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/-

The length of a rectangular park is 14
m more than its breadth
1 = b + 14m -------- (1)
Perimeter P = 200 m
2 (1+ b) = 200
2 (1+ 2b) = 200
2 (b + 14) + 2b = 200 [by (1)]
2b + 28 + 2b = 200
4b = 200-28
\(4b=172\Rightarrow b=\cfrac { 172 }{ 4 } =43\)
b = 43 m
1 = b + 14 m
= 43 m + 14 m = 57 m
1 = 57 m
Area = Ib sq. units
= ( 57 \(\times\) 43) m²
\(\therefore\) I = 57 m, A = 2451 m²

Rahim walks = 50m
Peter walks = rectangular path with length 40m and breadth 30m
Given: Both of them walk 2 rounds each.
Rahim:
= square path of
P = 4 \(\times\) side
= 4 \(\times\) 50 = 200m
\(\Rightarrow\) 2 rounds = 2 (200) = 400 m
Peter
P = 2(l + b)
= 2 (40 + 30) m \(\Rightarrow\) 2
(70) m = 140 m
\(\Rightarrow\) 2 rounds \(\Rightarrow\) 2 (140) m = 280 m
\(\Rightarrow\) Difference = 400 m - 280 m
= 120 m
Ans: Rahim, 120m

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) Given :
side (s) = 6 cm
To find: P and A
P = 4s
= 4 \(\times\) 6 = 24 cm
A = S² = 6² = 36 cm²
Ans: Perimeter = 24 cm, Area = 36 cm²
(ii) Given :
P = 100 m
To find: S and A
4s = 10
\(s=\cfrac { 100 }{ 4 } =25m\)
A = S²= 25² = 625 m²
Ans: side = 25 m, A = 625 m² .
(iii) Given :
A = 49 s. feet
To find: S and P
A = S²= 49
\(S=\sqrt { 49 } =7\) feet
P = 4S
= 4 \(\times\) 7 = 28 feet.
Ans: Side = 7 feet, Perimeter = 28 feet.

i) 2 cm² = 2 \(\times\)100 = 200 mm²
ii) 18 m² =18 \(\times\)10000 = 1,80,000 cm²
iii) 5 km² = 5 \(\times\) 1000000 = 50,00,000 m²

Base, b = 18 cm
Height, h = 12 cm
Area, \(A={1\over2}(b x h)\ sq\ units\)
\(={1\over2}(18\times12)\)
= 108 sq. cm. (or) 108 cm²

Perimeter of the equilateral triangle, P = 129
a + a + a = 129
3 \(\times\) a = 129
\(a={129\over3}\)
= 43 cm
The side of the equilateral triangle is 43 cm.

Length of the rectangle, l = 12
Breadth of the rectangle, b = 7
Area of the rectangle A = (l \(\times\)b) sq. units.
= 12 \(\times\) 7 = 84 sq. cm

Perimeter of the square, P = 8 cm
4 \(\times\) S = 8
\(S={8\over4}\)
= 2 cm
The side of the stamp is 2 cm.

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

l = 12 cm
b = 10 cm
P = 2 (l + b) units
= 2(12 + 10)
= 2× 22
= 44 cm
Perimeter of the rectangle is 44 cm.

From the figure,
Complete squares = 9
half squares = \(\cfrac { 13 }{ 2 } \)
\(\therefore\) Approximate area = Complet esquares + half squares
= \(9+\cfrac { 13 }{ 2 } \)
= \(\cfrac { 18+13 }{ 2 } =\cfrac { 31 }{ 2 } \)
= 15.5 sq.units

Total area of the shaded portion = Area of square + 2 (Area of triangle) + Area of rectangle
Square: side = 6 cm
Triangles: Same Triangles \(\Rightarrow\)b = 3, h = 4
Rectangle : b = 6 cm, I = 9 cm
\(\therefore\)Total area = s² + 2 (\(\cfrac { 1 }{ 2 } bh\) ) + lb
= 6² + 2 (\(\cfrac { 1 }{ 2 } \)\(\times\) 3 \(\times\) 24)+ 6 \(\times\) 9
= 36 + 12 + 54
= 102 cm²

Square of side = 10 m
Breadth of rectangle = 8 m
Square:
Area of the square = s² sq. units
= 102 = 100m²
P = 4s = 4(10) = 40 m
Rectangle:
Area of the rectangle = I\(\times\) b sq. units
b = 8 m
P = 40 m
2 (l+ b) = 40
2(l + 8) = 40
2l + 16 = 40
2l = 40 -16
\(l=\cfrac { 24 }{ 2 } =12\)
Area of the rectangle = (12 \(\times\) 8) m2
Difference = Area of the square - Area of the rectangle
= 100 - % = 4m²
Square plot has the greater area and 4 m²

Side of square = s unit
The side of square is made one - fourth
\(\Rightarrow \cfrac { 1 }{ { 4 }^{ 2 } } =\cfrac { 1 }{ 16 } \) times
\(\therefore\) Area of the new square is reduced to\(\cfrac { 1 }{ 6 } \) times to that of original

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

length = 40 cm
breadth = 20 cm
Side of the square = 10 cm
Area of the rectangle = lb
= ( 40 \(\times\) 20)cm²
= 800 cm²
(side) s = 10cm
Area of a square (S2) = 102 = 100 cm²
No. of square = \(\cfrac { 800 }{ 100 } =8\)

Each side of shape = 3 cm
\(\therefore\) 20 equal sides perimeter = 20 
times 3cm
P = 20 \(\times\) 3 = 60 cm
\(\therefore\) P = 60 cm

(i) Given :
Perimeter = 12 times 4 cm
= 12 \(\times\) 4 = 48 cm
Area \(\Rightarrow\) Given  = 5 squares
= 5 \(\times\) area of square
= 5 \(\times\) (sider)²
= 5 \(\times\) 42
\(\therefore\)  A = 5 \(\times\) 16 = 80 Cm²
 \(\therefore\) P = 48 cm, A = 80 cm²
(ii) Given :
Perimeter = 4 times 4 cm + 4 times 5cm
= 4 (4) + 4 (5)
= 16 + 20 = 36 cm
Area = 4 triangles + 1 square
= \(4\left( \frac { 1 }{ 2 } bh \right) +1\left( s \right) ^{ 2 }\)
= \(4\left( \frac { 1 }{ 2 } \left( 5 \right) \left( 4 \right) \right) +1\left( { 3 }^{ 2 } \right) \)
= 4 \(\times\) 5 \(\times\) 2 + 9
= 40 + 9 = 49 cm²
(iii) Given :
Perimeter = (50 + 12 + l3 + 40 + 10 +10 + 10 + 5) cm
= 150 cm
= \(\left( lb+\cfrac { 1 }{ 2 } bh+{ s }^{ 2 } \right) { cm }^{ 2 }\)
= \(\left( 50\times 5 \right) +\cfrac { 1 }{ 2 } \times 12\times 5+{ 10 }^{ 2 }\)
= 250 + 30 + 100
= 380 cm²
\(\therefore\) P = 150 cm, A = 380 cm²

(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

A square of side = 2 cm
A rectangle of length = 15 cm
and
breadth= 10 cm
Perimeter of the combined shape = Perimeter of square + Perimeter of rectangle.

Perimeter of the combined shape = 10 cm + 15 cm + 10 cm + 2 cm + 2 cm + 2 cm + 13 cm
\(\therefore\)  P = 54 cm.

Perimeter = 40 cm
Two sides are 13 cm and 15 cm
To find: 3^rd side (c)
a + b + C = 40 \(\Rightarrow\) 13 + 15 + c = 40
\(\Rightarrow\) 28 +c = 40
c = 40 - 28 = 12cm
\(\therefore\) Third side = 12 cm

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

Area of a rectangle = 820 sq. cm
Width (b) = 20 cm
Length (l) =?
Perimeter =?
\(I=\cfrac { A }{ b } \)
= \(\cfrac { 820 }{ 20 } =41\\ \)
= =2(41+20)
= 2 (61) = 122 cm
\(\therefore\) I = 41 cm , P = 122 cm

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