2. Find the area of the isosceles trapezoid to the rigtt $$ a=2 \mathrm{~cm} \quad b=2 \mathrm{~cm} \quad c=4 \mathrm{~cm} $$

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Let the given trapezoid have leg length $$ a = 2 \, \text{cm} $$, one base $$ b = 2 \, \text{cm} $$ (the shorter base), and the other base $$ c = 4 \, \text{cm} $$ (the longer base). The area $$ A $$ of a trapezoid is given by

$$
A = \frac{1}{2}(b+c)h,
$$

where $$ h $$ is the height.

Since the trapezoid is isosceles, the height $$ h $$ can be found by dropping perpendiculars from the endpoints of the shorter base to the longer base. This forms two congruent right triangles. In each right triangle, the horizontal leg is

$$
\frac{c-b}{2}.
$$

Thus, the height $$ h $$ is given by the Pythagorean theorem:

$$
h = \sqrt{a^2 - \left(\frac{c-b}{2}\right)^2}.
$$

Substitute the given values:

$$
h = \sqrt{2^2 - \left(\frac{4-2}{2}\right)^2} = \sqrt{4 - \left(\frac{2}{2}\right)^2} = \sqrt{4 - 1} = \sqrt{3}.
$$

Now, compute the area:

$$
A = \frac{1}{2}(2+4) \cdot \sqrt{3} = \frac{1}{2} \cdot 6 \cdot \sqrt{3} = 3\sqrt{3}.
$$

Thus, the area of the isosceles trapezoid is

$$
3\sqrt{3} \, \text{cm}^2.
$$
The area of the isosceles trapezoid is $$ 3\sqrt{3} \, \text{cm}^2 $$.

Find the area of the isosceles trapezoid to the rigtt

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