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?

Respuesta :

The answer is [tex]x = \frac{-7+ \sqrt{193} }{4} [/tex].

There is very similar question on Brainly, in which dimensions of the living room are given (20 feet, 30 feet).
So, knowing this, we first need to calculate how much 6% of the living room is.
The area of the living room is 600 square feet, since it is rectangle-shaped:
P = a * b = 200 feet * 30 feet = 600 square feet.

6% of 600 square feet is 36 square feet:
600 square feet : 100% = A : 6%
A = 600 * 6 / 100 = 36 square feet

The area of the triangular-based cabinet (A) is [tex] A = \frac{a*h}{2} [/tex], where a is the base, and h is the height of the triangle.
It is given:
A = 36
a = 2x + 3
h = 3x + 6

Now, let's implement this to the formula [tex]A = \frac{a*h}{2} [/tex]:
[tex]36 = \frac{(2x+3)(3x+6)}{2} [/tex]
⇒ [tex]36*2=6 x^{2} +12x+9x+18[/tex]
[tex]72=6 x^{2} +21x+18[/tex]

Let's both sides divide by 3:
[tex]24=2 x^{2} +7x+6[/tex]
⇒ [tex]2 x^{2} +7x-18=0[/tex]

This is the quadratic formula ([tex]a x^{2} +bx+c=0[/tex]) for which
[tex]x = \frac{-b+\sqrt{b^{2} -4ac} }{2a} [/tex] or
[tex]x = \frac{-b-\sqrt{b^{2} -4ac} }{2a} [/tex].
We will use only [tex]x = \frac{-b+\sqrt{b^{2} -4ac} }{2a} [/tex], because the length cannot have negative value.


From the equation [tex]2 x^{2} +7x-18=0[/tex], we know that
a = 2
b = 7
c = -18

Thus,  [tex]x = \frac{-7+ \sqrt{193} }{4} [/tex]:
[tex]x= \frac{-b+ \sqrt{ b^{2} -4ac} }{2a} = \frac{-7+ \sqrt{ (7)^{2} -4*2*(-18)} }{2*2} = \frac{-7+ \sqrt{49+144} }{4} = \frac{-7+ \sqrt{193} }{4} [/tex]