ICSE Class 8 Maths Mensuration Chapter 01 Perimeter and Area

Perimeter and Area

You are already familiar with the terms perimeter and area and know how to calculate the perimeters and areas of squares and rectangles. Let us revise what you have learnt in your previous class and learn some new concepts. To begin with, let us recall what perimeter and area mean.

The perimeter (P) of a closed plane curve is the total length of the curve. The unit of perimeter is the unit of length.

The area (A) of a closed plane curve is the region enclosed by the plane curve. The area of a plane figure is measured in square units, such as square cm (cm²) and square m (m²).

Triangles

In this section we will learn to calculate the perimeter and area of a triangle.

Perimeter of a Triangle

The perimeter of a triangle ABC = the sum of the lengths of its sides = BC + CA + AB = a + b + c.

Example: The perimeter of a triangle ABC in which AB = 3.2 cm, BC = 5 cm and CA = 7.3 cm is the sum (3.2 + 5 + 7.3) cm = 15.5 cm.

Area of a Triangle

The area of a triangle is equal to half the product of its base and the corresponding height (or altitude).

Area of a triangle = \(\frac{1}{2} \times \text{base} \times \text{height}\)

If BC is the base, the area of triangle ABC = \(\frac{1}{2}BC \times AD\).

If AC is the base, area = \(\frac{1}{2} AC \times BD\).

If AB is the base, area = \(\frac{1}{2} AB \times CD\).

If BC is taken as the base of the triangle ABC, shown in the figure, the perpendicular AD drawn from A to the base BC is called the height of the triangle. Any side of a triangle may be taken as the base.

Example: If the base of a triangle is 12 cm and its height is 8 cm then its area = \(\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 12 \times 8 \text{ cm}^2 = 48 \text{ cm}^2\).

We can also find the area of a triangle by Heron's formula. According to this formula, if a, b and c are the three sides of a triangle then

area of the triangle = \(\sqrt{s(s-a)(s-b)(s-c)}\)

where 2s = the perimeter of the triangle = a + b + c.

Example: Find the area of a triangle of sides 26 cm, 28 cm and 30 cm.

Solution: Here, a = 26 cm, b = 28 cm and c = 30 cm.

\(s = \frac{a + b + c}{2} = \frac{26 \text{ cm} + 28 \text{ cm} + 30 \text{ cm}}{2} = 42 \text{ cm}\).

Thus, the area of the triangle = \(\sqrt{s(s-a)(s-b)(s-c)} = \sqrt{42(42-26)(42-28)(42-30)} \text{ cm}^2 = \sqrt{42 \times 16 \times 14 \times 12 \text{ cm}^2}\)

= \(\sqrt{(14 \times 3) \times 4^2 \times 14 \times (4 \times 3) \text{ cm}^2} = \sqrt{14^2 \times 3^2 \times 4^2 \times 2^2 \text{ cm}^2} = 14 \times 3 \times 4 \times 2 \text{ cm}^2 = 336 \text{ cm}^2\).

If ABC is a right-angled triangle in which angle B = 90°, its area = \(\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} BC \times AB\).

Area of a right-angled triangle = \(\frac{1}{2} \times \text{product of sides containing the right angle}\)

Example: Find the area of triangle ABC if AB = 5 cm, AC = 13 cm and angle B = 90°.

Solution: Using Pythagoras' theorem, \(AB^2 + BC^2 = AC^2\)

\(\Rightarrow BC = \sqrt{AC^2 - AB^2} = \sqrt{169 - 25} \text{ cm} = \sqrt{144} \text{ cm} = 12 \text{ cm}\).

Thus, the area of triangle ABC = \(\frac{1}{2} AB \times BC = \frac{1}{2} \times 5 \times 12 \text{ cm}^2 = 30 \text{ cm}^2\).

If ABC is an equilateral triangle, a = b = c.

\(s = \frac{a + a + a}{2} = \frac{3a}{2}\).

area of triangle ABC = \(\sqrt{\frac{3a}{2}\left(\frac{3a}{2} - a\right)\left(\frac{3a}{2} - a\right)\left(\frac{3a}{2} - a\right)} = \sqrt{\frac{3a}{2} \cdot \frac{a}{2} \cdot \frac{a}{2} \cdot \frac{a}{2}} = \sqrt{\frac{3a \cdot a \cdot a \cdot a}{2 \cdot 2 \cdot 2 \cdot 2}} = \frac{\sqrt{3}}{4} a^2\).

Area of an equilateral triangle = \(\frac{\sqrt{3}}{4} \times (\text{side})^2\)

Example: Find the area of an equilateral triangle of side 6 cm, correct to three places of decimal.

Solution: The side (a) of the equilateral triangle = 6 cm.

the area of the triangle = \(\frac{\sqrt{3}}{4} a^2 = \frac{1.732}{4} \times 6^2 \text{ cm}^2 = 1.732 \times 9 \text{ cm}^2 = 15.588 \text{ cm}^2\).

Solved Examples

Example 1: Find the area of a triangle of sides 29 cm, 36 cm and 25 cm. Also, find the length of the perpendicular from the vertex opposite the side of length 36 cm.

Solution: Let the triangle be ABC and let a = 29 cm, b = 36 cm and c = 25 cm. Then, \(s = \frac{a + b + c}{2} = \frac{29 \text{ cm} + 36 \text{ cm} + 25 \text{ cm}}{2} = 45 \text{ cm}\).

Thus, the area of triangle ABC = \(\sqrt{s(s-a)(s-b)(s-c)} = \sqrt{45 \times (45-29) \times (45-36) \times (45-25) \text{ cm}^2}\)

= \(\sqrt{45 \times 16 \times 9 \times 20 \text{ cm}^2} = \sqrt{(3^2 \times 5) \times 4^2 \times 3^2 \times (5 \times 2^2) \text{ cm}^2}\)

= \(\sqrt{3^2 \times 3^2 \times 5^2 \times 4^2 \times 2^2 \text{ cm}^2} = 3 \times 3 \times 5 \times 4 \times 2 \text{ cm}^2 = 360 \text{ cm}^2\).

Let BD be the perpendicular on AC from the point B. Then, the area of triangle ABC = \(\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} AC \times BD\)

\(\Rightarrow \frac{1}{2} AC \times BD = 360 \text{ cm}^2 \Rightarrow \frac{1}{2} \times 36 \times BD = 360 \text{ cm} \Rightarrow BD = \frac{360}{18} \text{ cm} = 20 \text{ cm}\).

Hence, the length of the perpendicular on AC from the point B = 20 cm.

Example 2: Find the perimeter of an equilateral triangle of area \(16\sqrt{3} \text{ m}^2\).

Solution: Let each side of the triangle = a m. Then its area = \(\frac{\sqrt{3}}{4} a^2 \text{ m}^2\).

From the question, \(\frac{\sqrt{3}}{4} a^2 = 16\sqrt{3} \Rightarrow \frac{a^2}{4} = 16 \Rightarrow a^2 = 64 \Rightarrow a = 8\).

each side of the equilateral triangle = 8 m.

the perimeter of the triangle = 3a = 3 \times 8 m = 24 m.

Teacher's Note

When you're designing a garden bed or cutting fabric for a sewing project, you use these area formulas to determine how much material you need or how the space will be divided.

This is a preview of the first 3 pages. To get the complete book, click below.