Square of side (s) = 5 m
Perimeter = 4s
=4 (5)m = 20m
Also, Given, Each side is to be fenced with 2 rows of wire.
= 2 (20) = 40 m 
1 m = Rs. 10
40 m = Rs. 10 \(\times\) 40
= Rs. 400/-

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

Given length = 36 cm
(i) a square 4 s = \(36\Rightarrow s=\cfrac { 36 }{ 4 } =9\) cm
Each side of a square = 9 cm
(ii) an equilateral triangle 3s = \(36\Rightarrow s=\cfrac { 36 }{ 3 } =12\)
Each side of the triangle = 12 cm

(i) Scalene triangle with sides are 7 m, 8 m, 10m
Perimeter = 7m + 8m + 10m = 25m
\(\therefore\) P = 25m
(ii) An isosceles triangle with equal sides 10 cm, 10 cm
3^rd side = 7 cm
P = 10 cm + 10 cm + 7 cm
\(\therefore\) P = 27 cm
(iii) Equilateral triangle with side 6 cm
\(\Rightarrow\) 6 cm, 6 cm,6 cm
P = ( 6 + 6 + 6) cm
\(\therefore\) P = 18 cm

(i) Given :
Base (b) = 20 cm
height (h) = 40 cm
To find: Area of right angled \(\triangle\)
Formula:
Area \((A)=\cfrac { 1 }{ 2 } \times b\times h\ sq.unts\)
= \(\cfrac { 1 }{ 2 } \times 20\times 40\)
A  = 400 cm²
(ii) Given :
Base (b) = 5 feet
Area (A) = 20 sq. feet
To find: Height (h)
\(h=\left( \cfrac { 2A }{ b } \right) \)
= \(\cfrac { 2\times 20 }{ 5 } =8\)
h = 8 feet
(iii) = Given :
Height (h) = 12m
Area (A) = 24 sq. m
Base (B) = \(\cfrac { 2A }{ h } \)
= \(\cfrac { 2\times 24 }{ 12 } =4m\)
\(\therefore \) b = 4m