Hello, pocketasian!

Are you forgetting the "half" in: .$\displaystyle A \:=\:\tfrac{1}{2}bh$ ?

Find the area of an isosceles triangle with two sides of length $\displaystyle 2x$ and one side of length $\displaystyle x.$ Code:

A
o
/|\
/ | \
/ | \
2x / | \ 2x
/ |h \
/ | \
/ | \
B o-------o-------o C
½x M ½x
: - - - x - - - :

We have isosceles triangle $\displaystyle ABC$ with $\displaystyle AB \,=\,AC\,=\,2x,$

. . and $\displaystyle BC = x.$

$\displaystyle M$ is the midpoint of $\displaystyle BC\!:\;BM \,=\,MC \,=\,\tfrac{1}{2}x$

$\displaystyle h \,=\,AM$ is the altitude (height} of $\displaystyle \Delta ABC.$

In either of the right triangles: .$\displaystyle h^2 + \left(\tfrac{x}{2}\right)^2 \:=\:(2x)^2 $

. . $\displaystyle h^2 + \frac{x^2}{4} \:=\:4x^2 \quad\Rightarrow\quad h^2 \:=\:\frac{15x^2}{4} \quad\Rightarrow\quad\boxed{ h \:=\:\frac{\sqrt{15}}{2}\,x}$

The base of the triangle is: .$\displaystyle \boxed{b \:=\:x}$

Therefore: .$\displaystyle \text{Area} \;=\;\tfrac{1}{2}bh \;=\;\frac{1}{2}(x)\left(\frac{\sqrt{15}}{2}\,x\ri ght) \;=\;\frac{\sqrt{15}}{4}\,x^2$