Mensuration | Math | Chapter 6 | Grade 5

Mensuration - Math Chapter-6, Grade 5:

Objectives:

Student will be able to:
Find the perimeter of different shapes.
Learn the difference between area and perimeter.
Use formula to determine area and perimeter.
Determine volume of a cube and cuboid by using formula..

Outcomes:

Students will have learnt to calculate perimeter, area and volume of regular objects with the help of formula very quickly in their real life practices..

Materials:

Multiplication chart, division chart, different models of triangle, square, rectangle or polygons, graph paper, tin box, books, rods, different units of weight, bottle marked with litre and mililitres, etc.

Things To Know:

➧ Perimeter is the total length of boundary of a plane figure.
➧ Perimeter is the sum of total length of sides of triangle or quadrilateral or polygons.
➧ The perimeter of a rectangle is, p = 2(l+b), where l = length and b = breadth.
➧ The perimeter of a square is, p = 4l, where l = length of a side.
➧ Area is a total surface covered by a plane figure.
➧ Area of a regular object is calculated with the help of it's sides (length or breadth).
➧ Area of irregular object is calculated with the help of graph.
➧ Volume is the space occupied by an object.
➧ Volume is calculated by counting the numbers of unit cube.
➧ Volume of a regular object is calculated by multiplying it's length, breadth and height, i.e. v = l x b x h.

Perimeter/button

Question: What is perimeter?
Answer: Perimeter of a plane figure is the sum of total length of sides/boundary of that figure (like, regular plane figure - triangle, square, rectangle, quadrilateral, etc and the irregular plane figure).

Also Check:

#Perimeter of a Triangle:

Here, triangle ABC is bounded by 3 sides - AB, BC and CA.
So, the total length of the boundary is AB + BC + CA = 2 cm + 3 cm + 4 cm = 9 cm.
∴ Perimeter of a triangle ABC = sum of the length of 3 sides = 9 cm.

#Perimeter of a Quadrilateral:

Here, Quadrilateral PQRS is bounded by 4 sides - PQ, QR, RS and SP.
∴ Perimeter of a quadrilateral PQRS = sum of length of four sides.
=PQ + QR + RS + SP
= 4 cm + 2 cm + 5 cm + 3 cm
= 14 cm
∴ Perimeter of a Quadrilateral PQRS = sum of the length of 4 sides = 14 cm.

#Perimeter of a Polygon:

Here, Polygon ABCDE is bounded by 5 sides - AB, BC, CD, DE and EA. It is a pentagon.
The total length of the boundary of this pentagon is = AB + BC + CD + DE + EA.
∴ Perimeter of a Pentagon ABCDE = sum of length of five sides.
= AB + BC + CD + DE + EA
= 3cm + 2cm + 4cm + 2cm + 3cm
= 14 cm.

Similarly,
Perimeter of a hexagon = sum of the length of 6 sides.
Perimeter of a heptagon = sum of the length of 7 sides.
Perimeter of a octagon = sum of the length of 8 sides.
Perimeter of a nonagon = sum of the length of 9 sides.
Perimeter of a decagon = sum of the length of 10 sides.
∴ Perimeter of a polygon = sum of the length of 'n' number of sides.

Remember in Greek word:
tri = three; quadri = four; penta = five; hexa = six; hepta = seven; octa = eight; nona = nine; deca = ten and (gon = side) alert-info

#Perimeter of a Rectangle:

Here, ABCD is a rectangle. In rectangle, it's opposite sides are equal. So, the length of AB = CD = l and BC = DA = b.
The perimeter of a rectangle ABCD = AB +BC + CD + DA = b + l + b + l = 2l + 2b = 2(l + b)
∴ The perimeter of a rectangle ABCD = 2(l + b).

#Perimeter of a Square:

Here, ABCD is a Square. In square, all sides are equal. So, the length of AB = BC = CD = DE = l.
The perimeter of a square ABCD = AB + BC + CD + DE = l + l + l + l = 4l

∴ Perimeter of a square = 4l.

Grade 5: Math Subject-Chapter:6 - Mensuration: Exercise 6.1 (With Solution)/button

1. Find the perimeter of the following plane figures.

Solution:
Here, Perimeter (P) = 3cm + 4cm + 5cm = 12cm.
∴ Perimeter (P) = 12cm.

Solution:
Here, Perimeter (P) = 3cm + 4.5cm + 3.5cm = 11cm.
∴ Perimeter (P) = 11cm.

Solution:
Here, Perimeter (P) = 3cm + 5cm + 4.4cm + 2.6cm = 15cm.
∴ Perimeter (P) = 15cm.

Solution:
Here, Perimeter (P) = 3cm + 4.2cm + 3cm + 6.8cm = 17cm.
∴ Perimeter (P) = 17cm.

Solution:
Here, Perimeter (P) = 9cm + 2cm + 3cm + 2cm + 3cm + 3cm + 2cm + 12cm = 36cm.
∴ Perimeter (P) = 36cm.

Solution:
Here, Perimeter (P) = 3cm + 6cm + 5cm + 4.4cm + 2.6cm = 21cm.
∴ Perimeter (P) = 21cm.

2. The length of 3 sides of a triangular field are 16.2 m, 18.8 m and 24.5 m. If you want to make wire fence around it, how long wire is needed for one round? If you want to make three round of wire fence, how long wire is needed?
Solution:
Here,
Three sides of a triangular lengths are 16.2 m, 18.8 m and 24.5 m.
We know that,
Perimeter of a Triangle = sum of lengths of three sides = 16.2 m + 18.8 m + 24.5 m = 59.5 m.
∴ 59.5 m wire is needed for one round.
Now,
∴ To make three round of wire fence, 59.5 m x 3 = 178.5 m long wire is needed.

3. The playground is quadrilateral in shape which has sides of 13.9 m, 16.4 m, 14.2 m and 23.5 m. A players want to run through the boundary of the playground. How much distance does he travel in 5 rounds?
Solution:
Here,
Four sides of a quadrilateral are 13.9 m, 16.4 m, 14.2 m and 23.5 m.
We know that,
Perimeter of a Quadrilateral = sum of lengths of four sides = 13.9 m + 16.4 m + 14.2 m + 23.5 m = 68 m.
∴ Perimeter (P) = 68 m.
Hence, A player travels 68 m in one round.
Again,
In 5 rounds, a player can travel 68 m x 5 = 340 m.

Grade 5: Math Subject-Chapter:6 - Mensuration: Exercise 6.2 (With Solution)/button

1. Find the perimeter of the following rectangles.

Solution:
Here,
Length (l) = 7 cm.
Breadth (b) = 5 cm.
Perimeter (P) = ?
We know that,
Perimeter (P) = 2(l + b)
= 2(7 + 5) cm
= 2 x 12 cm
= 24 cm.

Solution:
Here,
Length (l) = 5 cm.
Breadth (b) = 2 cm.
Perimeter (P) = ?
We know that,
Perimeter (P) = 2(l + b)
= 2(5 + 2) cm
= 2 x 7 cm
= 14 cm.

Solution:
Here,
Length (l) = 9 cm.
Breadth (b) = 4 cm.
Perimeter (P) = ?
We know that,
Perimeter (P) = 2(l + b)
= 2(9 + 4) cm
= 2 x 11 cm
= 24 cm.

2. Find the perimeter of the following squares.

Solution:
Here,
Length (l) = 5 cm.
Perimeter (P) = ?
We know that,
Perimeter (P) of a square = 4l
= 4 x 5 cm
= 20 cm

Solution:
Here,
Length (l) = 6.5 cm.
Perimeter (P) = ?
We know that,
Perimeter (P) of a square = 4l
= 4 x 6.5 cm
= 26 cm

Solution:
Here,
Length (l) = 8.8 cm.
Perimeter (P) = ?
We know that,
Perimeter (P) of a square = 4l
= 4 x 8.8 cm
= 35.2 cm

3. Find the perimeter of the following rectangles.

a) l = 4 cm and b= 3 cm
Solution:
Here,
l = 4 cm
b = 3 cm
Perimeter of a rectangle (P) = 2(l + b)
= 2(4+3) cm
= 2 x 7 cm
= 14 cm

b) l = 5.3 cm and b= 2.2 cm
Solution:
Here,
l = 5.3 cm
b = 2.2 cm
Perimeter of a rectangle (P) = 2(l + b)
= 2(5.3+2.2) cm
= 2 x 7.5 cm
= 15 cm

c) l = 4.4 cm and b= 2.6 cm
Solution:
Here,
l = 4.4 cm
b = 2.6 cm
Perimeter of a rectangle (P) = 2(l + b)
= 2(4.4+2.6) cm
= 2 x 7 cm
= 14 cm

d) l = 6.5 cm and b= 3.5 cm
Solution:
Here,
l = 6.5 cm
b = 3.5 cm
Perimeter of a rectangle (P) = 2(l + b)
= 2(6.5+3.5) cm
= 2 x 10 cm
= 20 cm

4. Find the perimeter of a following squares.

a) l = 4 cm
Solution:
Here,
l = 4 cm
We know that,
Perimeter of a square (P) = 4l
= 4 x 4 cm
= 16 cm

b) l = 7 cm
Solution:
Here,
l = 7 cm
We know that,
Perimeter of a square (P) = 4l
= 4 x 7 cm
= 28 cm

c) 5.4 cm
Solution:
Here,
l = 5.4 cm
We know that,
Perimeter of a square (P) = 4l
= 4 x 5.4 cm
= 21.6 cm

d) l = 6.3 cm
Solution:
Here,
l = 6.3 cm
We know that,
Perimeter of a square (P) = 4l
= 4 x 6.3 cm
= 25.2 cm

e) l = 4 x 1/2 cm
Solution:
Here,
l = 4 x 1/2cm = 4.5 cm
We know that,
Perimeter of a square (P) = 4l
= 4 x 4.5 cm
= 18 cm

5. A rectangular ground is 42 m long and 18 m broad.
a) Find it's perimeter.
b) Find it's area [hint: A = l x b]
c) What is the difference for these two values?
Solution:
Here,
l = 42 m
b = 18 m
a) Perimeter of a rectangular ground (P) = 2(l + b) = 2(42 m + 18 m) = 2 x 60 m = 120 m.
∴ It's perimeter is 120 m.
b) Area of a rectangular ground (A) = l x b = 42 m x 18 m = 756 m²
∴ It's area is 756 m²
c) Here perimeter of a rectangular ground is measured in meter i.e. 120m and the area of a rectangular ground is measured in square meter i.e. 756 m².

6. A square field is 9 m long. Find it's perimeter.
Solution:
Here,
l = 9 m
We know that,
The perimeter of a square (P) = 4l = 4 x 9 m = 36 m
∴ Perimeter (P) = 36 m.

7. A square ground is 30 m long. Find the length of wire required to fence it.
a) 1 round
b) 3 rounds
c) 6 rounds
Solution:
Here,
l = 30 m
Perimeter (P) = ?
We know that,
Perimeter of a square (P) = 4l = 4 x 30 m = 120 m.
∴ Perimeter (P) = 120 m.
Now,
a) ∴ The length of wire required to fence it = 120 m.
b) So, the length of wire required to fence it 3 rounds = 120 m x 3 = 360 m.
c) Hence, the length of wire required to fence it 6 rounds = 120 m x 6 = 720 m.

Additional Questions:/button

1. The breadth of a rectangle is 𝑥 cm. It's length is double of it's breadth. The perimeter of a rectangle is 42 cm. Make an equation and solve it to find the value of 𝑥. Then, find the length and breadth of the rectangle.
Solution:
Here,
b = 𝑥 cm
l = 2𝑥 cm [∵ length is double of it's breadth.]
P = 42 cm
We know that,
Perimeter of a rectangle (P) = 2(l + b)
Or, 42 cm = 2(2x cm + 𝑥 cm)
Or, 42 cm = 2 x 3𝑥 cm
Or, 42 cm = 6𝑥 cm
Or, 6𝑥 cm/1 = 42 cm/1 [∵ Dividing both sides by 1 or you may divide both sides by 6.]
Or, 𝑥 = 42 cm x 1/6 cm x 1 [∵ Cross multiplication]
Or, 𝑥 = 7
∴ 𝑥 = 7
Now,
Breadth of a rectangle (b) = 𝑥 cm = 7 cm.
Length of a rectangle (l) = 2𝑥 cm = 2 x 7 cm = 14 cm.

2. The length of a square is 𝑥 cm and it's perimeter is 32 cm. Make an equation and solve it to find the value of 𝑥. Then find the length of the square.
Solution:
Here,
l = 𝑥 cm
P = 32 cm
We know that,
Perimeter of a square (P) = 4l
Or, 32cm = 4 x 𝑥cm
Or, 4 x 𝑥cm = 32cm
Or, 𝑥 = 32 cm/4cm
∴ 𝑥 = 8
Now,
The length of a square (l) = 𝑥 cm = 8 cm.

3. A square field is 21 m long. Find it's perimeter. Also, find the length of a rope to fence it 4 times.
Solution:
Here,
Length of a square (l) = 21 m
Perimeter of a square (P) = ?
We know that,
Perimeter of a square (P) = 4l = 4 x 21m = 84m.
Now,
The length of a rope to fence it 1 time = 84m
∴ The length of a rope to fence it 4 times = 84m x 4 = 336m.

4. The breadth of a rectangle is x m and the length is 3 m more than the breadth. If the perimeter of this land is 30 m. Find the value of x and determine the length and breadth of this rectangle.
Solution:
Here,
The breadth of a rectangle (b) = 𝑥 m
The length of a rectangle (l) = (𝑥 m + 3𝑥) m = 4𝑥 m [∵ the length of a rectangle is 3m more than the breadth.]
Perimeter of a rectangle (P) = 2(l + b)
Or, 30 m = 2(𝑥 m + 4𝑥 m)
Or, 30 m = 2 x 5𝑥m
Or, 30 m = 10𝑥 m
Or, 10𝑥 m = 30 m
Or, 𝑥 = 30 m/10 m
∴ 𝑥 = 3
Now,
Length of a rectangle (b) = 𝑥 m = 3 m.
Breadth of a rectangle (l) = 4𝑥 m = 4 x 3 m = 12 m.

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