7th Standard CBSE Syllabus & Materials
7th Standard CBSE
CBSE 7th Social Science Theme E - Understanding Market - New Sample Question Papers Study Material - QB365 Set A
NEW7th Standard CBSE
CBSE 7th Social Science Theme E - From Barter to Money - New Sample Question Papers Study Material - QB365 Set A
NEW7th Standard CBSE
CBSE 7th Social Science Theme D - The Constitution of India- An Introduction - New Sample Question Papers Study Material - QB365 Set A
NEW7th Standard CBSE
CBSE 7th Social Science Theme D - From the Rulers to the Ruled : Types of Governments - New Sample Question Papers Study Material - QB365 Set A
NEW7th Standard CBSE
CBSE 7th Social Science Theme B - The Age of Reorganisation - New Sample Question Papers Study Material - QB365 Set A
NEW7th Standard CBSE
CBSE 7th Social Science Theme B - The Rise of Empires - New Sample Question Papers Study Material - QB365 Set A

Published on: 10/10/2019
Perimeter and Area
Download CBSE Class 7th Standard CBSE Mathematics question papers, sample papers, important questions, and previous year solved papers in PDF format. Get free study materials, NCERT solutions, and exam preparation resources for Class 7th Standard CBSE Mathematics
Questions + Answers key
Take MCQ Mathematics Test

1.
Find the height of the wall whose length is 4 m and which can be covered by 2400 tiles of size 25 cm by 20cm.
2.
A copper wire, when bent in the form of a square encloses an area of 121 cm2 . If the same wire is bent in the form of a circle, find the area enclosed by it.
3.
A square park has each side 50 cm. At each corner of the park, there is a flower bed in the form of a circle whose radius 7 m, as shown in the figure, find the area of remaining part of two park.

4.
A rectangular lawn is 30 m by 20 m. If has two roads each 2 m wide running in the middle of it, one parallel to the length and other parallels to the breadth. Find the area of the roads.
5.
In the following figure, find the area of the given figure.

6.

(a) Taj Mahal stands on a square platform that is 95.40 m on each side. What is the area of this square platform in square metres?
(b) The floor area of the main building is 3214 m2. What is the area of the part of the platform that is not covered by the main building?
7.
Find the ratio of areas of \(\Delta\)MNO, \(\Delta\)MOP and \(\Delta\)MPQ in the following figure

8.
In the following figures, perimeter of \(\Delta\)ABC = perimeter of \(\Delta\)PQR. Find the area of \(\Delta\)ABC.

9.
In the following figure, ABCD is a square with AB = 15 cm. Find the area of the square BDFE.

10.
In the given figure, find the area of parallelogram ABCD, if the area of shaded triangle is 9 cm2.

11.
A wall of a room is of dimensions 5 m x 4m. lt has a window of dimensions 1.5 m x 1 m and a door of dimensions 2.25 m x 1 m. Find the area of the wall, which is to be painted.
12.
A rectangular metal plate is 7 cm long and 5 cm wide. Find the cost of the plate at the rate of Rs 75 per sq cm.
1.
Area of a tile = 25 x 20 cm2 = 500 cm2
Area of 2400 tiles = 2400 x 500 cm2
= 1200000 cm2
= \(\frac { 1200000 }{ 10000 } \)m2
[\(\therefore\) 10000 cm2 = 1 m2]
= 120 m2
Let the height of the wall be h metres then,
Area of the wall = 4h m2
Since 2400 tiles completely cover the wall.
\(\therefore\) Area of the wall = Area of 2400 tiles
\(\Rightarrow\) 4h = 120
\(\Rightarrow\) \(\frac { 4h }{ 4 } \)= \(\frac { 120 }{ 4 } \)
\(\Rightarrow\) h = 30
[Dividing both sides by 4]
Hence the height of the wall is 30 metre.
2.
Area enclosed the copper wire
In square shape = (side)2
\(\therefore\)(side)2 = 121 cm2
\(\Rightarrow\) side = \(\sqrt { 121 } \) = 11 cm.
Hence length of wire = 11 x 4
= 44 cm
Now this length = Circumference of the circle
\(\Rightarrow\) 2\(\pi\)r = 44
\(\Rightarrow\) 2 x \(\frac { 22 }{ 7 } \) x r = 44
\(\Rightarrow\) r = \(\frac { 44 }{ 2\times 22 } \times 7\)
Thus, r = 7 cm
Hence, area enclosed by the wire when it is bent in circular shape
= \(\pi\)r2
= \(\frac { 22 }{ 7 } \)x (7)2
= \(\frac { 22 }{ 7 } \)x 7 x 7
= 154 m2
3.
2346 m2
4.
96 m2
5.
115.5 cm2
6.
(a) Given, length of the side of the square platform
= 95.40 m
\(\therefore\) Area of square platform = Length x Length
= 95.40 x 95.40 = 9101.16 sq m
Hence, the area of square platform is 9101.16 sq m.
(b) Given, floor area of the main building'
= 3214 m2
Now, area of the part of the platform that is not covered by the main building
= Area of square platform - Floor area of the main building
= 9101.16 - 3214 = 5887.16 sq.
7.
\(\therefore\) Area of a triangle = \(\frac{1}{2}\) x Base x Height
\(\therefore\) Area of \(\Delta\)MNO = \(\frac{1}{2}\) x NO x MO
= \(\frac{1}{2}\) x 4 x 5 = 10 cm2
Now, Area of \(\Delta\)MPO = \(\frac{1}{2}\) x OP x MO
= \(\frac{1}{2}\) x 2 x 5 = 5 cm2
and Area of \(\Delta\)MPQ = \(\frac{1}{2}\) x PQ x MO = \(\frac{1}{2}\) x 6 x 5
= 15 cm2
\(\therefore\) Ratio of areas of the triangles = \(\Delta\)MNO: \(\Delta\)MPO: \(\Delta\)MPQ
= 10 cm2: 5 cm2: 15 cm2
= 2: 1: 3.
8.
Given, perimeter of \(\Delta\)ABC = perimeter of \(\Delta\)POR
So, perimeter of \(\Delta\)POR = 14 + 6 + 10 = 30 cm
Now, perimeter of \(\Delta\)ABC,
30 = AB + BC + AC \(\Rightarrow\) 30 = AB + 5 + 13
\(\Rightarrow\) 30 = AB + 18 \(\Rightarrow\) AB = 30 - 18 = 12
\(\therefore\) Area of the \(\Delta\)ABC = \(\frac{1}{2}\) x Base x Height
= \(\frac{1}{2}\) x 5 x 12 = 5 x 6 = 30 cm2.
9.
Given, ABCD is a square and AB = 15 cm
\(\therefore\) Diagonal of square ABCD = \(\sqrt{2}\) a
= \(\sqrt{2}\) x 15 = 15\(\sqrt{2}\) cm
\(\therefore\) Area of the square BDFE = (Side)2 = (15\(\sqrt{2}\))2
= 15 x 15 x \(\sqrt{2}\) x \(\sqrt{2}\) = 225 x 2 = 450 cm2.
10.
Given, area of shaded triangle = 9 cm2 and base of the triangle = 3 cm
Area of a triangle = \(\frac{1}{2}\) x Base x Height
\(\Rightarrow 9=\frac{1}{3}\times 3\times h \Rightarrow \frac{18}{3}=h\Rightarrow\)h = 6 cm.
\(\therefore\) Area of parallelogram = Height x Base of parallelogram
= 6 x (3 + 4) = 6 x 7 = 42 cm2.
11.
A wall of a room is of dimensions 5 m x 4 m.
Length of the room = 5 m
Breadth of the room = 4 m
\(\therefore\) Area of the room = 5 x 4= 20 m2
Length of the window = 1.5 m
Breadth of the window = 1 m
\(\therefore\) Area of the window = 1.5 x 1 = 1.5 m2
Length of the door = 2.25 m
Breadth of the door = 1 m
\(\therefore\) Area of the door = 2.25 x 1 = 2.25 m2
The area of the wall to be painted = Area of the room - Area of the window - Area of the door
= 20 - 1.5 - 2.25
= 20 - 3.75 = 16.25 m2.
12.
Length of rectangular metal plate = 7 cm
Breadth of rectangular metal plate = 5 cm
\(\therefore\) Area of a rectangle = Length x Breadth
= 7 x 5 = 35 cm2
\(\therefore\) Rate of the plate = Rs 75 per sq cm.
Then, total cost = 35 x 75
= Rs 2625.
7th Standard CBSE Syllabus & Materials
7th Standard CBSE
CBSE 7th Social Science Theme B - New Beginnings : Cities and States - New Sample Question Papers Study Material - QB365 Set A
NEW7th Standard CBSE
CBSE 7th Social Science Theme A - Climates of India - New Sample Question Papers Study Material - QB365 Set A
NEW7th Standard CBSE
CBSE 7th Social Science Theme A - Geographical Diversity of India - New Sample Question Papers Study Material - QB365 Set A
NEW7th Standard CBSE
CBSE 7th Science Earth, Moon and the Sun - New Sample Question Papers Study Material - QB365 Set A
CBSE 7th Standard CBSE Subjects
CBSE Standards