Mathematics, 22.06.2019 12:00 ChessieGiacalone

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?

Leonhard wants to place a triangular-based cabinet in the corner of his rectangle-shaped living room

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Answer from: k211

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.

Answer from: Quest

for the friend request.

when a figure, such as triangle (xyz) is reflected over the line y = x to create x'y'z' the x-coordinates and y-coordinates of the original image (pre-image) will be reversed for the image. (e.g., (x, y) > (y, x)) this means that the pre-image position of the x- and y-coordinates switch places which in turn provides a mirror like image of the original xyz.

i notice that when i draw a line segment from x to x', that the points are parallel from each other -- they are exactly across from each other and do not intersect with any other points.

yes, think i would see the same characteristic if you drew the line segment connecting y with the reflecting line and then y' with the reflecting line because of the essence of reflections. because it is a mirror image, each point was effected the same way which leads me to conclude that they same would apply to y and y'.

Answer from: Quest

y = 6x -3y -12

y +3y +12 = 6x

=> . x = { 2y / 3 +2 } this is equation of line in terms of x

=> . if line meets the x-axis the co-ordinates of meeting point would be ( x , 0 )

=> . x = 2 ( 0 ) / 3 + 2

=> . x = 2

so, the point is ( 2 , 0 )

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Leonhard wants to place a triangular-based cabinet in the corner of his rectangle-shaped living room...