MCQExams
0:0:1
CBSE
JEE
NTSE
NEET
Practice
Homework
×
CBSE Questions for Class 7 Maths Perimeter And Area Quiz 8 - MCQExams.com
CBSE
Class 7 Maths
Perimeter And Area
Quiz 8
Find the area of a parallelogram with a base of $$200$$ cm and height of $$2.5$$ cm.
Report Question
0%
$$500$$ $$cm^{2}$$
0%
$$510$$ $$cm^{2}$$
0%
$$520$$ $$cm^{2}$$
0%
$$300$$ $$cm^{2}$$
Explanation
Area of a parallelogram $$=$$ base $$\times$$ height
$$=200 \times 2.5$$
$$= 500$$ $$cm^{2}$$
Find the base of parallelogram if its area is $$\displaystyle 80{ cm }^{ 2 }$$ and altitude is $$10$$ cm.
Report Question
0%
$$6$$ cm
0%
$$8$$ cm
0%
$$10$$ cm
0%
None of the above
Explanation
Given, area of parallelogram $$=80\ cm^2$$
Altitude, $$a=10\ cm$$
Let, the base be $$b$$
For a parallelogram,
$$Area=Base\times Altitude$$
$$\Rightarrow 80=b\times 10$$
$$\Rightarrow b=\dfrac{80}{10}$$
$$\Rightarrow b=8$$
Hence, the base of the parallelogram is $$8\ cm$$
A circular lawn with a radius of $$5$$ meters is surrounded by a circular walkway that is $$4$$ meters wide (see figure). What is the area of the walkway?
Report Question
0%
$$48\ \pi \ m^2$$.
0%
$$60\ \pi \ m^2$$.
0%
$$56\ \pi \ m^2$$.
0%
$$42\ \pi \ m^2$$.
Explanation
The area of the walkway is the area of the entire image (walkway + lawn) minus the area of the lawn. To find the area of each circle, use the formula:
Large circle: $$A = \pi r^2 = \pi (9)^2 = 81\pi$$
Small circle: $$A = \pi r^2 = \pi(5)^2 =25\pi$$
Thus, required area = $$81\pi-25\pi= 56\pi m^2$$.
Find the area of a parallelogram with a base of $$34$$ meters and a height of $$8$$ meters.
Report Question
0%
262 $$m^{2}$$
0%
272 $$m^{2}$$
0%
282 $$m^{2}$$
0%
292 $$m^{2}$$
Explanation
Area of a parallelogram = base $$\times$$ height
= $$34 \times 8$$
= $$272$$ $$m^{2}$$
Area $$= \dfrac{1}{2}\times base \times height$$
Which of the following is true?
Report Question
0%
Calculation of area needs attitude.
0%
Calculation of area by the given formula doesn't need attitude.
0%
Base can be $$AB,\ BC$$ or $$AC$$
0%
None of the above
Explanation
Area of $$\Delta ABC=\dfrac { 1 }{ 2 } \times base\times altitude$$
To calculate the area we need the altitude which is not given but we can choose any side as a base from
$$AB, BC$$ or $$AC$$.
Therefore, statements $$A$$ and $$C$$ are true.
A parallelogram has an area of $$60$$ $$cm^{2}$$ and a base of $$12cm$$. Find the height.
Report Question
0%
$$3$$ cm
0%
$$4$$ cm
0%
$$5$$ cm
0%
$$6$$ cm
Explanation
Area of a parallelogram = base $$\times$$ height
$$60 = 12 \times height$$
height $$=60 \div 12$$
height $$=5$$ $$cm$$
Calculate the area of a parallelogram with a base of $$ 12$$ m and height of $$5$$ m.
Report Question
0%
59 $$m^{2}$$
0%
60 $$m^{2}$$
0%
61 $$m^{2}$$
0%
62 $$m^{2}$$
Explanation
Area of a parallelogram = base $$\times$$ height
= $$12 \times 5$$
= $$60$$ $$m^{2}$$
If P represents the area and W represents the circumference of the circle, then P in terms of W is
Report Question
0%
$$\displaystyle \frac{2\pi }{W}$$
0%
$$\displaystyle \frac{4\pi^{2} }{W}$$
0%
$$\displaystyle \frac{2\pi^{2} }{W}$$
0%
$$\displaystyle \frac{W^{2} }{4\pi }$$
Explanation
Since P is the area so P = $$\displaystyle \pi r^{2}$$ ....(1)
W is the perimeter thus W = $$\displaystyle 2\pi r\Rightarrow r=\frac{W}{2\pi }$$
Put the value of $$r$$ in (1)
$$\displaystyle \Rightarrow P=\pi \left (\frac{W}{2\pi } \right )^{2}$$
$$\displaystyle P=\pi \frac{W^{2}}{4\pi ^{2}}$$
$$\displaystyle \Rightarrow P= \frac{W^{2}}{4\pi }$$
Find the perimeter of the figure outlined by solid line in terms of $$x$$.
Report Question
0%
$$5x+3\pi x$$
0%
$$5x+6\pi x$$
0%
$$10x+3\pi x$$
0%
$$10x+6\pi x$$
Explanation
Perimeter of the figure $$=$$ Length of slanting sides of triangle + Circumference of semicircle
Circumference of semicircle $$=\pi r = \dfrac {22}{7} \times 3x = \dfrac {66x}{7}= 3 \pi x$$
For the two right angle triangle,
By Pythagoras theorem
$$(Hyp)^2 = (4x)^2+ \left(\dfrac{6x}{2}\right)^2$$
$$=16x^2+9x^2$$
$$=25x^2$$
Hyp $$= 5x$$ units
Length of two slanting sides of triangle $$=2 \times 5x = 10x$$
Perimeter of the figure $$=(3 \pi x + 10x)$$ units
Find the area and inner circumference of the ring.
Report Question
0%
$$9.48$$ sq. cm, $$3.14$$ cm
0%
$$9.42$$ sq. cm, $$6.48$$ cm
0%
$$9.42$$ sq. cm, $$2.14$$ cm
0%
None of these
Explanation
Area of a circular ring = $$\pi(R^2-r^2)$$
$$=$$ $$3.14(2^2-1^2)$$
$$= 3.14 \times 3$$
$$= 9.42$$ sq. cm
Inner circumference $$=$$ $$2\pi r$$
$$= 2 \times 3.14 \times 1$$
$$= 6.28$$ cm
Calculate the area of a circular ring whose outer and inner radii are $$12$$ and $$10\ cm.$$
Report Question
0%
$$118.16$$ $$cm^2$$
0%
$$128.16$$ $$cm^2$$
0%
$$138.16$$ $$cm^2$$
0%
$$148.16$$ $$cm^2$$
Explanation
Area of a circular ring $$=\pi(R^2-r^2)$$
$$=3.14(12^2-10^2)$$
$$= 3.14 \times 44$$
$$= 138.16$$ $$cm^2$$
Find the width of the ring.
Report Question
0%
$$2\ cm$$
0%
$$3\ cm$$
0%
$$4\ cm$$
0%
$$5\ cm$$
Explanation
Width of the ring, $$d =$$ outer radius $$-$$ inner radius
$$d = R - r$$
$$=10 - 5 = 5\ cm$$
The diameter of a circle is $$1$$. Calculate the area of the circle.
Report Question
0%
$$\dfrac {\pi}{8}$$
0%
$$\dfrac {\pi}{4}$$
0%
$$\dfrac {\pi}{2}$$
0%
$$\pi$$
0%
$$2\pi$$
Explanation
formula to calculate area of a circle is $$\pi { r }^{ 2 }$$.
Given, diameter $$d=1$$
Radius $$r=$$ $$\dfrac { d }{ 2 } =\dfrac { 1 }{ 2 } $$
$$\Rightarrow$$ Area of circle $$=$$ $$\pi { \left( \dfrac { 1 }{ 2 } \right) }^{ 2 }=\dfrac { \pi }{ 4 } $$ sq. units.
Find the area of the triangle given.
Report Question
0%
$$3$$ sq units
0%
$$6$$ sq units
0%
$$9$$ sq units
0%
None of the above
Find the area of the shaded portion
Report Question
0%
$$4 cm^{2}$$
0%
$$5 cm^{2}$$
0%
$$6 cm^{2}$$
0%
$$7 cm^{2}$$
Explanation
We know that the area of a triangle is given by the formula $$A=\dfrac{1}{2} bh$$
, where
b
is the length of the base and
h
is the triangles height.
$$ar (ADBE)=ar(ABC)-ar(DEC)$$
$$=\dfrac{1}{2}\times 2\times 6-\dfrac{1}{2}\times 1\times 2$$
$$=6-1=\boxed{5cm^2}$$
Sonali pounded a stake into the ground, when she attached a rope to both the stake and her dog's collar, the dog could reach $$9$$ feet from the stake in any direction. Find the approximate area of the lawn, in square feet, the dog could reach from the stake. (
$$\displaystyle \pi $$ =3.14)
Report Question
0%
$$28$$
0%
$$57$$
0%
$$113$$
0%
$$254$$
Explanation
Dog can reach $$9$$ feet from the stake in any direction, so radius of the circular ground will be $$9$$.
We know formula for area of circle is $$\pi { r }^{ 2 }$$
Hence area will be $$3.14\times { 9 }^{ 2 }$$(value of pi is given)
$$=3.14\times 81=254.34$$
The radius of a circle whose circumference is $$\pi$$ is
Report Question
0%
$$\dfrac{1}{2}$$
0%
$$1$$
0%
$$2$$
0%
$$\pi$$
0%
$$2 \pi $$
Explanation
Given that the circumference of circle is $$\pi $$.
We know that the circumference of circle is $$2\pi r$$, where $$r$$ is radius of circle.
$$\therefore$$ Circumference $$=\pi= 2 \pi r$$
$$\therefore r=\dfrac {1}{2}$$
In the figure given above, $$\overline {AC}$$ is a diameter of the large circle and B lies on $$\overline {AC}$$ so that $$\overline {AB}$$ is a diameter of the small circle. If $$AB = 1$$ and $$BC = 2$$, Calculate the area of the shaded region.
Report Question
0%
$$\dfrac {\pi}{4}$$
0%
$$\pi$$
0%
$$2\pi$$
0%
$$\dfrac {9\pi}{4}$$
0%
$$9\pi$$
Explanation
To find the area of the shaded region, we have to find the area of the bigger circle and the area of the smaller circle.
For large circle, $$AC$$ is diameter measuring $$3$$.
$$AC=AB+BC=1+2=3$$
Area of larger circle is $$=$$ $$\pi { R }^{ 2 }$$
$$\pi { \left( \dfrac { 3 }{ 2 } \right) }^{ 2 }=\pi \dfrac { 9 }{ 4 } $$
Area of smaller circle is $$=\pi { r }^{ 2 }$$
$$\pi { \left( \dfrac { 1 }{ 2 } \right) }^{ 2 }=\pi \dfrac { 1 }{ 4 } $$
Area of shaded region is $$=$$ Area of larger circle $$-$$ Area of smaller circle
$$=\dfrac { 9\pi }{ 4 } -\dfrac { \pi }{ 4 } $$
$$=\dfrac { 9\pi -\pi }{ 4 } =\dfrac { 8\pi }{ 4 } =2\pi $$
The area of the parallelogram $$ABCD$$ is
Report Question
0%
$$20$$
0%
$$10$$
0%
$$5\sqrt{3}$$
0%
$$10\sqrt{3}$$
0%
$$18$$
Explanation
From the given diagram, we conclude that the height of the parallelogram $$ABCD$$ is $$4$$ and its base is given by $$5$$.
We know that, Area of parallelogram $$=\text{base} \times \text{height}$$
$$\therefore$$ Area of parallelogram $$ABCD$$ is given by $$\text{base}\times \text{height}=4\times 5=20$$
Hence, the answer is $$20$$.
A circle of radius $$x$$ has an area twice that of a square of side $$ a.$$ The equation used to find the radius of the circle is
Report Question
0%
$$\pi a^2 = 2x^2$$
0%
$$\pi x^2 = 2a^2$$
0%
$$\pi x^2 = 4a^2$$
0%
$$\pi a^2 = 4x^2$$
0%
$$4\pi x^2 = a^2$$
Explanation
The area of square whose side length is $$a$$ is $${a}^{2}$$
So the area of circle is $$2{a}^{2}$$
Since, the radius of circle be $$x$$ , then the area is $$\pi {x}^{2} = 2{a}^{2}$$
The figure shows a rectangle $$ABCD$$.
Calculate the unshaded area.
Report Question
0%
$$4x+3y$$
0%
$$8x+5y$$
0%
$$12xy$$
0%
$$22xy$$
Explanation
For the bigger rectangle,
Length of the rectangle $$AD=BC =6x$$ cm
and breadth of the rectangle $$AB=CD=4y$$ cm
So, area of rectangle $$=6x\times 4y\ sq.cm$$
$$=24xy\ sq.cm$$
For the shaded
rectangle
Length of shaded
rectangle
$$=2x$$ cm
and breadth of shaded
rectangle
$$=y$$ cm
So, area of shaded
rectangle
$$=2x\times y \ sq.cm$$
$$=2xy\ sq.cm$$
Hence, the area of unshaded part $$=(24 xy-2 xy)\ sq.cm=22 xy\ sq.cm$$
In figure, what is the approximate area of parallelogram DAWN?
Report Question
0%
11.57
0%
13.64
0%
14.63
0%
17.25
0%
20.00
Explanation
area of parallelogram $$=ab\sin \theta$$
$$AD\times DN \sin 47^o$$
$$4\times 5\times \sin 47^o$$
$$20\times 0.731$$
$$14.63$$
What will it cost to carpet a room with indoor/outdoor carpet if the room is 10 feet wide and 12 feet long? The carpet costs 12.51 per square yard.
Report Question
0%
$166.80
0%
$175.90
0%
$184.30
0%
$189.90
A lawn $$30$$ m long and $$16$$ m wide is surrounded by a path $$2$$ m wide. What is the area of the path?
Report Question
0%
$$200$$ sq. m
0%
$$280$$ sq. m
0%
$$300$$ sq. m
0%
$$320$$ sq. m
Explanation
Given, length of lawn $$=16$$ m, width $$=2$$ m
We need to find the area of the path.
Figure is shown according to question having sides $$=16+2+2 , 30+2+2 = 20m, 34m$$
Thus area of path $$= (34\times 20 - 30\times 16)m^{2}$$
$$= (680 - 480) $$
$$= 200$$ sq. m
What is the area of a circle with a diameter of $$16$$?
Report Question
0%
$$8$$ $$\pi$$
0%
$$16$$ $$\pi$$
0%
$$64$$ $$\pi$$
0%
$$128$$ $$\pi$$
0%
$$256$$ $$\pi$$
Explanation
Given, diameter $$=d=16$$
We need to find area of circle.
Area of circle $$=\pi r^2$$
Therefore, radius $$=r= \dfrac {16}{2} = 8 $$
Area of the circle $$ = \pi r^2 = \pi (8)^2 = 64\pi $$
Two similar parallelograms have corresponding sides in the ratio $$1:k$$. What is the ratio of their areas?
Report Question
0%
$$1:3{k}^{2}$$
0%
$$1:4{k}^{2}$$
0%
$$1:{k}^{2}$$
0%
$$1:2{k}^{2}$$
The area of MCM college is $$213$$ hectares. Find its area in meters.
Report Question
0%
$$2130$$
0%
$$21300$$
0%
$$213000$$
0%
$$2130000$$
Convert $$200 cm^2$$ in $$mm^2$$.
Report Question
0%
$$2000$$ $$mm^2$$
0%
$$20000$$ $$mm^2$$
0%
$$200000$$ $$mm^2$$
0%
$$2000000$$ $$mm^2$$
A wire of length 36 cm is bent in the form of a semicircle. What is the radius of the semicircle?
Report Question
0%
$$9\ cm$$
0%
$$8\ cm$$
0%
$$7\ cm$$
0%
$$6\ cm$$
Explanation
Length of wire is in the form of semicircle.
Given, length of wire $$=36\ cm$$
The length of the wire will cover the diameter as well as the circumference of the semicircle.
$$\therefore \ $$Length of the wire $$= \dfrac {1}{2} (2\pi r) + 2r\\$$
$$\Rightarrow 36 = (\pi + 2)r\\$$
$$\Rightarrow 36 = \left (\dfrac {22}{7} + 2\right ) r = \dfrac {36}{7}\times r\\$$
$$\Rightarrow r = 7\ cm$$.
Hence, the radius of the semicircle is $$7\ cm$$.
In the following figure, what is the area of the shaded circle inside of the square?
Report Question
0%
512
0%
256
0%
16
0%
50.24
0%
12.57
0:0:1
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
0
Answered
0
Not Answered
0
Not Visited
Correct : 0
Incorrect : 0
Report Question
×
What's an issue?
Question is wrong
Answer is wrong
Other Reason
Want to elaborate a bit more? (optional)
Practice Class 7 Maths Quiz Questions and Answers
<
>
Support mcqexams.com by disabling your adblocker.
×
Please disable the adBlock and continue.
Thank you.
Reload page
