Question

Leonhard wants to place a triangular-based cabinet in the corner of his rectangle-shaped living room. The triangular base has a length of 2x + 3 feet and a height of 3x + 6 feet. What value of x causes the cabinet to take up 6% of the living room floor?

Answer 1

Answer:

A. $x=\frac{-7+\sqrt{193}}{4}$

Step-by-step explanation:

We have, the dimensions of the living room are 30 ft and 20 ft.

Thus, area of the living room = length × width = 30 × 20 = 600 ft².

Now, it is given that the cabinet takes 6% of the living room i.e. 6% of 600 ft² = 0.06 × 600 = 36 ft².

As, the triangle has dimensions (2x+3) ft and (3x+6) ft.

So, the area of the triangle = $\frac{1}{2}\times base\times height$

i.e. Area of cabinet = $\frac{1}{2}\times (2x+3)\times (3x+6)$

i.e. Area of cabinet = $\frac{3}{2}\times (2x+3)\times (x+2)$

i.e. Area of cabinet = $\frac{3}{2}\times (2x^2+7x+6)$

Since, the cabinet takes 6% of the living room, we have,

$\frac{3}{2}\times (2x^2+7x+6)$ = 36

i.e. $2x^2+7x+6=36\times \frac{2}{3}$

i.e. $2x^2+7x+6=24$

i.e. $2x^2+7x-18=0$

Further, as the solution of a quadratic equation $ax^{2}+bx+c=0$ is given by $x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$

On comparing we have, a=2, b=7, c= -18.

Thus, $x=\frac{-7\pm \sqrt{7^{2}-4\times 2\times (-18)}}{2\times 2}$

i.e. $x=\frac{-7\pm \sqrt{49+144}}{4}$

i.e. $x=\frac{-7\pm \sqrt{193}}{4}$

i.e. $x=\frac{-7+\sqrt{193}}{4}$ and $x=\frac{-7-\sqrt{193}}{4}$

So, according to the options, we have,

A. $x=\frac{-7+\sqrt{193}}{4}$ is the correct value of x.