A circle is inscribed in a triangle with sides 28 cm, 45 cm and 53 cm. What is the area of the triangle excluding the area of the circle? (Use π = 3.14)

Correct option is B

​$$\textbf{Given:} \\\text{Sides of triangle} = 28 \text{ cm},\;45 \text{ cm},\;53 \text{ cm} \\\pi = 3.14 \\\textbf{Concept Used:} \\\text{Area of triangle using Heron's formula and area of incircle} \\\textbf{Formula Used:} \\s = \frac{a+b+c}{2} \\\text{Area of triangle} = \sqrt{s(s-a)(s-b)(s-c)} \\\text{Area of triangle} = r \times s \\\text{Area of circle} = \pi r^2 \\\textbf{Solution:} \\a = 28,\; b = 45,\; c = 53 \\s = \frac{28 + 45 + 53}{2} \\s = \frac{126}{2} \\s = 63 \\\text{Check type of triangle:} \\28^2 + 45^2 = 784 + 2025 = 2809 \\53^2 = 2809 \\\text{Triangle is right-angled} \\\text{Area of triangle} = \frac{1}{2} \times 28 \times 45 \\= 14 \times 45 \\= 630 \\\text{Using } \text{Area} = r \times s: \\630 = r \times 63 \\r = 10 \\\text{Area of inscribed circle} = \pi r^2 \\= 3.14 \times 10^2 \\= 314 \\\text{Required area} = \text{Area of triangle} - \text{Area of circle} \\= 630 - 314 \\= 316 \\\textbf{Final Answer:} \\316$$​​