PERIMETER OF 2D SHAPES

Find the perimeter of the following:

Problem 1 :

Solution :

Given, side a = 5 cm

Perimeter of a triangle = 3a

= 3 × 5

= 15 cm

So, perimeter of a triangle is 15 cm.

Problem 2 :

Solution :

Given, length l = 4 cm and width = 7 cm

Perimeter of a rectangle = 2(l + w)

= 2(4 + 7)

= 2(11)

= 22 cm

So, perimeter of a rectangle is 22 cm.

Problem 3 :

Solution :

Given, a = 5 cm, b = 12 cm

By using Pythagorean Theorem,

c² = a² + b²

c = √(a² + b²)

c = √(5² + 12²)

c = √(25 + 144)

c = √169

c = 13 cm

Perimeter of a triangle = a + b + c

= 5 + 12 + 13

= 30 cm

So, perimeter of a triangle is 30 cm.

Problem 4 :

Solution :

Let a = 6 cm, b = 4 cm, c = 6 cm and d = 10 cm

Perimeter of a Trapezoid = a + b + c + d

= 6 + 4 + 6 + 10

= 26 cm

So, Perimeter of a Trapezoid is 26 cm.

Problem 5 :

Solution :

Let a = 5 cm, b = 12 cm

By using Pythagorean Theorem,

c² = a² + b²

c = √ (a² + b²)

c = √(5² + 12²)

c = √(25 + 144)

c = √169

c = 13 cm

Perimeter of a triangle = a + b + c

= 5 + 12 + 13

= 30 cm

So, perimeter of a triangle is 30 cm.

Problem 6 :

Solution :

Let a = 4 cm, b = 5 cm

By using Pythagorean Theorem,

c² = a² + b²

c = √(a² + b²)

c = √(4² + 5²)

c = √(16 + 25)

c = √41

c = 6.4 cm

Perimeter of shapes = a + b + c + d

= 4 + 2 + 7 + 6.4

= 19.4 cm

So, perimeter of shapes is 19.4 cm.

Problem 7 :

Kim decides to get fit by running laps of a nearby shopping complex. Its dimensions are shown in the diagram alongside.

How many laps does he have to complete to run 13 km in total?

Solution :

Let a = 400 m and b = 300 m

By using Pythagorean Theorem,

c² = a² + b²

c = √a² + b²

c = √400² + 300²

c = √160000 + 90000

c = √250000

c = 500 m

Perimeter of shapes = 700 + 700 + 500 + 400 + 300

= 2600 m

1 km = 1000 m

13 km = 13000 m

= 13000/2600

= 5 laps