Based on the description of your triangle, you can draw the following figure:

You can do this because you know:

1. The two equivalent sides are given.
2. Since a triangle is $\displaystyle 180$ degrees, you have only $\displaystyle 180-120$ or $\displaystyle 60$ degrees left for the two angles of equal size. Therefore, those two angles must be $\displaystyle 30$ degrees and $\displaystyle 30$ degrees.

Now, based on the properties of an isosceles triangle, you can draw the following as well:

Based on your standard reference $\displaystyle 30-60-90$ triangle, you know:

$\displaystyle \frac{h}{4}=\frac{1}{2}$

Therefore, $\displaystyle h$ is $\displaystyle 2$.

This means that $\displaystyle x$ is $\displaystyle 2\sqrt{3}$ and the total base of the triangle is $\displaystyle 4\sqrt{3}$.

Now, the area of the triangle is:

$\displaystyle \frac{1}{2}bh$ or $\displaystyle \frac{1}{2}*2*4\sqrt{3}=4\sqrt{3}\:cm^2$

Use the formula for area of a triangle:

$\displaystyle A=\frac{1}{2}(base)(height)$

$\displaystyle A=\frac{1}{2}(4)(7)=14$

1. Find the height of the triangle:

$\displaystyle height=2\cdot base$

$\displaystyle height=2(10)=20$

2. Use the formula for area of a triangle:

$\displaystyle A=\frac{1}{2}(b)(h)$

$\displaystyle A=\frac{1}{2}(10)(20)=100$

Based on the description of this question, you can draw your triangle as such. We can do this thanks to the nature of an isosceles triangle:

Now, you know that the area of a triangle is defined as:

$\displaystyle A=0.5BH$

So, for our data, we can say:

$\displaystyle 32=0.5 * 4h^2=2h^2$

Solving for $\displaystyle h$, we get:

$\displaystyle 16=h^2$

Thus, $\displaystyle h=4$.

Now, for our little triangle on the right, we can draw:

Using the Pythagorean Theorem, we know that the other side is:

$\displaystyle \sqrt{8^2+4^2}=\sqrt{64+16}=\sqrt{80}$

This can be simplified to:

$\displaystyle \sqrt{16*5}=4\sqrt{5}$

Now, we know that this side is the "equal" side of the isosceles triangle. Therefore, we can know that the total perimeter is:

$\displaystyle 2*4\sqrt{5}+4*4=16+8\sqrt{5}\:cm$

Based on the description of this question, you can draw your triangle as such.  We can do this thanks to the nature of an isosceles triangle:

Now, you know that the area of a triangle is defined as:

$\displaystyle A=0.5BH$

So, for our data, we can say:

$\displaystyle 160=0.5 * 5h^2=2.5h^2$

Solving for $\displaystyle h$, we get:

$\displaystyle 160=2.5h^2$

$\displaystyle 64=h^2$

Thus, $\displaystyle h=8$.

Now, for our little triangle on the right, we can draw:

Using the Pythagorean Theorem, we know that the other side is:

$\displaystyle \sqrt{8^2+20^2}=\sqrt{64+400}=\sqrt{464}$

This can be simplified to:

$\displaystyle \sqrt{16*29}=4\sqrt{29}$

Now, we know that this side is the "equal" side of the isosceles triangle. Therefore, we can know that the total perimeter is:

$\displaystyle 2*4\sqrt{29}+5*8=40+8\sqrt{29}$

Based on the description of your triangle, you can draw the following figure:

You can do this because you know:

1. The two equivalent sides are given.
2. Since a triangle is $\displaystyle 180$ degrees, you have only $\displaystyle 180-120$ or $\displaystyle 60$ degrees left for the two angles of equal size. Therefore, those two angles must be $\displaystyle 30$ degrees and $\displaystyle 30$ degrees. 

Now, based on the properties of an isosceles triangle, you can draw the following as well:

Based on your standard reference $\displaystyle 30-60-90$ triangle, you know:

$\displaystyle \frac{h}{30}=\frac{1}{2}$

Therefore, $\displaystyle h$ is $\displaystyle 15$.

This means that $\displaystyle x$ is $\displaystyle 15\sqrt{3}$, and the total base of the triangle is $\displaystyle 30\sqrt{3}$.

Now, the area of the triangle is:

$\displaystyle \frac{1}{2}bh$ or $\displaystyle \frac{1}{2}*15*30\sqrt{3}=225\sqrt{3}\:units^2$ 

Based on the description of your triangle, you can draw the following figure:

You can do this because you know:

1. The two equivalent sides are given.
2. Since a triangle is $\displaystyle 180$ degrees, you have only $\displaystyle 180-120$ or $\displaystyle 60$ degrees left for the two angles of equal size. Therefore, those two angles must be $\displaystyle 30$ degrees and $\displaystyle 30$ degrees.

Now, based on the properties of an isosceles triangle, you can draw the following as well:

Based on your standard reference $\displaystyle 30-60-90$ triangle, you know:

$\displaystyle \frac{h}{15}=\frac{1}{2}$

Therefore, $\displaystyle h$ is $\displaystyle 7.5$.

This means that $\displaystyle x$ is $\displaystyle 7.5\sqrt{3}$ and the total base of the triangle is $\displaystyle 15\sqrt{3}$.

Therefore, the perimeter of the triangle is:

$\displaystyle 15+15+15\sqrt{3}=30+15\sqrt{3}\:cm$

Because the interior angles of a triangle add up to $\displaystyle \small 180^o$, and two of Triangle A's interior angles measure $\displaystyle \small 83^o$, we must simply add the two given angles and subtract from $\displaystyle \small 180^o$ to find the missing angle:

$\displaystyle \small 83^o+83^o=166^o$

$\displaystyle \small 180^o-166^o=14^o$

Therefore, the missing angle (and the smallest) from Triangle A measures $\displaystyle \small 14^o$. If the two triangles are similar, their interior angles must be congruent, meaning that the smallest angle is Triangle B is also $\displaystyle \small 14^o$.

The side measurements presented in the question are not needed to find the answer!

We must first find the length of the congruent sides in Triangle A. We do this by setting up a right triangle with the base and the height, and using the Pythagorean Theorem to solve for the missing side ($\displaystyle \small c$). Because the height line cuts the base in half, however, we must use $\displaystyle 3cm$ for the length of the base's side in the equation instead of $\displaystyle 6cm$. This is illustrated in the figure below:

Using the base of $\displaystyle \small 3cm$ and the height of $\displaystyle \small 4cm$, we use the Pythagorean Theorem to solve for $\displaystyle \small c$:

$\displaystyle \small 3^2+4^2=c^2$

$\displaystyle \small 9+16=c^2$

$\displaystyle \small 25=c^2$

$\displaystyle \small c=5$

Therefore, the two congruent sides of Triangle A measure $\displaystyle \small 5cm$; however, the question asks for the two congruent sides of Triangle B. In similar triangles, the ratio of the corresponding sides must be equal. We know that the base of Triangle A is $\displaystyle 6cm$ and the base of Triangle B is $\displaystyle 12cm$. We then set up a cross-multiplication using the ratio of the two bases and the ratio of $\displaystyle \small c$ to the side we're trying to find ($\displaystyle \small x$), as follows:

$\displaystyle \small \frac{6}{12}=\frac{5}{x}$

$\displaystyle \small 6x=60$

$\displaystyle \small x=10$

Therefore, the length of the congruent sides of Triangle B is $\displaystyle 10cm$.

In order to prove triangle congruence, the triangles must have SAS, SSS, AAS, or ASA congruence. Here, we have one common side (S), and no other demonstrated congruence. Hence, we cannot guarantee that side $\displaystyle AB$ is not one of the two congruent sides of $\displaystyle \Delta ABD$, so we cannot state congruence with $\displaystyle \Delta ABC$.