(i)

(ii)
length = 2 (breadth)
I = 2b --------- (1)
Formula: P = 2(1 + b)
= 2 (2b+b) = 2 (3b)
P = 6b units \(\Rightarrow\) P = 6 \(\times\) breadth.
\(\therefore\) Perimeter of a rectangle.
(iii) Side is reduced to half.
\(a=\left( a-\cfrac { a }{ 2 } \right) =\left( \cfrac { 2a-a }{ 2 } \right) =\cfrac { a }{ 2 } \)
 \(\therefore\) Perimeter of a square = 4a
=\(4\left( \cfrac { a }{ 2 } \right) \)
P = 3a units
(iv) All sides are equal in length a = b = c
\(\therefore\) Perimeter of a triangle= ( a + b + c) units
= a + a + a = 3a
\(\therefore\) P = 3a units.

base (b) = 25 m and height (h) = 20 m
Area (A) = \(\left( \cfrac { 1 }{ 2 } \times base\times height \right) \)sq.units
= \(\cfrac { 1 }{ 2 } \times 25\times 20\)
A = 250 m²
A = 250m²
1 sq. m = Rs. 45 
250 sq. m = 45 \(\times\) 250
\(\therefore\)Cost ofthe field = Rs. 11, 250

Perimeter (P) = 40 m
Side (8) =?
Area (A) =?
4S = P
\(4S=40\Rightarrow S=\cfrac { 40 }{ 4 } =10m\)
A = S² = 10² = 100m²
\(\therefore\) Side (S) = 10 m, Area (A) = 100 m²

Sides are = 6 feet, 8 feet and 10 feet
Perimeter = ( a + b + c) feet
= (6+8+10) = 24 feet.
Area = \(\cfrac { 1 }{ 2 } \)bh.sq.feet
= \(\cfrac { 1 }{ 2 } \times 6\times 8\)
= 3 \(\times\) 8 = 24 feet
\(\therefore\) P = 24 feet, A = 24 sq. feet

Length (l) = 6 m
breadth (b) = 4 m
Perimeter of the rectangle (P) = 2 (1+b) units
= 2 (6 + 4)
= 2(10) = 20m
Area(A) = l\(\times\)b = (6 \(\times\) 4)m²
= 24m
\(\therefore\) P = 20 m, A = 24 m²