Question 1209961
<pre>
You don't need all those angles you show in the 2nd link.  You just need to
learn about two kinds of special right triangles, the 45-45-90 right triangle
and the 30-60-90 right triangle.  That's what this kite is made up of, 2
of each.  There are two 30-60-90 right triangles on the left, and two 45-45-90
right triangles on the right:

{{{drawing(400,5600/17,-10,7,-7,7,

line(-5*sqrt(3),0,5,0),
line(-5*sqrt(3),0,0,5),
line(-5*sqrt(3),0,0,-5),
line(0,-5,5,0),
line(5,0,0,5),
line(0,-5,0,5),
locate(-1.5,2.6,matrix(1,2,8,cm)),

locate(-7.6,1.05,30^o),
locate(.2,4.05,45^o),
rectangle(0,0,.3,-.3) )}}}

You should learn the following information about these
two special right triangles.  They occur very often.

The sides of the standard 45-45-90 right triangle:
The two legs are 1 unit long each.
The hypotenuse is {{{sqrt(2)}}} units long.
The triangle is both a right triangle and also an isosceles triangle.

The sides of the standard 30-60-90 right triangle:
The shorter leg is 1 unit long. 
The longer leg is {{{sqrt(3)}}} units long.
The hypotenuse is 2 units long.

Again, I emphasize that you need to learn the above information about 
those two standard right triangles. They occur often in these
geometry problems.

Let the longer leg of your 30-60-90 right triangle be x.

You are given that the shorter leg of the 30-60-90 triangle
on the upper left is given as 8 cm.

So make the proportion:

{{{matrix(1,6,shorter,leg, of,my,30-60-90,triangle)/
matrix(1,6,longer,leg, of,my,30-60-90,triangle)}}}{{{""=""}}}{{{matrix(1,7,shorter,leg, of,the,standard,30-60-90,triangle)/
matrix(1,7,longer,leg, of,the,standard,30-60-90,triangle)}}}

{{{8/x}}}{{{""=""}}}{{{1/sqrt(3)}}}

cross-multiply:

{{{8*sqrt(3)}}}{{{""=""}}}{{{1*x}}}

{{{x}}}{{{""=""}}}{{{8*sqrt(3)}}}

So the longer leg of each of the 30-60-90 right triangles on the left
have base {{{8*sqrt(3)}}} and given height 8.  So the area of each of 
those two 30-60-90 right triangles is:

{{{A}}}{{{""=""}}}{{{expr(1/2)*base*height}}}
{{{A}}}{{{""=""}}}{{{expr(1/2)*8*sqrt(3)*8}}}
{{{A}}}{{{""=""}}}{{{32*sqrt(3)}}}

Now do the same for each of the two 45-45-90 right triangles on the right.
There's less work to do here, because you know that in a 45-45-90 triangle
both legs are the same length. So we know that the base and height are
the same length.  So they are both 8 cm.  

{{{A}}}{{{""=""}}}{{{expr(1/2)*base*height}}}
{{{A}}}{{{""=""}}}{{{expr(1/2)*8*8}}}
{{{A}}}{{{""=""}}}{{{32}}}

Since the kite is made up of two 30-60-90 right triangles plus
two 45-45-90 right triangles the total area is:

{{{2*(32*sqrt(3))}}}{{{""+""}}}{{{2*32}}}

{{{64*sqrt(3)}}}{{{""+""}}}{{{64}}}{{{cm^2}}}

and if you like, you can factor out 64, and get

{{{64(sqrt(3) + 1)}}}{{{cm^2}}}

That's about {{{matrix(1,2,174.8512517,cm^2)}}}

Edwin</pre>