## help_people one year ago 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

1. help_people

@Nishant_Garg

2. help_people

you here?

3. anonymous

Hmmm consider the drawing |dw:1442693956662:dw| 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 |dw:1442694478789:dw| 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 |dw:1442694716457:dw| 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$

4. anonymous

|dw:1442695223716:dw|

5. help_people

ok

6. anonymous

or you can simply do $(8+2x)(7+2x)=58$$\implies 4x^2+30x+56=58 \therefore 4x^2+30x-2=0$

7. help_people

ty! can oyu help w/ a few mor