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CBSE Questions for Class 7 Maths Perimeter And Area Quiz 5 - MCQExams.com
CBSE
Class 7 Maths
Perimeter And Area
Quiz 5
The area of the cross-section;
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133.5 $$cm^{3}$$
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187.5 $$cm^{3}$$
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138.5 $$cm^{3}$$
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183.5 $$cm^{3}$$
Explanation
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
$$=AB\times AD+\dfrac{1}{2}\times CE\times EF$$
$$=10\times 12+\dfrac{1}{2}\times 6\times 4.5$$
$$=133.5cm^2$$
Find the area of a ring shaped region enclosed between two concentric circles of radii $$20$$ cm and $$15$$ cm.
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550 $$cm^{2}$$
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425 $$cm^{2}$$
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496 $$cm^{2}$$
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810 $$cm^{2}$$
Explanation
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
The following figure shows a square cardboard $$ABCD$$ of side $$28\ cm.$$ Four identical circles of the largest possible size are cut from this card as shown below.
Find the area of the remaining cardboard.
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$$168$$ $$cm^{2}$$
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$$158$$ $$cm^{2}$$
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$$148$$ $$cm^{2}$$
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$$138$$ $$cm^{2}$$
Explanation
Let the radius of each circle be $$r.$$
Then,
$$4r=28\ cm$$
$$r=7\ cm$$
Therefore,
Area of $$4$$ circle $$=4\times \pi (7)^2$$
$$=4\times \dfrac{22}{7}\times 7\times 7$$
$$=616\ cm^2$$
Area of square $$=28^2$$
$$=784\ cm^2$$
Area of remaining cardboard $$=$$ Area of square $$-$$ Area of $$4$$ circles
$$=784-616$$
$$=168\ cm^2$$
If the difference between the circumference and diameter of a circle is $$30 cm$$ then what is the radius of the circle?
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$$7 cm$$
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$$15 cm$$
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$$22 cm$$
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$$2 cm$$
Explanation
Let r be the radius of the circle
Then $$\displaystyle 2\pi r-2r=30\Rightarrow 2r\left ( \pi -1 \right )=30$$
$$\displaystyle \Rightarrow r\left ( \frac{22}{7}-1 \right )=15\Rightarrow r\times \frac{15}{7}=15\Rightarrow r=\frac{15\times 7}{15}=7\:cm$$
The area of a circular plot is $$3850$$ square meters. What is the circumference of the plot ?
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$$240 \text{ m}$$
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$$210 \text{ m}$$
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$$220 \text{ m}$$
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$$260 \text{ m}$$
Explanation
If $$r$$ is the radius of the plot then area is given as $$\pi r^2$$
as per the question,
$$\displaystyle\pi r^{2}=3850$$
$$\Rightarrow r^{2}=\dfrac{3850\times 7}{22}=1225$$
$$\displaystyle \Rightarrow r= 35\text{ m}$$
since circumference $$=2\pi r$$
$$=2\times \dfrac{22}{7}\times 35\text{ m}$$
$$=220\text{ m}$$
In the given figure, the area enclosed between two concentric circles is $$808.5\ cm^2$$. The circumference of the outer circle is $$242\ cm$$. Calculate t
he radius of the inner circle
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$$32\ cm$$
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$$35\ cm$$
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$$23\ cm$$
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$$48\ cm$$
Explanation
Let the radius of the inner circle be $$r$$ and the 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, $$\pi(R^2-r^2)=808.5$$
$$\Rightarrow {\left(\cfrac{77}{2}\right)}^2-r^2=\cfrac{808.5 \times 7}{22}$$
$$\Rightarrow \cfrac{5929}{4}-r^2=257.25$$
$$\Rightarrow r^2=1225$$
$$\Rightarrow r=35$$
Thus, Radius of the inner circle $$=35\ cm$$
The area enclosed between two concentric circles is $$770$$ $$cm^2$$. If the radius of the outer circle is $$21$$ $$cm$$, calculate the radius of the inner circle.
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$$7$$ $$cm$$
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$$14$$ $$cm$$
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$$2.1$$ $$cm$$
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$$35$$ $$cm$$
Explanation
Let radius of the inner circle be $$r$$ and radius of the outer circle be $$R=21cm$$
Area enclosed between the two concentric circles $$=\pi(R^2-r^2)=770cm^2$$
$$\Rightarrow 770=\pi(R^2-r^2)$$
$$\Rightarrow 770=\dfrac{22}{7}(21^2-r^2)$$
$$\Rightarrow \dfrac{770\times 7}{22}=441-r^2$$
$$\Rightarrow 245=441-r^2$$
$$\Rightarrow r^2=196$$
$$\Rightarrow r=\sqrt{196}$$
$$\Rightarrow r=14$$
Thus, radius of the inner circle $$=14$$ $$cm$$
A circular ground whose diameter is $$140$$ meters is to be fenced by wire three times around its circumference. Find the length of wire needed.
$$[$$use $$\displaystyle \pi = \frac {22}{7}$$$$]$$
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$$440m$$
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$$1320m$$
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$$660m$$
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None of these
Explanation
Diameter of circular ground $$=140m$$.
Radius of circular ground $$=70m$$.
Circumference of circular ground $$=2\pi r$$
$$\displaystyle =2\times \frac {22}{7}\times 70$$
$$=440m$$
Length of wire needed to fenced three times
$$=440\times 3$$
$$=1320m$$
Find the the perimeter of the figure given correct to one decimal place.
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$$56.0m$$
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$$56.6m$$
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$$57.7m$$
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$$57.9m$$
Explanation
The perimeter of the given figure $$= AB + CD + EF + GH + 4 \times$$
( Perimeter of a quadrant of the circle ) $$....(1)$$
Now, $$AB = EF = 20 m - (2 m + 2 m) = 16 m$$
Similarly, $$CD = GH = 10 m - (2 m + 2 m) = 6 m$$
Now, the perimeter of the quadrant having radius $$2 \,\,m\,\,$$
$$ = \dfrac { 2\pi r }{ 4 }$$
$$= \dfrac { \pi r }{ 2 }$$
$$ =\dfrac { \pi \times 2 }{ 2 }$$
$$ =\pi \,\, $$
So, the perimeter of the $$4$$ quadrants having radius $$ 2\,\, m \,\,= \,\,4\pi\,\, $$
Put all the respective values in $$(1)$$, we get
Perimeter of the given figure $$= 16 m + 6 m + 16 m + 6 m + 4$$
$$\pi \,\,\, $$
$$= 44 m + (4 \times 3.14 )$$
$$ = 44 + 12.56 $$
$$= 56.56 m \sim 56.6 m$$
Hence, the perimeter of the given figure correct to one decimal place is 56.6 m.
The inner circumference of a circular track $$14\ m$$ wide is $$440\ m$$. The radius of the outer circle is
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$$70\ m$$
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$$56\ m$$
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$$77\ m$$
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$$84\ m$$
Explanation
Let $$R $$ and $$r $$ be tha radii of the outer and inner circle.
So by given $$R-r = 14m$$
Inner circumference $$= \displaystyle 2\pi r=2\times \frac{22}{7}\times r=\frac{44r}{7}$$
Given $$\displaystyle \frac{44r}{7}=440\Rightarrow r=\frac{440\times 7}{44}=70\ m$$
$$\displaystyle \therefore $$ Radius of outer circle $$= 70\ m + 14\ m = 84\ m$$
What is the perimeter of the given figure correct to one decimal place?
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$$56.6$$ m
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$$46.5$$
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$$56.7$$
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$$23.7$$
Explanation
Perimeter(P)=AB + EF + CD + GH + 4 x Arc AH
$$\displaystyle =2\times AB+2\times EF+$$ Circumference of circle with radius 2
$$\displaystyle =2\times (20-4)m+2\times (10-4)m+2\times \frac{22}{7}\times 2$$
= 32 m + 12 m + 12.57 m =56.6m
The circumference of a circle is $$44$$ m then the area of the circle is
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$$\displaystyle 6084.5\:m^{2}$$
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$$\displaystyle 276.5\:m^{2}$$
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$$\displaystyle 154\:m^{2}$$
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$$\displaystyle 44\:m^{2}$$
Explanation
Circumference of circle: $$44$$ m
Let the radius of circle be $$r$$.
Therefore, $$2\pi r = 44$$
$$\Rightarrow r = 7$$ m
Area of circle:
$$\pi r^2 = \dfrac{22}7\times7^2 = 154\ m^2$$
The circumference of a circular field is $$528\ m$$. Then
its radius is
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$$84\ m$$
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$$80\ m$$
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$$76\ m$$
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$$78\ m$$
Explanation
Let $$r$$ be the radius of the circle
Circumference of the circle $$=528$$
$$\Rightarrow 2\pi r=528$$
$$\Rightarrow r=\dfrac{528\times 7}{22\times 2}$$
$$\Rightarrow r=84\ m$$
A circular park has a path of uniform width around it. The difference between outer and inner circumference of the circular path is $$132\ m$$. Its width is _____ $$\displaystyle \left(\pi=\frac{22}{7}\right)$$
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$$22\ m$$
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$$20\ m$$
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$$21\ m$$
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$$24\ m$$
Explanation
Given $$\displaystyle 2\pi r_{2}-2\pi r_{1}=132$$
$$\displaystyle \Rightarrow r_{2}-r_{1}=\frac{132}{2\pi }=\frac{132\times 7}{44}=21\ m$$
Find the area of the parallelogram whose base is $$17\ cm$$ and height $$0.8\ m$$?
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$$\displaystyle 13.6\:cm^{2}$$
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$$\displaystyle 1360\:cm^{2}$$
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$$\displaystyle 13.6\:m^{2}$$
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$$\displaystyle 1360\:m^{2}$$
Explanation
Base $$= 17\ cm$$
Height $$= 0.8\ m =0.8 \times 100 = 80\ cm$$
Area of parallelogram
$$= b \times h$$
$$= 17\times 80$$
$$= 1360\ cm^2$$
If the diameter of a circle is $$14$$ cm, then its circumference is
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$$66\;cm$$
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$$44\;cm$$
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$$33\;cm$$
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$$55\;cm$$
Explanation
Since, diameter $$= d=14\;cm$$
$$\displaystyle C=\pi d=\frac{22}{7}\times 14=44\;cm$$
The length and breadth of a rectangle are in the ratio $$3:2$$. If the sides of the rectangle are extended on each side by $$1 m$$, the ratio of length to breadth becomes $$10:7$$. Find the area of the original rectangle in square meters.
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$$2350\ m^{2}$$
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$$1150\ m^{2}$$
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$$54\ m^{2}$$
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$$1000\ m^{2}$$
Explanation
Let length of rectangle$$=3x$$
breadth of rectangle $$=2x$$
Now,
$$ \cfrac { 3x+1 }{ 2x+1 } =\cfrac { 10 }{ 7 } \\ 21x+7=20x+10\\ x=3$$
$$\therefore$$ length$$=3\times 3=9m,$$ breadth$$=2\times 3=6m$$
Area of original rectangle
$$=(9\times 6){ m }^{ 2 }\\ =54\ { m }^{ 2 }$$
The area of a circle is $$\displaystyle 2464\:m^{2}$$, then the diameter is
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$$56$$ m
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$$154$$ m
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$$176$$ m
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None of the above
Explanation
Since, Area of circle $$= 2464m^2$$
$$\therefore \displaystyle \frac{\pi d^{2}}{4}=2464$$
$$\Rightarrow \displaystyle d^{2}=\dfrac{2464\times 4}{\dfrac{22}{7}}=3136$$
$$ \Rightarrow \displaystyle d=\sqrt{3136}=56m$$
In the given figure, $$P$$ is the center of the circle. The area and perimeter respectively, of the figure are
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$$3256 cm^{2}$$, $$234.2 cm$$
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$$2492 cm^{2}$$, $$229.2 cm$$
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$$2942 cm^{2}$$, $$234.2 cm$$
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$$3256 cm^{2}$$, $$229.2 cm$$
Explanation
Area of figure$$=$$area of circle + area of rectangular - area of quarter circle.
$$ =\pi { r }^{ 2 }+l\times b-\cfrac { 1 }{ 4 } \times \pi { r }^{ 2 }\\ =\cfrac { 3 }{ 4 } \pi { r }^{ 2 }+l\times b=\cfrac { 3 }{ 4 } \times 3.14\times 20\times 20+50\times 40\\ =942+2000\\ =2942c{ m }^{ 2 }$$
&
Perimeter of figure
$$=2\pi r-\cfrac { 2\pi r }{ 4 } +2(l+b)-2r\\ =\cfrac { 3\pi r }{ 2 } +2(l+b)-2r\\ =\cfrac { 3\times 3.14\times 20 }{ 2 } +2(50+40)-2\times 20\\ =94.2 +180-40\\ =234.2cm$$
A field is in the form of a parallelogram whose base is $$420\ m$$ and altitude is $$36 m$$. Find the cost of watering at $$10$$ paise per sq. m.
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Rs. $$15.120$$
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Rs. $$1512$$
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Rs. $$151.20$$
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None
Explanation
Base $$= 420\ m$$
Height $$= 36\ m$$
Area $$= b \times h = 420 \times 36$$
$$= 15,120 \displaystyle \ m^{2}$$
The cost of watering per sq m
$$= 10$$ paise
Cost of watering the field $$= 15120 \times 0.1$$
Rs. $$1512$$
The degree measure of the circumference of the circle is always
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$$ \displaystyle 0^{\circ} $$
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$$ \displaystyle 90^{\circ} $$
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$$ \displaystyle 180^{\circ} $$
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$$ \displaystyle 360^{\circ} $$
Explanation
$$\Rightarrow$$ 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 $$360^o.$$
The ratio between the length and breadth of a rectangular garden is $$5:3$$. If the perimeter of the garden is $$160$$ meters, what will be the area of $$5\text{ m}$$ wide road around its outside?
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$$600\displaystyle \text{ m}^{2}$$
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$$1200\displaystyle \text{ m}^{2}$$
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$$900\displaystyle \text{ m}^{2}$$
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$$1000\displaystyle \text{ m}^{2}$$
Explanation
Let the length and breadth of the rectangular garden be $$5x$$ meter and $$3x$$ meters.
Given $$2(5x + 3x) = 160 $$
$$ \Rightarrow 16x = 160 $$
$$ \Rightarrow x = 10$$
$$\therefore $$ Length of garden $$= 50 \text{ m}$$ and Breadth of garden $$= 30 \text{ m}$$
Area of $$5 \text{ m}$$ wide road around its outside
$$=$$ Area of outer rectangle $$-$$ Area of inner rectangle
$$[\because $$ length of outer rectangle $$= 50+2\times 5$$ and breadth of outer rectangle $$=30+2\times 5 ]$$
Area $$= (50 + 10) \times (30 + 10) - 50 \times 30$$
$$= 60 \times 40 - 50 \times 30$$
$$= (2400 - 1500)$$
$$ = 900 \text{ m}^{2} $$
$$\therefore$$ Area of road is $$900\text{ m}^2$$
Circumference of a circle is equal to
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$$\pi\;r$$
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$$2\pi\;r$$
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$$\displaystyle \frac{\pi r}{2}$$
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$$\displaystyle2+ \frac{\pi r}{2}r$$
Explanation
$$c=\pi d=2\pi r$$
A wire is bent into the shape as shown. It is made up of $$5$$ semicircles. What is the length of the wire? (Take $$\pi =3.14$$)
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$$27\ cm$$
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$$42.39\ cm$$
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$$45\ cm$$
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$$27.92\ cm$$
Explanation
Let the length of the wire be $$l.$$
The length of the wire is equal to the arc length of the semi-circles drawn on the diameters $$AB,\ BC,\ CD,\ EF$$ and $$ED.$$
$$l= \pi \times 1.5 +\pi \times 3 +\pi \times 4.5 +\pi \times 3 +\pi \times 1.5 $$
$$= \pi \times \left ( 1.5+3+4.5+3+1.5 \right )$$
$$ = 3.14 \times 13.5 $$
$$= 42.39\ cm$$
The number of revolutions a wheel of diameter $$40 cm$$ makes in travelling a distance of $$176 m$$ is: $$\displaystyle \left ( \pi =\frac{22}{7} \right )$$
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$$140$$
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$$150$$
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$$160$$
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$$166$$
Explanation
Diameter of wheel$$=40cm,$$ Radius $$=20$$cm
Distance covered in $$1$$ revolution$$=2\pi r=2\times \cfrac { 22 }{ 7 } \times 20\\ =\cfrac { 880 }{ 7 } cm$$
total distance$$ =176m=17600cm$$
$$ \therefore $$ No. of revolutions$$=\cfrac { 17600 }{ 880/7 } =140$$
Find the area of the figure shown above
.
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$$9$$ sq cm
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$$10$$ sq cm
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$$11$$ sq cm
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$$15$$ sq cm
Explanation
Area $$=$$ (Area of rectangle $$ABDC$$) $$\times 2+$$ Area of rectangle $$ DGEF$$
$$=2\times \left( 3\times 1 \right) +5\times 1$$
$$=6+5$$
$$=11$$ sq cm
Samuel wanted to implant some vertical
stones along the boundary of his plot at a distance of $$10$$ m each. If length of the plot is $$30$$ m and the breadth is $$15$$ m, then the number of stones used is:
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$$450$$
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$$45$$
0%
$$9$$
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$$10$$
Explanation
Perimeter of plot$$=2(l+b)=2(30+15)=90m$$
Distance between two stones$$=10m$$
Therefore, number of stones$$=\cfrac { 90 }{ 10 } =9$$
Area of the shaded figure is:
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$$2400\ sq. m$$
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$$48\ sq. m$$
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$$50\ sq. m$$
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$$98\ sq. m$$
Explanation
We know that area of rectangle of sides $$l$$ & $$b$$ = $$ l \times b $$
Area of the shaded region
$$=$$ Area of $$A+$$ Area of $$B$$
$$=8\times 6+10\times 5$$
$$=48+50=98\ sq.\ m$$
The adjacent sides of a parallelogram are $$8\ m$$ and $$5\ m$$. The distance between the longer sides is $$4\ m$$. What is the distance between the shorter sides?
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$$6.4\ m$$
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$$6\ m$$
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$$8.6\ m$$
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$$None\ of\ these$$
Explanation
Area of the parallelogram $$=$$ Longer side $$\times$$ Distance between them
$$= 8 m \times 4 m = 32 \displaystyle\ m^{2}$$
Also, area of the parallelogram $$=$$ Shorter side $$\times$$ Distance between shorter sides
$$ \displaystyle \Rightarrow 32\ m^{2}=5\ m \times$$ Distance between shorter sides
$$ \displaystyle \Rightarrow $$ Distance between shorter sides $$ \displaystyle \frac{32\ m^{2}}{5\ m}$$
$$= \displaystyle 6.4\ m$$
The base of a parallelogram is three times its height. If the area of the parallelogram is $$75$$ sq cm, then its height is
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$$5 cm$$
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$$\displaystyle 5\sqrt{2}cm$$
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$$\displaystyle 3\sqrt{2}cm$$
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$$15 cm$$
Explanation
let height of ||g be 'b', then base of ||g be 3b.
area of ||g$$=$$base$$\times $$height
$$75c{ m }^{ 2 }=3b\times b\\ { b }^{ 2 }=25$$
$$b=5$$cm, therefore height $$=5$$cm
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