Question

Tabitha wants to hang a painting in a gallery. The painting and frame must have an area of 58 square feet. The painting is 7 feet wide by 8 feet long. Which quadratic equation can be used to determine the thickness of the frame, x?

Image of a frame 7 feet wide by 8 feet long, with an of x on the bottom and right sides

x2 + 15x − 2 = 0
x2 + 15x + 58 = 0
4x2 + 30x − 2 = 0
4x2 + 30x + 58 = 0

Answer 1

@Nishant_Garg

Answer 2

you here?

Answer 3

Hmmm consider the drawing

We are given the combined area(in respective units)
\[A_{outer}=58\]
and we can find the inner area
\[A_{inner}=7 \times 8=56\]
then the area of the thickness is
\[A_{thickness}=58-56=2\]
Now to find an expression for the area of the thickness in terms of x, we can do

This way we have divided out middle section into 4 trapeziums, out of which the left right trapezium are the same and up the down are the same, so they can be just doubled,
Now for first pair of trapezium
So we have
\[A_{trap1}=\frac{a+b}{2}.h=\frac{8+8+2x}{2}.x=\frac{16+2x}{2}.x=(8+x).x=x^2+8x\]
So we have
\[2 \times A_{trap1}=2x^2+16x\]
Similarly for other pair of trapezium we can do
\[2 \times A_{trap2}=(a+b).h=(7+7+2x).x=(14+2x)x=2x^2+14x\]
So total Area of thickness or the middle section
\[A_{thickness}=2x^2+16x+2x^2+14x=4x^2+30x\]
But we know that
\[A_{thickness}=2\]\[\therefore 4x^2+30x=2 \implies 4x^2+30x-2=0\]