7th Grade > Mathematics
AREA AND PERIMETER MCQs
Perimeter And Area
Total Questions : 99
| Page 3 of 10 pages
Answer: Option B. -> 121 cm2
:
B
Area of the square = Side ×Side
Given: Side = 11 cm
∴ Area = 11 cm × 11 cm = 121 cm2
:
B
Area of the square = Side ×Side
Given: Side = 11 cm
∴ Area = 11 cm × 11 cm = 121 cm2
Answer: Option B. -> 80 cm
:
B
There are 4 squares of side 8 cm.
When we keep the squares next to each other, the length of the side will be 8×4 = 32 cm and the breadth will be 8 cm.
Hence, it will form a rectangle of length 32 cm and breadth of 8 cm.
∴ perimeter of the rectangle
= 2(length + breadth)
= 2(32 + 8)
= 80 cm
:
B
There are 4 squares of side 8 cm.
When we keep the squares next to each other, the length of the side will be 8×4 = 32 cm and the breadth will be 8 cm.
Hence, it will form a rectangle of length 32 cm and breadth of 8 cm.
∴ perimeter of the rectangle
= 2(length + breadth)
= 2(32 + 8)
= 80 cm
:
Given that
Perimeter of the square = 40 cm
Perimeter of a square = 4 × Side
⇒40cm=4×Side
Side=40cm4=10cm
Answer: Option A. -> 3 m
:
A
Area of a parallelogram = Base × corresponding height .
Therefore, Base = Area of a parallelogramCorresponding height
⇒ Base = 248= 3 m.
:
A
Area of a parallelogram = Base × corresponding height .
Therefore, Base = Area of a parallelogramCorresponding height
⇒ Base = 248= 3 m.
Answer: Option B. -> 70 m
:
B
To find out theminimum length of the wire to be bought by Puja so that the entire farm is covered, we have to find out the perimeter of the entire farm.
From the above figure, the length of the side AF = (Length of side BC) - (Length of side ED) = (20-5) m = 15m.
Length of side AB = (Length of side DC + Length of side FE) = (5+10) m = 15m
As we know, the perimeter of a plane figure is the length of its boundary.
Perimeter of the farm = (15+20+5+5+10+15) m = 70 m.
So, the minimum length of the wire to be bought by Puja is 70 m.
:
B
To find out theminimum length of the wire to be bought by Puja so that the entire farm is covered, we have to find out the perimeter of the entire farm.
From the above figure, the length of the side AF = (Length of side BC) - (Length of side ED) = (20-5) m = 15m.
Length of side AB = (Length of side DC + Length of side FE) = (5+10) m = 15m
As we know, the perimeter of a plane figure is the length of its boundary.
Perimeter of the farm = (15+20+5+5+10+15) m = 70 m.
So, the minimum length of the wire to be bought by Puja is 70 m.
Answer: Option A. -> 25 m2
:
A
Given: The metal sheet in the form ofa square of side 5m.
The area of asquare ofside'a' is given by a2.
So, area of the sheet =52=25m2
:
A
Given: The metal sheet in the form ofa square of side 5m.
The area of asquare ofside'a' is given by a2.
So, area of the sheet =52=25m2
Answer: Option B. -> 25 cm
:
B
When a circular discis cut into four equal parts, the perimeter of each part is more than 2πr4. This is because the length of two radii are also added to the perimeter in each part.
Perimeter of each part = 2πr4+ r + r
=πr2 + 2r
=22×77×2+ 2r
= 11+2×7
= 11 + 14
= 25 cm
:
B
When a circular discis cut into four equal parts, the perimeter of each part is more than 2πr4. This is because the length of two radii are also added to the perimeter in each part.
Perimeter of each part = 2πr4+ r + r
=πr2 + 2r
=22×77×2+ 2r
= 11+2×7
= 11 + 14
= 25 cm
Answer: Option A. -> 6 cm2
:
A
EF divides the rectangle ABCDinto two trapeziums.
Area of trapezium ABFE
=12×(Sumofparallelsides)×(Distancebetweenthem
=12×(AE+BF)×AB
=12×(1+2)×4
=12×3×4=6cm2
Area of trapezium CDEF =12×(FC+ED)×CD
=12×(1+2)×4
=12×3×4=6cm2
Area of both the trapeziums are equal.
∴ EF divides the rectangle ABCD in two equal parts and the area of both the parts are 6cm2.
:
A
EF divides the rectangle ABCDinto two trapeziums.
Area of trapezium ABFE
=12×(Sumofparallelsides)×(Distancebetweenthem
=12×(AE+BF)×AB
=12×(1+2)×4
=12×3×4=6cm2
Area of trapezium CDEF =12×(FC+ED)×CD
=12×(1+2)×4
=12×3×4=6cm2
Area of both the trapeziums are equal.
∴ EF divides the rectangle ABCD in two equal parts and the area of both the parts are 6cm2.