The perimeter of rectangle ABCD as 2(AB + BC).
If the perimeter is denoted by P, length as l and breadth as b. Then,
P = 2(l + b)
Example 1: Find the perimeter of a rectangle whose length is 20 cm and breadth is 8 cm.
Perimeter of rectangle = 2(20 + 8)
= 2 x 28 = 56 cm
Example 2: The perimeter of a rectangle is 46 cm, length is 15 cm, find the breadth of the rectangle.
46 = 2(15 + b)
23 = 15 + b
b = 8 cm
Area of rectangle = (length × breadth)
Area = l × b
Example 3: Find the area of a rectangle whose length is 5 cm and width is 3 cm.
Area = 5 x 3 = 15 cm2
Example 4: Find the length of a rectangle whose area is 400 cm2 and the width is 16 cm.
Area = Length x Width
400 = Length x 16
Length = 25 cm
Example 5: A rectangular park of length 30 m and breadth 24 m is surrounded by a 4 m wide path. Find the area of the path.
Let ABCD be the park and EFGH be the path (outer rectangle) surrounding it.
Length of rectangle EFGH = (30 + 4 + 4) m = 38 m
Breadth of rectangle EFGH = (24 + 4 + 4) m = 32 m
Area of path = area of rectangle EFGH - area of rectangle ABCD
= (38 × 32 - 30 × 24) m2
= (1216 - 720) m2
= 496 m2
Example 6: There are two rectangular paths in the middle of a park. Find the cost of paving the paths with concrete at the rate of Rs.15 per m2. It is given that AB = CD = 50 m, AD = BC = 40 m and EF = PQ = 2.5 m.
Area of paths = Area of PQRS + Area of EFGH - area of square MLNO
= (40 × 2.5 + 50 × 2.5 - 2.5 × 2.5) m2
= 218.75 m2
So, cost of paving the concrete at the rate of Rs.15 per m2 = 218.75 × 15
= Rs.3281.25