Nope, the area is the product of the two sides
$\displaystyle A = \sqrt{5}(2+\sqrt{10}) = 2\sqrt{5} + 5\sqrt{2}$
In case you were wondering about the last term:
$\displaystyle \sqrt{5}\sqrt{10} = \sqrt{50} = \sqrt{25}\sqrt{2} = 5\sqrt{2}$
Alternatively
$\displaystyle \sqrt{5}\sqrt{10} = \sqrt{5}\sqrt{5}\sqrt{2} = 5\sqrt{2}$
Yeah that's right.
To solve it I used the fact that the area of a rectangle is $\displaystyle length \times width$ and I also used the distributive property which says $\displaystyle a(b+c) = ab+ac$
As the sides are $\displaystyle 2+\sqrt{10}$ and $\displaystyle \sqrt{5}$ to find the area I multiplied them.
Surds act like any other number when multiplied and using the distributive property above for this case: $\displaystyle a = \sqrt{5}$ and $\displaystyle b= 2$ and $\displaystyle c=\sqrt{10}$
Therefore the area is $\displaystyle ab+ac = 2\sqrt{5} + \sqrt{5}\sqrt{10}$
As a law of surds, for positive p,q: $\displaystyle \sqrt{p}\sqrt{q} = \sqrt{pq} and so \sqrt{5}\sqrt{10} = \sqrt{50}$
You could leave the answer as $\displaystyle 2\sqrt{5}+\sqrt{50}$ but it is more usual to simplify.
To simplify a surd get the prime factors of the number inside it- in this case 50.
$\displaystyle 50 = 2 \times 5 \times 5$
Using the rule directly above $\displaystyle \sqrt{50} = \sqrt{2}\sqrt{5}\sqrt{5}$
because $\displaystyle \sqrt{a}\sqrt{a} = a$ we can simplify the above to give $\displaystyle \sqrt{2}\sqrt{5}\sqrt{5} = 5\sqrt{2}$ which is fully simplified.
I then combined this with the first part (ab) and got the final answer of
$\displaystyle A = 2\sqrt{5}+5\sqrt{2}$