Respuesta :
ANSWER:
Width of border is 1.5 inches
SOLUTION:
Given,
Width of the photo graph [tex]\left(\mathrm{w}_{1}\right)[/tex] = 4.25 inches
Height of the photo graph [tex]\left(\mathrm{h}_{1}\right)[/tex] = 7.75 inches
And it is mounted in a frame with border of x inches, so the dimensions after mounting will be
Width after mounting [tex]\left(\mathrm{W}_{2}\right)[/tex] = (4.25 + x + x) [ since, two sides of photograph will have borders
Height after mounting [tex]\left(\mathrm{h}_{2}\right)[/tex] = (7.75 + x + x) [ since, two sides of photograph will have borders
Area of border = 45 square inches
As we all know that, photograph will be in rectangular shape,
Area of photograph = [tex]\mathrm{w}_{1} \times \mathrm{h}_{1}[/tex]
[tex]= 4.25 \times 7.75[/tex]
[tex]=32.9375 \mathrm{in}^{2}[/tex]
Area of photograph after mounting = [tex]\mathrm{w}_{2} \times \mathrm{h}_{2}[/tex]
[tex]= (4.25+2x) \times (7.75+2x)[/tex]
[tex]= (4.25 \times 7.75) + (4.25 + 7.75) \times 2x + (2x \times 2x)[/tex]
[tex]=4 x^{2}+24 x+32.9375 \text { in }^{2}[/tex]
Now, area of border = Area of photograph after mounting - Area of photograph before mounting
[tex]45=4 x^{2}+24 x+32.9375-32.9375[/tex]
[tex]45=4 x^{2}+24 x[/tex]
[tex]x^{2}+6 x-\frac{45}{4}=0[/tex] [by dividing the equation with 4]
[tex]x^{2}+6 x-\frac{15 \times 3}{2 \times 2}=0[/tex]
[tex]x^{2}-\left(\frac{-15}{2}+\frac{3}{2}\right) x-\frac{15 \times 3}{2 \times 2}=0[/tex]
[tex]x\left(x-\frac{3}{2}\right)+\frac{15}{2}\left(x-\frac{3}{2}\right)=0[/tex]
[tex]\left(x-\frac{3}{2}\right)\left(x+\frac{15}{2}\right)=0[/tex]
[tex]x=\frac{-15}{2} \text { or } \frac{3}{2}[/tex]
As length can’t be negative, x = [tex]\frac{3}{2} = 1.5[/tex]
Hence , the width x is 1.5 inches
