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CBSE Questions for Class 7 Maths Perimeter And Area Quiz 6 - MCQExams.com
CBSE
Class 7 Maths
Perimeter And Area
Quiz 6
Expenditure incurred in cultivating a square field at the rate of Rs. $$170$$ per hectare is Rs. $$680$$. What would be the cost of fencing the field at the rate of Rs. $$3$$ per meter?
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Rs. $$2400$$
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Rs. $$3600$$
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Rs. $$3000$$
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Rs. $$2000$$
Explanation
Total cost of cultivation$$=$$Rs.$$680$$
Cost per hectare$$=$$Rs. $$170$$
$$\therefore $$Area of cultivation$$=\cfrac { 680 }{ 170 } =4$$ hectare$$=4\times { 10 }^{ 4 }m^2$$
Now, field's shape is square, therefore area $$=side \times side$$
$$\Rightarrow 4\times { 10 }^{ 4 }={ (side) }^{ 2 }$$
$$\Rightarrow $$ side$$=\sqrt { 4\times { 10 }^{ 4 } } =200m$$
Perimeter of field$$=4\times 200=800m$$
$$\therefore $$ Cost of fencing $$800m=$$ Rs.$$\left( 3\times 800 \right) =$$ Rs.$$2400$$
Front side wall of a house consists of a rectangle of $$6$$ m $$\times$$ $$4$$ m surrounded by a semicircle with base $$4$$ m It has two isosceles triangles make with vertical sides of the rectangle The net area of the wall in sq m is :
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$$\displaystyle 4\left ( 15+{\pi } \right )$$
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$$\displaystyle 4\left ( 18+\pi \right )$$
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$$\displaystyle 4\left ( 24+\pi \right )$$
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$$\displaystyle 64+4\pi$$
Explanation
Net area of wall$$=2\times $$ area of triangle + area of rectangle+area of semi circle
$$= 2 \times \dfrac{1}{2}\times base \times height + length \times breadth + \dfrac {1}{2} πr^2 $$
$$=2\times \cfrac { 1 }{ 2 } \times 6\times 6+6\times 4+\pi \times { (2) }^{ 2 }$$
$$=36+24+4\pi $$
$$=60+4\pi $$
$$=4\left( 15+\pi \right) m^2$$
One side of the square is $$4000$$ meters long. Find is its area in $$km^2$$
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$$1.6$$ $$km^2$$
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$$16$$ $$km^2$$
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$$160$$ $$km^2$$
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$$1600$$ $$km^2$$
Area _____ increases if we increase the perimeter of square or rectangle.
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always
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never
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may or may not be
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none of these
Convert:
$$58.23$$ $$cm^2 = $$ ____ $$mm^2$$
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$$5.823$$
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$$582.3$$
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$$5823$$
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$$58230$$
What is the area of a figure formed by a square of side 8 cm and an isosceles triangle with the base as one side of the square and the other equal sides as 5 cm?
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$$ \displaystyle 70 \ cm^{2}$$
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$$ \displaystyle 76 \ cm^{2}$$
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$$ \displaystyle 82 \ cm^{2}$$
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$$ \displaystyle 64 \ cm^{2}$$
Explanation
Construction: Draw EF perpendicular to AB which will act as a median.
From figure:
Area of given figure$$=ar\triangle ABE+ar \ \square ABCD$$
Now, $$ar\triangle ABE=2\times ar\triangle AFE$$ {$$\because$$ its an isosceles triangle, therefore, EF will act as the median}
$$\triangle AFE$$ is a right-angled triangle, so by Pythagoras
theorem
$$EF=\sqrt { A{ E }^{ 2 }+A{ F }^{ 2 } } $$
$$=\sqrt { { \left( 5 \right) }^{ 2 }+{ \left( 4 \right) }^{ 2 } } =\sqrt { 9 } =3cm$$
$$\therefore ar\triangle AFE=\cfrac { 1 }{ 2 } \times $$ base $$\times $$ height
$$=\cfrac { 1 }{ 2 } \times 4\times 3=6cm^2$$
$$\therefore ar\triangle ABE=2\times 6=12cm^2$$
Also, area of square $$ABCD = (8\times 8) cm^2$$
Now, area of figure$$=12+{ (8) }^{ 2 }=76cm^2$$
The perimeter of an equilateral triangle and a square are same then $$\displaystyle \frac{area\, of \Delta }{area\, of\ \square }=$$
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$$\displaystyle \frac{4}{3}$$
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$$1$$
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$$1.5$$
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$$<1$$
Explanation
Let side of equilateral $$\triangle =a$$
Side of square$$=b$$
Since, perimeter of triangle = perimeter of square
$$\therefore 3a=4b\Rightarrow \cfrac { a }{ b } =\cfrac { 4 }{ 3 } $$
$$\therefore \cfrac { ar\triangle }{ ar\triangle } =\cfrac { \cfrac { \sqrt { 3 } }{ 4 } \times { a }^{ 2 } }{ { b }^{ 2 } } =\cfrac { \sqrt { 3 } }{ 4 } \times { \left( \cfrac { a }{ b } \right) }^{ 2 }=\cfrac { \sqrt { 3 } }{ 4 } \times \cfrac { 16 }{ 9 } =\cfrac { 4 }{ 3\sqrt { 3 } } <1$$
A square and a rectangular field with measurements as given in the following figure have the same perimeter. Which field has a larger area?
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Square has larger area, square $$=3600\;m^2$$.
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Rectangle has larger area, square $$=3600\;m^2$$.
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Square has larger area, square $$=1600\;m^2$$.
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Square has larger area, square $$=3800\;m^2$$.
Explanation
Let $$x$$ be the breadth of the rectangle.
It is given that the perimeter of a rectangle = perimeter of the square.
$$\therefore\;2(80+x)=4\times60$$
$$\Rightarrow\; 80+x=120$$
$$\Rightarrow\;x=120-80=40$$
i.e. breadth of the rectangle $$=40\;m$$
Now, Area of the square $$=(60\times60)\;m^2$$
$$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=3600\;m^2$$.
and the area of the rectangle $$=(40\times80)\;m^2$$.
$$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=3200\;m^2$$.
Hence, the square field has a larger area.
The radius of a circle is $$14 m$$, then the circumference of a circle is
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$$616 m$$
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$$88 m$$
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$$154 m$$
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none of these
Explanation
$$r =14 m$$
Circumference = $$2\pi r$$
$$=\, 2\, \times\, \displaystyle \dfrac {22}{7}\, \times14\, =\, 88\, m$$
The area of a parallelogram is $$120$$ $$cm^{2}$$ and its altitude is $$10$$ cm. The length of the base is
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$$24$$ cm
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$$12$$ cm
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$$8$$ cm
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$$4$$ cm
Explanation
$$Area of parallelogram$$
= $$Base \times height = 120 cm^{2}$$.
$$Base=\cfrac{120}{10}$$
$$\therefore $$ Base = $$12 cm$$
If a wire is bent into the shape of a square, the area of the square is $$81$$ sq cm. When the wire is bent into a semicircular shape, what is the area of the semicircle? $$ \displaystyle \left ( \pi =\frac{22}{7} \right )$$
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$$ \displaystyle 77 \ cm^{2}$$
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$$80 \ cm^2$$
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$$75 \ cm^2$$
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$$70 \ cm^2$$
Explanation
Let $$a$$ be the length of each side of the square.
Then, $$ \displaystyle a^{2}=81\Rightarrow a=9cm$$
$$ \displaystyle \therefore $$ Length of wire $$=$$ Perimeter of square
$$= 4\times 9 = 36 cm $$
When the wire is bent into a semi-circular shape then, perimeter of semi circle $$= \displaystyle \pi r+2r=36 $$
$$ \Rightarrow r=\cfrac{36}{\pi +2}=\cfrac{36}{\dfrac{22}{7}+2}$$
$$=\cfrac{36\times\:7}{\left ( 22+14 \right )}$$
$$=\cfrac{36\times\:7}{36}cm$$
$$=7cm $$
Now,
Area of the semicircle $$ \displaystyle =\frac{1}{2}\pi r^{2}=\frac{1}{2}\times\frac{22}{7}\times\:7\times\:7cm^{2}$$
$$ \displaystyle =77cm^{2}$$
Find the area of the unshaded portion of figure, the shaded portions being semi-circular regions.
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$$100.5\;m^2$$.
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$$101.5\;m^2$$.
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$$201.5\;m^2$$.
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$$101.8 \;m^2$$.
Explanation
Area of unshaded portion $$=$$ Area of rectangle - Area of circle
$$=$$ $$l\times b-\pi { r }^{ 2 }$$
$$=$$ $$14$$$$\times 10-\dfrac { 22 }{ 7 } \times 3.5\times 3.5$$
$$=$$ $$140 - 38.5$$
$$=$$ $$101.5$$ $${ m }^{ 2 }$$
The circumference of a circle is $$44 m$$, then the area of the circle is
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$$6084.5\, m^2$$
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$$276.5\, m^2$$
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$$154\, m^2$$
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$$44\, m^2$$
Explanation
Circumference of circle $$(C) = 2\pi r\, =\, 44$$
$$\therefore r\, =\, \displaystyle \frac {44}{2\, \times\, \displaystyle \frac {22}{7}}\, = \dfrac {44 \times 7}{2 \times 22} = 7\, m$$
Area of a circle $$=\, \pi r^2$$
$$=\, \displaystyle \frac {22}{7}\, \times\, 7\, \times\,7\, =\, 154\, m^2$$
The area of a circle is $$2464 m^2$$, 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 these
Explanation
By given, Area =$$2464m^2$$
$$\displaystyle \frac {\pi d^2}{4}\, =\, 2464$$
$$d^2\, =\, \displaystyle \frac {2464\, \times\, 4}{\displaystyle \frac {22}{7}}\, =\, 3136$$
$$d\, =\, \sqrt {3136}\, =\, 56$$m
It is given that two scalene triangles are not congruent, then we can conclude that both are always equal in area.
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True
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False
The inner diameter of a circular building is $$54 cm$$ and the base of the circular wall occupies a space of $$\displaystyle 2464cm^{2}$$. The thickness of the wall is:
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$$1 cm$$
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$$2 cm$$
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$$4 cm$$
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$$5 cm$$
Explanation
Let the outer radius of the wall be $$R$$ cm.
Base of the wall is circular in shape with R as its radius.
Given, $$ \pi { R }^{ 2 } = 2464 $$
$$ => { R }^{ 2 } = \cfrac {2464 \times 7}{22} \\\ \ \ \ \ \ \ \ \ \ \ = 784 $$
$$ => R = 28 cm $$
Inner Radius, $$ r= \cfrac {54}{2} = 27 cm $$
Hence, thickness $$ = $$ Outer radius $$ - $$ Inner Radius
$$ = 28 - 27 \\= 1 cm $$
$$ABCD$$ is a square field of side $$80\;m$$. If $$EF$$ is perpendicular to $$DC$$, find the area of $$\Delta EDC$$ and the area of the shaded portion respectively.
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$$3200\;m^2,\;3200\;m^2$$
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$$3200\;m^2,\;3500\;m^2$$
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$$3600\;m^2,\;3200\;m^2$$
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$$3200\;m^2,\;3600\;m^2$$
Explanation
ar $$\Delta EDC=\dfrac { 1 }{ 2 } DC\times EF$$
$$=$$ $$\dfrac { 1 }{ 2 } \times 80\times 80$$
$$=$$ $$3200$$ $${ m }^{ 2 }$$
area of shaded portion $$=$$ Area of square - $$3200$$ $${ m }^{ 2 }$$
$$=$$ $${ \left( 80\times 80 \right) }-3200$$
$$=$$ $$3200$$ $${ m }^{ 2 }$$
Mrs. Kaushik has a square plot with the measurement as shown in the figure. she wants to construct a house in the middle of the plot. A garden is developed around the house. Find the total cost of developing a garden around the house at the rate of $$Rs.\;55\;per\;m^2$$.
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$$Rs.\;17,875$$
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$$Rs.\;17,675$$
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$$Rs.\;12,875$$
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$$Rs.\;17,825$$
Explanation
Area of the garden = Area of the outer square - Area of the inner rectangle
$$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=25\times25\;m^2-20\times15\;m^2$$
$$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=(625-300)\;m^2=325\;m^2$$
Cost of developing a garden @ $$Rs.\;55\;per\;m^2=Rs.\;(55\times325)=Rs.\;17,875$$.
In figure, $$AB$$ and $$CD$$ are two perpendicular diameters of a circle with centre $$O$$. If $$OA$$$$=$$$$7cm$$, find the area of the shaded region. [Use $$\pi=\displaystyle\frac{22}{7}$$]
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$$31.4cm^2$$
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$$42.5cm^2$$
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$$54.2cm^2$$
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$$66.5cm^2$$
Explanation
Area of shaded region $$=$$ area of larger circle $$-$$ area of smaller circle $$-$$ area of the triangle
$$=\pi { R }^{ 2 }-\pi { r }^{ 2 }-\dfrac { 1 }{ 2 } \times DC\times BO$$
$$=\pi { R }^{ 2 }-\pi { r }^{ 2 }-\dfrac { 1 }{ 2 } \times DC\times BO$$
$$=\dfrac { 22 }{ 7 } \left( { 7 }^{ 2 }-{ \left( 3.5 \right) }^{ 2 } \right) -\dfrac { 1 }{ 2 } \times 14\times 7$$
$$=\dfrac { 22 }{ 7 } \left( 49-12.25 \right) -49$$
$$=(\dfrac { 22 }{ 7 } \times 36.75)-49$$
$$= 115.5 - 49$$
$$= 66.5$$ $${ cm }^{ 2 }$$
If the diameter of a semicircular protractor is $$14cm$$ then find its perimeter.
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$$36cm$$
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$$76cm$$
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$$112cm$$
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$$162cm$$
Explanation
Diameter $$=14cm$$
Radius=$$\displaystyle\frac{Diameter}{2}=\frac{14cm}{2}=7cm$$
Length of the semicircular part $$=\pi r=\displaystyle\frac{22}{7}\times (7)=22cm$$
Total perimeter=length of semicircular part +Diameter
$$=22cm+14cm=36cm$$
Thus, the perimeter of the protractor is $$36cm$$
The area of the shaded portion in the given figure is
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$$7.5\pi$$ sq.units
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$$6.5\pi$$ sq.units
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$$5.5\pi$$ sq.units
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$$4.5\pi$$ sq.units
Explanation
$$\textbf{Step 1 : Find the required area}$$
$$\text{Area of the outer semi-circle =} \dfrac{\pi}{2}\times 5^2$$
$$\text{Area of the inner semi-circle =} \dfrac{\pi}{2}\times (5-1)^2$$
$$\therefore \text{ Area of shaded region = Area of outer semicircle - Area of inner semicircle}$$
$$=\dfrac{\pi}{2}\times 5^2-\dfrac{\pi}{2}\times 4^2$$
$$= \dfrac{\pi}{2}(25-16)$$
$$= \text{4.5 $\pi$ unit}^2$$
$$\boldsymbol{\textbf{Hence, Area of shaded portion is 4.5 $\pi$} \textbf{ sq. unit}}$$
Find the area of a circle whose circumference is $$22cm$$.
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$$35.2cm^2$$
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$$38.5cm^2$$
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$$41.7cm^2$$
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$$47.6cm^2$$
Explanation
Let $$r$$ be the radius of the circle.
Then, Circumference$$=22cm$$
$$\displaystyle \Rightarrow 2\pi r=22$$
$$\displaystyle\Rightarrow 2\times \frac{22}{7}\times r=22$$
$$\Rightarrow r=\displaystyle\frac{7}{2}cm$$
$$\therefore$$ Area of the circle$$=\displaystyle \pi r^2$$
$$=\displaystyle\frac{22}{7}\times\frac{7}{2}\times \frac{7}{2}cm^2$$
$$=38.5 \, cm^2$$
If the radius of circle is $$\displaystyle \pi $$, then its area will be:
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$$\displaystyle \pi $$
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$$\displaystyle { \pi }^{ 2 }$$
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$$\displaystyle { \pi }^{ 3 }$$
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$$\displaystyle 3\pi $$
Explanation
Area of circle $$=$$ $$\pi { r }^{ 2 }=\pi \times { \left( \pi \right) }^{ 2 }={ \pi }^{ 3 }$$
Which of the following has least area?
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$$A1$$
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$$A2$$
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$$A3$$
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None of these
Richa wants to add a rectangular stone in her ring of length $$2$$ $$cm$$ and breadth $$1.5$$ $$cm$$. Find the area of stone in millimetrrs.
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$$0.3$$ $$mm^2$$
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$$3$$ $$mm^2$$
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$$30$$ $$mm^2$$
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$$300$$ $$mm^2$$
The inner circumference of a circular tracks is $$220\ m$$. The track is $$7$$ $$m$$ wide everywhere. Calculate the cost of putting up a fence along the outer circle at the rate of Rs. $$2$$ per metre. Use $$\pi=\displaystyle\frac{22}{7}$$
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Rs. $$947$$
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Rs. $$726$$
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Rs. $$612$$
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Rs. $$528$$
Explanation
Circumference of inner side = $$220 m$$
$$\Rightarrow 2\pi r=220\Rightarrow r=\dfrac { 220\times 7 }{ 44 } =35m$$
Now, width of track $$=7m$$
$$\therefore $$ Outer radius $$=35+7 = 42 m$$
Therefore outer circumference $$=2\pi R=2\times \dfrac { 22 }{ 7 } \times 42=264m$$
$$\therefore $$ Cost of fencing $$=$$ Rs.
$$\left( 264\times 2 \right) $$ $$=$$ Rs. $$528$$
The radii of two circles are $$8$$cm and $$6$$cm respectively. Find the radius of the circle having its area equal to the sum of the areas of the two circles.
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$$32cm$$
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$$23cm$$
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$$15cm$$
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$$10cm$$
Explanation
Let the radius of required circle is $$r$$ cm.
So, $$\pi r^2=\pi r^2_1+\pi r^2_2$$
$$r^2=r^2_1+r^2_2$$
$$r^2=(8)^2+(6)^2$$
$$r^2=100$$
$$r=10cm$$
Abhay made a straight cut through a circular rubber band and then laid the rubber band flat, as shown in the figure.
Which measure corresponds to the length of the cut rubber band ?
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Chord
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Circumference
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Diameter
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Radius
Explanation
Circumference of the rubber band is equal to the length of the cut rubber band .
Hence option B is answer
In the figure shown here, $$BDEC$$ is a rectangle. $$ABC$$ and $$DEF$$ are straight lines. The area, in cm$$^2$$ , of the whole figure is ______ .
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$$120$$
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$$128$$
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$$136$$
0%
$$142$$
Explanation
Area of the whole figure= area of rectangle BDEC+ area of triangle ABD+ area of triangle ECF
$$=DE\times EC+\dfrac { 1 }{ 2 } \times AB\times BD+\dfrac { 1 }{ 2 } \times CE\times EF$$
$$=8\times 10+\dfrac { 1 }{ 2 } \times 8\times 3+\dfrac { 1 }{ 2 } \times 8\times 11$$
$$=80+12+44$$
$$=136{ cm }^{ 2 }$$
The areas of two concentric circles forming a ring are 154 sq. cm and 616 sq. cm. The breadth of the ring is
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$$21 \ cm$$
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$$56 \ cm$$
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$$14 \ cm$$
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$$7 \ cm$$
Explanation
Breadth of the ring is equal to the difference between the radius of the outer circle and the radius of the inner circle
Let the radius of the outer circle be $$r_2$$ and that of the inner circle be $$r_1$$
We know that, the area of a circle is $$\pi r^2$$
Given that the area of outer circle is $$616 \ \displaystyle cm^{2}$$
$$\displaystyle \therefore \ \pi r_{2}^{2}=616 \ cm^{2}$$
$$\displaystyle \Rightarrow r_{2}^{2}=\frac{616\times 7}{22}\quad \quad \left[\because \ \pi=\dfrac{22}{7}\right]$$
$$=196 \ cm^2$$
$$\displaystyle \therefore \ r_{2}=14 \ cm $$
Also, the area of the inner circle is $$\displaystyle 154 \ cm^{2}$$
$$\displaystyle \therefore \ \pi r_{1}^{2}= 154 \ cm^2$$
$$\displaystyle \Rightarrow r_{1}^{2}=\frac{154\times 7}{22}\quad \quad \left[\because \ \pi=\dfrac{22}{7}\right]$$
$$=49 \ cm^2$$
$$\displaystyle \therefore r_{1}=7 \ cm.$$
$$\displaystyle \therefore \ $$ The required breadth $$\displaystyle = r_{2}-r_{1}$$
$$=14-7$$
$$=7 \ cm$$
Hence, the breadth of the ring is $$7 \ cm$$.
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