3 Tutor System
Starting just at 265/hour

# Find the area of a triangle two sides of which are 18 cm and 10 cm and the perimeter is 42 cm.

Let the sides of a triangle a = 18 cm, b = 10 cm and c

We have, perimeter = 42 cm

So, a + b + c = 42

By substituting the values,

$$\Rightarrow$$ 18 + 10 + c = 42
$$\Rightarrow$$ 28 + c = 42cm
$$\Rightarrow$$ c = 42 - 28 cm
$$\Rightarrow$$ c = 14cm

Now, we know that,
$$s = \frac{a + b + c}{2}$$

$$\therefore$$ $$s = \frac{18 + 10 + 14}{2} = \frac{42}{2} = 21cm$$

Now, Area of triangle
= $$\sqrt{21(21 - 18)(21 - 10)(21 - 14)}$$
(Since, Heron's formula [area = $$\sqrt{s(s - a)(s - b)(s - c)}$$])

= $$\sqrt{21 × 3 × 11 × 7}$$
= $$\sqrt{7 × 3 × 3 × 11 × 7}$$
= $$21\sqrt{11} {cm}^2$$