Find the area of a triangle, given that its sides are 8 cm and 11 cm long, and its perimeter is 32 cm.                                                                                                                                                                                                                                          [Page No. 142]

Third side = $32 - 8 - 11 = 13 \text{ cm}$

Sides: $a = 8$, $b = 11$, $c = 13$

Semi-perimeter: $s = \frac{32}{2} = 16$

Heron's formula: $\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}$

$$A = \sqrt{16(16-8)(16-11)(16-13)}$$

$$= \sqrt{16 \times 8 \times 5 \times 3}$$

$$= \sqrt{1920} = 8\sqrt{30} \text{ cm}^2$$

Caution
Students should use the semi-perimeter $= \frac{(a+b+c)}{2}$ in Heron's formula, not the full perimeter, and should check the triangle inequality.