Answer:
x = 2[tex]\sqrt{5} [/tex]
Step-by-step explanation:
Okay first you need to find the side of the other triangle (the side that also is a hypotenuse for the other triangle).
So we can use pythagorean's theorum
[tex]a^{2} +b^{2} =c^{2} [/tex] with c being the hypotenuse
[tex]a^{2} +2^{2} =7^{2} [/tex]
a^2 + 4 = 49 subtract 4 from both sides
a^2 = 45 then do the sqaure root
a = [tex]\sqrt{45} [/tex] which can be simplified to 3[tex]\sqrt{5} [/tex]
so... 3[tex]\sqrt{5} [/tex] or [tex]\sqrt{45} [/tex] is our hypotenuse for the other triangle
[tex]5^{2} +b^{2} = \sqrt{45} ^{2}
[/tex]
25 + b^2 = 45 subtract 25 from both sides
b^2 = 20 then do the square root
b (or x in this problem) = [tex]\sqrt{20} [/tex]
or when simplified, 2[tex]\sqrt{5} [/tex]
