What is the area of the triangle given below?

A$\sqrt{\mathbf{2}}\mathbf{+}\mathbf{10}\sqrt{\mathbf{5}}$

B$\mathbf{17}\mathbf{+}\mathbf{5}\sqrt{\mathbf{10}}$

C$\mathbf{12}\sqrt{\mathbf{2}}\mathbf{+}\mathbf{8}\sqrt{\mathbf{5}}$

D$\mathbf{8}\mathbf{.5}\mathbf{+}\mathbf{2}\mathbf{.5}\sqrt{\mathbf{10}}$

Two radicals can be only added if the radicand of the terms are same. If the radicand of the terms are same, then add the radicals by adding their coefficients.

That is, $a\sqrt{x}+b\sqrt{x}+c\sqrt{x}=(a+b+c)\sqrt{x}$

Two radicals can be only multiplied if the index of the terms are same. If the index of the terms are same, then the coefficient of the radicals are multiplied together and the radicands are multiplied together under a common radical symbol.

That is, $\left(a\sqrt{x}\right)\left(b\sqrt{y}\right)=(a\times b)\sqrt{x\times y}$

Area of a right angle triangle is half of the product of its perpendicular height and base.

Let the length of the perpendicular height be ‘$h$’ and the length of the base be ‘$b$’,

Then the area of the triangle ($A$) is given as:

$Area\left(A\right)=\frac{1}{2}\times h\times b\dots \left(1\right)$

From the figure,

The height(h) of the triangle is $2\sqrt{2}+\sqrt{5}$units.

The base($b$) of the triangle is $3\sqrt{2}+\sqrt{5}$units.

Substitute $2\sqrt{2}+\sqrt{5}$as $h$and $3\sqrt{2}+\sqrt{5}$as $b$in (1) to get the required area.

$\begin{array}{c}Area\left(A\right)=\frac{1}{2}\times h\times b\\ =\frac{1}{2}\times (2\sqrt{2}+\sqrt{5})\times (3\sqrt{2}+\sqrt{5})\\ =\frac{1}{2}\left[2\sqrt{2}(3\sqrt{2}+\sqrt{5})+\sqrt{5}(3\sqrt{2}+\sqrt{5})\right]\\ \left[\text{Using\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}(a+b)(c+d)=a(c+d)+b(c+d)\right]\\ \\ =\frac{1}{2}\left[\left(2\sqrt{2}\right)\left(3\sqrt{2}\right)+\left(2\sqrt{2}\right)\left(\sqrt{5}\right)+\left(\sqrt{5}\right)\left(3\sqrt{2}\right)+\left(\sqrt{5}\right)\left(\sqrt{5}\right)\right]\\ [\text{Using\hspace{0.17em}\hspace{0.17em}}a(b+c)=a\left(b\right)+a\left(c\right)]\\ \\ =\frac{1}{2}\left[(2\times 3)(\sqrt{2}\times \sqrt{2})+\left(2\right)(\sqrt{2}\times \sqrt{5})+\left(3\right)(\sqrt{5}\times \sqrt{2})+{\left(\sqrt{5}\right)}^2\right]\\ =\frac{1}{2}\left[\left(6\right){\left(\sqrt{2}\right)}^2+\left(2\right)\left(\sqrt{2\times 5}\right)+\left(3\right)\left(\sqrt{5\times 2}\right)+{\left(\sqrt{5}\right)}^2\right]\\ =\frac{1}{2}\left[6\left(2\right)+2\left(\sqrt{10}\right)+3\left(\sqrt{10}\right)+5\right]\\ \left[\text{Using\hspace{0.17em}\hspace{0.17em}}{\left(\sqrt{x}\right)}^2=x\right]\\ \\ =\frac{1}{2}\left[12+2\sqrt{10}+3\sqrt{10}+5\right]\\ =\frac{1}{2}\left[12+5+2\sqrt{10}+3\sqrt{10}\right]\\ =\frac{1}{2}\left[17+(2+3)\sqrt{10}\right]\\ =\frac{1}{2}\left[17+5\left(\sqrt{10}\right)\right]\\ =\frac{1}{2}(17+5\sqrt{10})\\ =\frac{1}{2}\left(17\right)+\frac{1}{2}\left(5\sqrt{10}\right)\\ =8.5+2.5\sqrt{10}\end{array}$

Therefore, the area of the given triangle is $8.5+2.5\sqrt{10}$square units.

Option $D$is correct.