Question

(1) From the measures given in the figure, find the area of ∆ABC.

(2) Using Pythagoras theorem, find the length of side AC.

(3) Find the area of ∆ABC using Heron's formula.

Answer 1

(1) Height of the right triangle is 12 cm.
Base of the right triangle is 5 cm.
Area of a right triangle = $\frac{1}{2}$ (Base × Height)
= $\frac{1}{2}$ × 12 cm × 5cm
= 30 sq cm
∴ Area of the right triangle is 30 sq cm.

(2) Height of the right triangle = AB = 12 cm
Base of the right triangle = BC = 5 cm
By Pythagoras theorem,
AC² = BC² + AB²
AC² = 144 + 25
AC = 13 cm
∴ The length of the third side AC is 13 cm .

(3) Lengths of the sides of the triangle are 12 cm, 5 cm and 13 cm.
Semi-perimeter of a triangle = $s=\frac{a+b+c}{2}$
Semi-perimeter $=\frac{5+12+13}{2}=15\mathrm{cm}$
Area of a triangle = $\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$
$$\begin{gathered} =\sqrt{15\left(15-5\right)\left(15-12\right)\left(15-13\right)} \\ =\sqrt{15\left(10\right)\left(3\right)\left(2\right)} \\ =\sqrt{900}=30 \end{gathered}$$
∴ Area of the right triangle is 30 sq cm.