ICSE Class 8 Maths Chapter 28 Perimeter and Area of Plane Figures

Chapter 28: Perimeter and Area of Plane Figures

Perimeter of Plane Figures

The perimeter of a closed plane figure is the length of its boundary i.e. the sum of lengths of its sides.

Perimeter of a Triangle

If \(a\), \(b\) and \(c\) are the lengths of the sides of any triangle ABC, then its

perimeter \(= a + b + c\).

Perimeter of Quadrilateral

If \(a\), \(b\), \(c\) and \(d\) are the lengths of the four sides of any quadrilateral ABCD, then its

perimeter \(= a + b + c + d\).

Perimeter of a Rectangle

If \(l\) and \(b\) are the length and breadth of any rectangle ABCD, then its

perimeter \(= 2 (l + b)\).

Perimeter of a Square

If \(a\) is the length of a side of any square ABCD, then its

perimeter \(= 4a\).

Length of Diagonal of a Rectangle

If \(d\) is the length of a diagonal and \(l\), \(b\) are the length and the breadth of any rectangle ABCD, then

\[d = \sqrt{l^2 + b^2}\]

In the adjoining diagram, \(\angle B = 90°\). From \(\triangle ABC\), by Pythagoras theorem, we get

\(AC^2 = AB^2 + BC^2\)

\(\Rightarrow\) \(d^2 = l^2 + b^2\)

\(\Rightarrow\) \(d = \sqrt{l^2 + b^2}\).

In fact, a rectangle has two diagonals and length of each diagonal \(= \sqrt{l^2 + b^2}\).

Length of Diagonal of a Square

If \(d\) is the length of a diagonal and \(a\) is the length of a side of any square ABCD, then

\[d = \sqrt{2} a\]

In the adjoining diagram, \(\angle B = 90°\). From \(\triangle ABC\), by Pythagoras theorem, we get

\(AC^2 = AB^2 + BC^2\)

\(\Rightarrow\) \(d^2 = a^2 + a^2 = 2a^2\)

\(\Rightarrow\) \(d = \sqrt{2} a\).

In fact, a square has two diagonals and length of each diagonal \(= \sqrt{2} a\).

Teacher's Note

Understanding perimeter helps in everyday tasks like calculating how much fencing is needed around a garden or how much border trim is required for a picture frame.

Area of Plane Figures

The area of a closed plane figure is the measure of the region (surface) enclosed by its boundary.

Area of a Triangle

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

Any side of a triangle can be taken as its base and the length of perpendicular (altitude) drawn from the opposite vertex to the base is called its (corresponding) height. In the adjoining diagram, BC is the base of \(\triangle ABC\). From A, draw AD perpendicular to BC, then the length of the line segment AD is the height corresponding to the base BC.

If \(a\), \(b\) and \(c\) are the lengths of the sides of any triangle ABC; then its

\[\text{area} = \sqrt{s(s-a)(s-b)(s-c)}\]

where \(s = \text{semi-perimeter} = \frac{a + b + c}{2}\).

This is known as Heron's Formula.

Area of a right angled triangle: Let ABC be a triangle in which \(\angle B = 90°\),

then its area \(= \frac{1}{2} \times BC \times AB\)

\(= \frac{1}{2}\) (product of sides containing right angle).

Area of an equilateral triangle: Let ABC be an equilateral triangle with side \(a\), then

\(s = \text{semi-perimeter} = \frac{a + a + a}{2} = \frac{3a}{2}\).

By Heron's formula,

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

\[= \sqrt{\frac{3a}{2}\left(\frac{3a}{2} - a\right)\left(\frac{3a}{2} - a\right)\left(\frac{3a}{2} - a\right)}\]

\[= \sqrt{\frac{3a}{2} \cdot \frac{a}{2} \cdot \frac{a}{2} \cdot \frac{a}{2}} = \sqrt{\frac{3}{4}a^2}\]

Hence, area of an equilateral triangle with side \(a\) \(= \frac{\sqrt{3}}{4} a^2\).

Teacher's Note

Calculating triangle area is essential in construction and design, such as determining how much paint is needed for a triangular roof section or how much material is required for a triangular garden bed.

Example 1

If the base of a right-angled triangle is 6 units and the hypotenuse is 10 units, find its area.

Solution

Let \(\triangle ABC\) be a right-angled triangle at B i.e. \(\angle B = 90°\) and its base \(BC = 6\) units and hypotenuse \(AC = 10\) units.

From \(\triangle ABC\), by Pythagoras theorem, we get

\(AC^2 = BC^2 + AB^2\) \(\Rightarrow\) \(10^2 = 6^2 + AB^2\)

\(\Rightarrow\) \(AB^2 = 10^2 - 6^2 = 100 - 36 = 64\)

\(\Rightarrow\) \(AB = \sqrt{64}\) units \(= 8\) units.

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

\(= \left(\frac{1}{2} \times 6 \times 8\right)\) sq. units \(= 24\) sq. units.

Example 2

Calculate the area of a triangle whose sides are 29 cm, 20 cm and 21 cm. Hence find the length of the altitude corresponding to the longest side.

Solution

Since the sides of the triangle are 29 cm, 20 cm and 21 cm.

\(\therefore\) \(s = \text{semi-perimeter} = \frac{29 + 20 + 21}{2}\) cm \(= 35\) cm.

\(\therefore\) Area of triangle \(= \sqrt{s(s-a)(s-b)(s-c)}\)

\(= \sqrt{35(35-29)(35-20)(35-21)}\) cm\(^2\)

\(= \sqrt{35 \times 6 \times 15 \times 14}\) cm\(^2\)

\(= \sqrt{5 \times 7 \times 2 \times 3 \times 3 \times 5 \times 7 \times 2}\) cm\(^2\)

\(= (5 \times 7 \times 2 \times 3)\) cm\(^2\) \(= 210\) cm\(^2\)

The longest side of the triangle is 29 cm, let \(h\) cm be the length of the corresponding altitude, then

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

\(\Rightarrow\) \(210 = \frac{1}{2} \times 29 \times h\) \(\Rightarrow\) \(h = \frac{420}{29} = 14\frac{14}{29}\).

Hence, the length of the altitude corresponding to the longest side \(= 14\frac{14}{29}\) cm.

Example 3

Find the area of an equilateral triangle of side 8 m correct to three significant figures. Use \(\sqrt{3} = 1.732\).

Solution

Given side of an equilateral triangle \(= a = 8\) m.

Area of the equilateral triangle \(= \frac{\sqrt{3}}{4} a^2\)

\(= \left(\frac{\sqrt{3}}{4} \times 8 \times 8\right)\) m\(^2\) \(= 16\sqrt{3}\) m\(^2\)

\(= (16 \times 1.732)\) m\(^2\) \(= 27.712\) m\(^2\)

\(= 27.7\) m\(^2\), correct to 3 significant figures.

Example 4

If the area of an equilateral triangle is \(49\sqrt{3}\) cm\(^2\), find its perimeter.

Solution

Let a side of the equilateral triangle be \(a\) cm, then its area \(= \frac{\sqrt{3}}{4} a^2\) cm\(^2\).

According to given information, \(\frac{\sqrt{3}}{4} a^2 = 49\sqrt{3}\)

\(\Rightarrow\) \(a^2 = 4 \times 49\) \(\Rightarrow\) \(a = \sqrt{4 \times 49} = 14\).

\(\therefore\) The perimeter of the triangle \(= 3a\) cm \(= (3 \times 14)\) cm \(= 42\) cm.

Example 5

If the base of an isosceles triangle is 10 cm and its perimeter is 36 cm, find the area of the triangle.

Solution

Let ABC be an isosceles triangle with BC as its base and AB, AC its equal sides, then \(BC = 10\) cm. Let the length of each of its two equal sides be \(x\) cm, then perimeter of \(\triangle ABC = (x + x + 10)\) cm \(= 36\) cm (given)

\(\Rightarrow\) \(2x + 10 = 36\) \(\Rightarrow\) \(2x = 36 - 10 = 26\)

\(\Rightarrow\) \(x = 13\).

\(\therefore\) Thus, sides \(a\), \(b\) and \(c\) of the triangle ABC are 10 cm, 13 cm and 13 cm respectively.

\(s = \text{semi-perimeter} = \frac{1}{2}\) of 36 cm \(= 18\) cm

\(\therefore\) \(s - a = (18 - 10)\) cm \(= 8\) cm,

\(s - b = (18 - 13)\) cm \(= 5\) cm

and \(s - c = (18 - 13)\) cm \(= 5\) cm

\(\therefore\) The area of \(\triangle ABC\) \(= \sqrt{s(s-a)(s-b)(s-c)}\)

\(= \sqrt{18 \times 8 \times 5 \times 5}\) cm\(^2\) \(= \sqrt{144 \times 25}\) cm\(^2\)

\(= 12 \times 5\) cm\(^2\) \(= 60\) cm\(^2\).

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