Samacheer Class 6 Maths - Perimeter and Area

l = 6 m, b = 4 m Perimeter of the rectangle
= 2 (l + b) units
= 2 (6 + 4) m
= 2 (10) m
= 20 m
Area of the rectangle = l × b sq units
= 4 × 6 m²
= 24 m²

a = 8 cm
The perimeter of a square
= 4a units
= 4 × 8 cm
= 32 cm
Area of the square = a × a sq units
= 8 × 8 cm²
= 64 cm²

Perimeter of the triangle
= (a + b + c) units
= (6 + 8 + 10) feet
= 24 feet
Area of the triangle = \(\frac{1}{2}\) × b × h sq units
\(\frac{1}{2}\) × 6³× 8 feet square = 24 sq. feet

(i) Perimeter of the triangle
= (a + b + c) units
= (7 + 8 + 10) m
= 25
(ii) Perimeter of the triangle
= (10 + 10 + 7) cm
= 27 cm
(iii) Perimeter of the triangle
= (6 + 6 + 6) cm
= 18 cm

Given Area = 820 cm²
Width = 20 cm
Area of the rectangle
= l × b sq. units
820 = l × 20
\(\frac{820}{20}\) = l
41 = l
length l = 41 cm
Perimeter = 2(l + b) units
= 2(41 + 20) cm
= 2(61) cm
= 122 cm

perimeter = 40 m
4a = 40 m
a = \(\frac{40}{4}\)
Side a = 10 m
Area = a × a sq units
= 10 × 10 m²
= 100 m²

Let the third side be C
perimeter = (a + b + c) units
40 = 13 + 15 + C
40 = 28 + C
C = 40 – 28
C = 12 units
C = 12 cm

b = 25 m, h = 20 m
Area of the triangle = \(\frac{1}{2}\) × bh sq.units
= \(\frac{1}{2}\) × 25 × 20 m²
= 250 m²
Cost of levelling 1 m² = Rs 45
Cost of levelling 250 m² = Rs 45 × 250
= Rs. 11250

Perimeter of the combined shape = Lengths of the outer boundaries
= (15 + 10 + 2 + 2 + 2 + 13 + 10) cm = 54 cm
Perimeter = 54 cm

Objective Type Questions

1. A. Perimeter remains the same but the area changes
2. B. Area remains the same but the perimeter changes
3. C. There will be a change in both area and perimeter
4. D. Both the area and perimeter remains the same.

(a) Perimeter remains the same but the area changes
Posted in Class 6 on January 23, 2025 January 24, 2025
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Given the length of the wire = 36 cm
i) When a square is formed out of it
The perimeter of the square = 36 cm
4 × side = 36
side = \(\frac{36}{4}\) = 9 cm
Side of the square
ii) When an equilateral triangle is formed out of it, its perimeter = 36 cm
i.e., side + side + side = 36 cm .
3 × side = 36 cm
side = \(\frac{36}{3}\) = 12 cm
One side of an equilateral triangle = 12 cm

Distance covered by Rahim
= 50 × 4 m
= 200 m
If he walks 2 rounds, distance covered = 2 × 200 m
= 400 m
Distance covered by peter
= 2 (40 + 30) m
= 2(70)m
= 140 m
If he walks 2 rounds, distance covered = 2 × 140 m
= 280 m
∴ Rahim covers more distance by (400 – 280) = 120 m

Let the length be b + 14 m
breadth = b
perimeter = 200
2 (l + b) = 200
2 (b + 14 + b) = 200
2 (2b + 14) = 200
28 + 4b = 200
4b = 200 – 28
4b = 172 m
b = \(\frac{172}{4}\)
b = 43 m
Length = b + 14
= 43 + 14
Length l = 57 m
Area = l × b units
= 57 × 43 m²
= 2451 m²

a = 5 m
The perimeter of the garden
= 4 a units
= 4 × 5 m
= 20 m
For 1 row
Amount needed to fence l m= Rs 10
Amount needed to fence 20 m
= Rs 10 × 20
= Rs 200
For 2 rows
Total amount needed = 2 × Rs 200 = Rs 400

Challenge Problems

Number of equal sides in the shape = 20
One of its side = 3 cm
Perimeter = length of one side × Number of equal sides
∴ Perimeter = (3 × 20) cm = 60 cm
∴ Perimeter = 60 cm

Area of rectangle = (length × breadth) units 2
Length = 40 cm
Breadth = 20 cm
∴ Area = (40 × 20) cm 2 = 800 cm 2
Area of rectangle = 800 cm 2
Area of square = (side × side) units 2
side = 10 cm
∴ Area of square = (10 × 10) cm 2 = 100 cm 2
Required number of squares = \(\frac{\text { Area of Rectangle }}{\text { Area of } 1 \text { square }}=\frac{800 \mathrm{cm}^{2}}{100 \mathrm{cm}^{2}}\) = 8
8 squares can be formed.

Given perimeter of a rectangle = 64 cm
Also given length is three times its breadth.
Let the breadth of the rectangle = b cm
∴ Length = 3 × b cm
Perimeter = 64 m
i.e., 2 × (l + b) = 64 m
2 × (3b + b) = 64 m
2 × 4b = 64m
4b = \(\frac{64}{2}\) = 32 m
b = \(\frac{32}{4}\) = 8 m
l = 3 × b = 3 × 8 = 24 m
∴ Breadth of the rectangle = 8 m
Length of the rectangle = 24 m

Length of the string to be made into rectangle = 48 cm
∴ Perimeter of the rectangle = 48 cm
2 × (l + b) = 48 cm
l + b = \(\frac{48}{2}\)
l + b = 24 cm
Possible pairs of length and breadth are (1, 23), (2, 22) (3, 21), (4, 20), (5, 19),
(6, 18), (7, 17), (8, 16), (9, 15), (10, 14), (11, 13), (12, 12)
Number of different rectangles = 12.

Perimeter of A = s + s + s + s units = 4 s units
Perimeter of B = (2s + 2s + 2s + 2s) units
= 8s units = 2 (4s) units.
∴ The perimeter of B is twice the perimeter of A

Let the side of square is s units then area = (s × s) units 2
If the side of the new square is made one fourth then side = \(\left(\frac{1 \times s}{4}\right)\) units
Then area = \(\left(\frac{1 \times s}{4} \times \frac{1 \times s}{4}\right)\) units 2 = \(\frac{s \times s}{16}=\frac{1}{16}\) (s × s) units 2
Area of the new square is reduced to \(\frac{1}{16}\) times to that of original area.

a = 10 m, b = 8 m
Perimeter of the square plot
= 4 a units
= 4 × 10 m
= 40 m
Perimeter of the rectangular plot
40 = 2 (l + b) units
40 = 2 (l + 8) m
40 = 2 l + 16
2 l = 40 – 16
2 l = 24
l = \(\frac{24}{2}\)
l = 12 m
Area of the square plot
= a × a sq units
= 10 × 10 m²
= 100 m²
Area of the rectangular plot
= l × b sq units
= 8 × 12 m²
= 96 m²
Square plot has the greater area by 100 m² – 96 m² – 4 m²