Answer
$\sqrt{x-7}=3$,
$\sqrt {x}=4$,
and $1+\sqrt {16}=5$.
Work Step by Step
$a=\sqrt{x-7}$ and $b=\sqrt x$.
The hypotenuse is
$h=1+\sqrt x$
Use the Pythagorean theorem.
$\Rightarrow h^2=a^2+b^2$.
Substitute all values.
$\Rightarrow (1+\sqrt x)^2=(\sqrt{x-7})^2+(\sqrt x)^2$.
Use the special formula $(A+B)^2=A^2+2AB+B^2$
We have $A=1$ and $B=\sqrt x$
$\Rightarrow (1)^2+2(1)(\sqrt x)+(\sqrt x)^2=x-7+x$
Simplify.
$\Rightarrow 1+2\sqrt x+x=2x-7$
Add $-1-x$ on both sides.
$\Rightarrow 1+2\sqrt x+x-1-x=2x-7-1-x$
Simplify.
$\Rightarrow 2\sqrt x=x-8$
Square both sides.
$\Rightarrow (2\sqrt x)^2=(x-8)^2$
Use the special formula $(A+B)^2=A^2+2AB+B^2$
We have $A=x$ and $B=8$
$\Rightarrow 4x=(x)^2-2(x)(8)+(8)^2$
Simplify.
$\Rightarrow 4x=x^2-16x+64$
Subtract $4x$ from both sides.
$\Rightarrow 4x-4x=x^2-16x+64-4x$
Simplify.
$\Rightarrow 0=x^2-20x+64$
Rewrite the middle term $-20x$ as $-16x-4x$.
$\Rightarrow 0=x^2-16x-4x+64$
Group the terms.
$\Rightarrow 0=(x^2-16x)+(-4x+64)$
Factor each group.
$\Rightarrow 0=x(x-16)-4(x-16)$
Factor out $(x-16)$.
$\Rightarrow 0=(x-16)(x-4)$
Set each factor equal to zero.
$\Rightarrow x-16=0$ or $x-4=0$
Simplify.
$\Rightarrow x=16$ or $x=4$
$x=4$ does not satisfy $a=\sqrt{x-7}$.
The sides of the triangle are:
$a=\sqrt{x-7}=\sqrt{16-7}=\sqrt{9}=3$,
$b=\sqrt {x}=\sqrt {16}=4$.
$h=1+\sqrt {16}=1+4=5$.
