First find the missing length of the left triangle with the pythagorean theorem a² + b²= c² → 4² + 6² = c² → 16 + 36 = c² → 52 = c² → √52 = c
Now if you turn your head you can see a right triangle made of the two smaller triangles. The √52 is one of the legs. We know that the hypotenuse is 6 + y so we can substitute these values into the pythaorean theorem as well to find the blank side length of the smaller triangle to the right.
√52² + b² = (6 + y)² → 52 + b² = (6 + y)(6 + y) I know this seems like a lot but this is how I'd do it. You need to multiply (foil method) the (6+y)(6+y) to get 36 + 6y + 6y + y² which simplifies to 36 + 12y + y². Here's what we have: to find the hypotenuse of the smaller triangle to the right we need to simplify 52 + b² = 36 + 12y + y²; subtract 52 from both sides
b² = y² + 12y - 16 (I rearranged to make it easier to see); square root both sides
b = [tex]\sqrt{y² + 12y - 16}[/tex] this is the hypotenuse of the smaller triangle to the right
Then we can use the pythagorean theorem for the small triangle to the right to eventually find the value of y.
4² + y² = [tex]\sqrt{y²}+12y - 16 }[/tex] ²
16 + y² = y² + 12y - 16; now solve, add 16 to both sides
32 + y² = y² + 12y; subtract y² from both sides
32 = 12y; divide both sides by 12
approximately 2.6 = y but since it says do not approximate, I would put y = 32/12
I hope that helps!!!
Ignore the weird A under the radical, I don't know why that's there. It's supposed to be just y²
