Find the dimensions of the prayer hall discussed in Section 4.1.
Step-by-Step Solution
Key Concept: Model the length and breadth of the rectangular hall with variables, use the relation Area = length × breadth, and the given condition that length exceeds breadth by 5 m. This leads to a quadratic equation whose positive root gives the required dimensions.
1. Let the breadth of the hall be $x$ metres.
2. Then the length is $x+5$ metres (5 m more than the breadth).
3. Given that the area of the hall is $1200\,\text{m}^2$, we have
$$x(x+5)=1200.$$
4. Expand and bring all terms to one side:
$$x^{2}+5x-1200=0.$$
5. Solve the quadratic equation using the quadratic formula:
$$x=\frac{-5\pm\sqrt{5^{2}-4\cdot1\cdot(-1200)}}{2\cdot1}
=\frac{-5\pm\sqrt{25+4800}}{2}
=\frac{-5\pm\sqrt{4825}}{2}.$$
6. Since a length cannot be negative, take the positive root:
$$x=\frac{-5+\sqrt{4825}}{2}\approx\frac{-5+69.46}{2}\approx32.23\text{ m.}$$
7. Hence the breadth $=x\approx32.23\,\text{m}$.
8. The length $=x+5\approx32.23+5=37.23\,\text{m}.$
9. For practical purposes, the dimensions are taken as approximately $32\,\text{m}$ (breadth) and $37\,\text{m}$ (length).
Correct Answer: Breadth $\approx 32.2\,\text{m}$, Length $\approx 37.2\,\text{m}$ (approximately $32\,\text{m} \times 37\,\text{m}$).