Math - Geometry Lesson Plans : Perimeter & Area of Triangles

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Mensuration - Formulae for Perimeter and Area of Triangles

The perimeter of a plane figure is the sum of its sides.
The unit of perimeter is meter (m) or centimeter (cm).
The area of a plane figure is the surface enclosed by its sides.
The unit of area is square meter (m^{2}) or square centimeter (cm^{2}).

A triangle is a polygon which has three sides.
An equilateral triangle has equal sides and equal angles.
An isosceles triangle has two equal sides and two equal angles.
A scalene triangle has three unequal sides and three unequal angles.
A right-angled triangle has one right angle (90°).
An acute-angled triangle has all angles less than 90°.
An obtuse-angled triangle has one angle greater than 90°.

Perimeter of a triangle = Sum of three sides
In the figure alongside of the ΔABC, perimeter = AB + BC + AC.

Example
Find the perimeter (in cm) of a triangle whose sides are 10 cm, 20 cm and 30 cm. Solution.
Perimeter of a triangle = Sum of three sides
= 10 + 20 + 30 = 60 cm.

Area of a triangle = ½ × Base × Height
Any side of the triangle may be considered its base.
Then, the length of the perpendicular from the opposite vertex is taken as the corresponding height or altitude.
In the figure alongside of the ΔABC, area = ½ × AC × BD.

Example
Find the area (in cm^{2}) of a triangle with base 14 cm and height 9 cm. Solution.
Area of a triangle = ½ × Base × Height = ½ × 14 × 9 = 63 cm^{2}.

Where a, b and c are the lengths of the sides of the triangle,
and s = ½ (a + b + c) is the semi-perimeter of the triangle.

Area of a triangle = √

s (s − a) (s − b) (s − c)

by Heron's Formula (or Hero's Formula)

Example
The lengths of the sides of a triangle are 3 cm, 4 cm and 5 cm. Find the area (in cm^{2}) of the triangle. Solution.
Semi-perimeter s = ½ (3 + 4 + 5) = 6 cm.

By Heron's Formula (or Hero's Formula), area of the triangle = √

6 × 3 × 2 × 1

= 6 cm^{2}.

Area of an equilateral triangle = √3 / 4 × Side^{2}
In equilateral Δ ABC, AB = BC = AC = a and AD = DC = ½ a

∴ Using the Pythagorean Theorem on Δ ADB, height = √

a^{2} − (a / 2)^{2}

= √3 / 2 × a

Area = ½ × Base × Height = ½ × a × √3 / 2 × a = √3 / 4 × a^{2}

Example
If the area of an equilateral triangle is 16√3 cm^{2}, find its perimeter (in cm). Solution.
Area of an equilateral triangle = √3/4 × side^{2}
∴ side^{2} = 4/√3 × area = 4/√3 × 16√3 = 64 or side = 8 cm.
Perimeter = 3 (side) = 3 (8) = 24 cm.

In isosceles Δ ABC, AB = BC = a, AC = b

∴ Using the Pythagorean Theorem on Δ ADB, height = √

4a^{2} − b^{2}

/ 2

Area = ½ × Base × Height = ½ × b × √

4a^{2} − b^{2}

/ 2 = ¼ b √

4a^{2} − b^{2}

Area of an isosceles triangle = ¼ b√

4a^{2} − b^{2}

Example
Find the area (in m^{2}) of an isosceles triangle, each of whose sides is 10 m and base 12 m. Solution.

= ¼ (12) √

4 × 10^{2} − 12^{2}

= 48 m^{2}.

Area of an isosceles triangle = ¼ b √

4a^{2} − b^{2}

Area of a right-angled triangle = ½ × Product of sides containing the right angle
In the figure alongside, Δ ABC is a right-angled triangle in which ∠ B = 90° whose area = ½ ×AC × BC.
Note that AB^{2} = AC^{2} + BC^{2} (By Pythagorean Theorem)

Example
The base of a right-angled triangle is 3 cm and its hypotenuse is 5 cm. Find the area (in cm^{2}) of the triangle. Solution.
In right-angled Δ ABC, base AC = 3 cm and hypotenuse AB = 5 cm. ^{2}.

By Pythagorean Theorem, height BC = √

5^{2} − 3^{2}

= 4 cm.

Area of the triangle = ½ × AC × BC = ½ × 3 × 4 = 6 cm