Respuesta :

Trancescient

Step-by-step explanation:

From Pythagorean theorem, one of the sides can be determined as x^2 + y^2 =8^2

or y = (8^2 - x^2)^(1/2)

we can write the perimeter P as

P = 2x + 2y ---> 20 = 2x + 2(8^2 - x^2)^(1/2)

Dividing by 2, we get

10 = x + (8^2 - x^2)^(1/2)

Move the x to the other side,

10 - x = (8^2 - x^2)^(1/2)

Take the square of both sides to get rid of the radical sign:

(10 - x)^2 = 8^2 - x^2

Move everything to the left and expand the quantity inside the parenthesis:

x^2 + (100 - 20x + x^2) - 64 = 0

2x^2 - 20x + 64 = 0

or

x^2 - 10x + 32 = 0

Now we can see that a = -10 and b = 32

ivycoveredwalls

Answer:

a = -10, b = 18

Step-by-step explanation:

The Pythagorean Theorem is, indeed, involved.  Use it to find an expression (you won't get a number!) for the height of the rectangle.

Using the right triangle, one leg has length  x  and hypotenuse length 8.  for a moment, label the height  h.  Then

[tex]x^2+h^2=8^3\\\\h^2=64-x^2\\\\h=\sqrt{64-x^2}[/tex]

This expression tells the height of the rectangle, so it is the length of the two vertical sides.  The top and bottom sides each have length x.

Perimeter = 20 says that the total length of all the sides is 20.  Set that up and do a heap of algebra!

[tex]x+x+\sqrt{64-x^2}+\sqrt{64-x^2}=20\\\\2x+2\sqrt{64-x^2}=20[/tex]

Divide by 2 (to simplify a bit).

[tex]x +\sqrt{64-x^2}=10[/tex]

Subtract x to get the square root by itself (you'll see why in the next step).

[tex]\sqrt{64-x^2}=10-x[/tex]

Square both sides of the equation.

[tex](\sqrt{64-x^2})^2=(10-x)^2\\\\\\64-x^2=100-20x+x^2\\\\64=100-20x+2x^2\\\\0=36-20x+2x^2[/tex]

Divide by 2 again (because you can)

[tex]0=18-10x+x^2[/tex]

Rearrange terms to match the order in the question.

[tex]x^2-10x+18=0[/tex]

The coefficient of  x  is  a = -10.  The constant is  b = 18.