can someone help me with this and explain me how to do it please :) ^

Answer:
x= 4
[tex]perpendicular (p) = \sqrt{x} + 1[/tex]
[tex]base(b) = 2 \sqrt{x} [/tex]
[tex]hypotenuse(h) = 2 \sqrt{x} + 1[/tex]
According to the Pythagoras Theoram,
p²+b²= h²
[tex] {( \sqrt{x } + 1) }^{2} + {(2 \sqrt{x} )}^{2} = {(2 \sqrt{x} + 1)}^{2} [/tex]
Answer:
4
Step-by-step explanation:
2√x, √x + 1, 2√x + 1 sides of triangle
√x= y
2y, y+1, 2y+1
(2y)²+(y+1)²= (2y+1)²
4y²+y²+2y+1= 4y²+4y+1
y²-2y=0
y=0 (discarded as side can't be zero) or y= 2
so √x=2 ⇒ x= 4