The area of a rectangle is 45 cm². two squares are constructed such that two adjacent sides of the rectangle are each also the side of one of the squares. the combined area of the two squares is 106 cm². find the lengths of the sides of the squares.

5 cm, 9 cm

Step-by-step explanation:

If x and y are the lengths of sides of the rectangle, then the problem statement tells you ...

Substituting y = 45/x into the second equation, you get

... x² + (45/x)² = 106

Multiplying by x² gives ...

... x⁴ -106x² +2025 = 0 . . . . . after subtracting 106x²

... (x² -53)² -784 = 0 . . . . . . . . complete the square

... x² = 53 ±√784 = 53 ±28 . . . solve for x²

... x = √81 or √25 = 9 or 5 . . . . solve for x (which must be positive)

y is the other of these two values, so ...

The lengths of the sides of the squares are 5 cm and 9 cm.

_____

Comment on the graph

The graph is symmetrical about the origin, so there are also two solutions at (x, y) = (-5, -9) and (-9, -5). Since x and y represent distances, these are extraneous solutions with respect to the original question.

-3

step-by-step explanation:

a drawing really does here:

so i draw the horizontal line y=2 and the vertical line x=-2 at this intersection i put b (this is where my right angle is going to be).   so point b is at (-2,2).   point a is 5 units long and is also on y=2 so this point has to be (3,2) or (-7,2) -doesn't matter i will go with (3,2).

now we know c is at (-2,y) we don't know the y yet bet we know that c is on x=-2.

we have the area of the triangle is .5bh=12.5

we know the base length is 5 (ab)

we have the height is (2-y) if you look at your

plug in

.5(5)(2-y)=12.5

solve for y to get the y-coordinate for c

2.5(2-y)=12.5

divide both sides by 2.5 giving:

2-y=5

now subtract 2 on both sides

-y=3

y=-3

160

step-by-step explanation:

idk

i hope i