = (4\sqrt{5} - 6)\ cm$$
Now area of triangle PLR
$$A =\dfrac {1}{2}\times \left ( 4\sqrt {5} \right )\times 8$$
$$= 16\sqrt{5}\ cm^2$$
Now, area of triangle PLQ
$$A =\dfrac {1}{2}\times \left ( 4\sqrt {5} - 6\right )\times 8$$
$$= (16\sqrt{5}-24) \,cm^2$$
Hence, area of triangle PLR $$=$$ Area of triangle PLQ $$ +$$ Area of triangle PQR
So, area of triangle PLR $$-$$ Area of triangle PLQ $$ =$$ Area of triangle PQR
$$ \Rightarrow 16 \sqrt {5} - (16 \sqrt {5} -24)$$,
$$ \Rightarrow $$ Area of triangle PQR $$= 24 \, cm^2$$
$$ \Rightarrow \dfrac {1}{2}\times PR\times QM = 24$$
$$ \Rightarrow \dfrac {1}{2} \times 12 \times QM =24$$
$$ \therefore QM = 4\,cm$$
A wire is bent to form a square of side $$22\, cm$$ If the wire is rebent to form a circle, its radius is.
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$$22\, cm$$
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$$14\, cm$$
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$$11\, cm$$
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$$7\, cm$$
Explanation
Option (b) is correct.
Given, side of a square is $$22\, cm$$ The wire has same length.
Hence, perimeter of square arid circumference of circle are equal.
ACQ,
Circumference of circle $$=$$ Perimeter of square
$$= 2 \times \pi \times r = 4\times (side)$$
$$ \Rightarrow 2\times \dfrac {22}{7}\times r = 4\times 22$$
$$\Rightarrow r =\dfrac {4\times 22 \times 7}{2\times 22}$$
$$ \Rightarrow r= 14\, cm$$
Therefore, $$14\, cm$$ is the radius of the circle.
Area of circular garden with diameter $$8\, m$$ is.
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$$12.56\, m^2$$
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$$ 25.12 \, m^2$$
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$$50.28 \, m^2$$
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$$200.96 \, m^2$$
Explanation
Given, diameter is $$8\;\text{m}$$
Hence, radius $$ =\dfrac{8}{2}\;\text{m}= 4\;\text{m}$$
Therefore, area of the circular garden
$$=\pi r^2$$
$$=\dfrac {22}{7}\times 4\times 4$$
$$= 50.28\;\text{m}^2$$
In Fig., EFGH is a parallelogram, altitudes FK and Fl are 8 cm and ,10 cm respectively. If $$EF =10\, cm$$, then area of EFGH is
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$$ 20\, cm^2$$
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$$ 32\, cm^2$$
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$$ 40\, cm^2$$
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$$ 80\, cm^2$$
Explanation
Option (c) is correct
Area of parallelogram (EFGH) $$=$$ Base $$\times$$ corresponding height
$$ = 10\times 4$$
$$ = 40\, cm^2$$
Area of a rectangle and the area of a circle are equal. If the dimensions of the rectangle are $$10\,cm \times 11\, cm$$, then radios of the circle is
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$$21\, cm$$
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$$10.5\, cm$$
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$$14\, cm$$
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$$7\, cm$$
Explanation
Option (d) is correct.
Given, Length of rectangle $$=l=14 \,cm$$
Breadth of rectangle $$= b =11\,cm$$
Area of circle $$=$$ Area of rectangle
$$\Rightarrow \pi r^2 = l\times b$$
$$\Rightarrow \dfrac {22}{7}\times r^2 = 14\times 11$$
$$\Rightarrow r^2 =\dfrac {14\times 11\times 7}{22}$$
$$\Rightarrow r^2 =\sqrt {49}$$
$$ \Rightarrow r= 7\, cm$$
Therefore, $$7\, cm$$ is the radius of the circle.
Ratio of circumference of a circle to its radius is always $$2 \pi : 1$$
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0%
True
0%
False
Explanation
We know that, for a circle with radius $$r$$,
the circumference $$=2\pi r$$
$$\therefore \ \text{Circumference : Radius}$$ $$= 2 \pi r : r =2\pi :1$$.
Hence, the given statement is true.
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