Determine the length and width of the rectangular region of the house

Home -> Solved problems -> Determine the length and width of the rectangular region of the house

Solution

\[\begin{aligned} \text { Total area } & =A_{triangle}+A_{rectangle} \\\\ & =x(3 x)\left(\frac{1}{2}\right)+3 x(x+1) \\\\ & =\frac{3}{2} x^2+3 x^2+3 x \\\\ & =\frac{9}{2} x^2+3 x \end{aligned}\]

We know that

\[\begin{aligned} \text { Total area } & =180 \end{aligned}\]

\[\begin{aligned}
\Rightarrow \frac{9}{2} x^2+3 x=180 \\\\
\frac{9}{2} x^2+3 x-180=0 \\\\
9 x^2+6 x-360=0 \\\\
3 x^2+2 x-120=0 \\\\
x=\frac{-2 \pm \sqrt{4-4(3)(-120)}}{2(3)} \\\\
x=\frac{-2 \pm 38}{6}=\frac{36}{6} \\\\
x=6
\end{aligned}
\]

By formulating the total area as the sum of the triangle’s area and the rectangle’s area, substituting the expression in terms of x and solving the quadratic equation, we found the meaningful positive solution and therefore the exact dimensions of the rectangular region.
This example emphasizes that many real‑life “house” problems reduce to quadratic modelling and algebraic reasoning—and mastering these techniques opens the door to deeper problem‑solving in mathematics.

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Home -> Solved problems -> Determine the length and width of the rectangular region of the house

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