Answer:
The side length of the canvas is [tex]10+10\sqrt{2}\ in[/tex]
Step-by-step explanation:
Let
x-----> the length side off the square shaped canvas
we know that
Applying the Pythagoras Theorem
[tex]x^{2} +x^{2}=(x+10)^{2}\\ \\2x^{2}=x^{2}+20x+100\\ \\x^{2}-20x-100=0[/tex]
The formula to solve a quadratic equation of the form [tex]ax^{2} +bx+c=0[/tex] is equal to
[tex]x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}[/tex]
in this problem we have
[tex]x^{2}-20x-100=0[/tex]
so
[tex]a=1\\b=-20\\c=-100[/tex]
substitute in the formula
[tex]x=\frac{20(+/-)\sqrt{-20^{2}-4(1)(-100)}} {2(1)}[/tex]
[tex]x=\frac{20(+/-)\sqrt{800}} {2}[/tex]
[tex]x=\frac{20(+/-)20\sqrt{2}} {2}[/tex]
[tex]x=\frac{20(+)20\sqrt{2}} {2}=10+10\sqrt{2}[/tex]
[tex]x=\frac{20(-)20\sqrt{2}} {2}=10-10\sqrt{2}[/tex] ---> is a negative number
The solution must be a positive number
therefore
The side length of the canvas is [tex]10+10\sqrt{2}\ in[/tex]
The side length of the canvas is best found by using the quadratic formula
because the equation is prime. Solving the equation produces two
approximate measurements, and one must be discarded for being
unreasonable.
I took the test and this was correct.
