Perimeter (P) = a + b + C
= (40 + 40 + 40) cm
= 120 cm
Removed portion = 6 cm
\(\therefore\) Perimeter of the remaining
portion = (120 - 6)
= 114 cm

Side of a square plot = 12 m
Perimeter of the square plot = (4 \(\times\) s) units
= 4 × 12 = 48m
Cost of fencing the plot at the rate of Rs. 15 per metre = 48 × 15 = Rs. 720

Perimeter of the black board, P = 6 m
Breadth of the black board, b = 1 m
length, l = ?
2 (l + b) = 6
2 (l + 1) = 6
\(l+1={6\over2}\)
= 3
l = 3 – 1
= 2 m
The length of the black board is 2 m.

a = 3 cm
b = 4 cm
c = 5 cm
P = (a + b + c) units
= 3 + 4 + 5 = 12 cm
Perimeter of the triangle is 12 cm.

s = 5 cm
P = (4 × s) units
= 4 × 5
= 20 cm
Perimeter of the square is 20 cm

Perimeter = 80 cm
No. of identical shapes = 9
Side of square = 4 cm
Possible shape:

\(\therefore\) P = 80cm

To prove: Square B = 2 (Square A)
Let the side of the square B = 4 cm
P = 4s = 4 \(\times\) 4 = 1.6.cm
Next, The side of the square A = 2 cm
P = 4s = 4 \(\times\) 2 = 8 cm
\(\therefore\) Square B = 16 = 2 (8) = 2 square A
\(\therefore\) Perimeter of square B is twice that of square A.

Length of the string = 48 cm
No. of possible pairs = 12
12 pairs,
The possible pairs of length and breadth of the rectangles 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)
12 different rectangles can be made.

l = 3b
P = 64 cm
2 (I + b) = 64
2(3b + b) = 64
2 (4b) = 64
\(8b=64\Rightarrow b=\cfrac { 64 }{ 8 } 8\)cm
I = 3 (8) = 24 crn
The sides of the rectangle I = 24 cm, b = 8 cm

(i) Given:
Length (1) = 5 cm
breadth (b) = 8 cm
Perimeter = 2 (I + b)
= 2 (5 + 8)
= 2 \(\times\)13 = 26 cm
Are a (A) = lb = 5 \(\times\) 8 = 40 cm²
(ii) Given :
Length (1) = 13 cm
Perimeter (P) = 54 cm
To find: b and A
\(b=\cfrac { P-2l }{ 2 } \)
= \(\cfrac { 54-2(13) }{ 2 } \)
= \(\cfrac { 54-26 }{ 2 } =\cfrac { 28 }{ 2 } =14\)
b = 14 cm
Area (A) = lb = 13 \(\times\) 14
A = 182 cm²
(iii) Given:
Breadth (b) = 15 cm
Perimeter (P) = 60 cm
To find: 1 and A
\(I=\cfrac { P-2b }{ 2 } \)
= \(\cfrac { 60-2(15) }{ 2 } =\cfrac { 60-30 }{ 2 } =\cfrac { 30 }{ 2 } \)
1 = 15 cm
A = lb = 15 \(\times\) 15 = 225
A = 225 cm²
(iv) Given :
Length (1) = 10m
Area (A) = 120 sq. m
To find: band P
\(b=\cfrac { A }{ l } \)
= \(\cfrac { 120 }{ 10 } =12\)
b = 12m
P = 2 (1 +  b)
= 2(10 + 12)
= 2 \(\times\) 22 = 44
P = 44 m
(v) Given:
Breadth (b) = 4 feet
Area (A) = 20 s. feet
To find: 1 and P
\(\therefore l=\cfrac { A }{ b } \)
= \(\cfrac { 20 }{ 4 } =5\)
1= 5 feet
P = 2(I + b)
= 2(5 + 4)
= 2(9) = 18
P = 18 feet