SOLUTION: Find the area of the kite I got the answer 156.16cm https://ibb.co/ch1SdtSh https://ibb.co/gM0Lsps7

Algebra ->  Surface-area -> SOLUTION: Find the area of the kite I got the answer 156.16cm https://ibb.co/ch1SdtSh https://ibb.co/gM0Lsps7      Log On


   



Question 1209961: Find the area of the kite
I got the answer 156.16cm
https://ibb.co/ch1SdtSh
https://ibb.co/gM0Lsps7

Found 3 solutions by Edwin McCravy, ikleyn, mccravyedwin:
Answer by Edwin McCravy(20055) About Me  (Show Source):
You can put this solution on YOUR website!
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:



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%282%29 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%283%29 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:

%22%22=%22%22

8%2Fx%22%22=%22%221%2Fsqrt%283%29

cross-multiply:

8%2Asqrt%283%29%22%22=%22%221%2Ax

x%22%22=%22%228%2Asqrt%283%29

So the longer leg of each of the 30-60-90 right triangles on the left
have base 8%2Asqrt%283%29 and given height 8.  So the area of each of 
those two 30-60-90 right triangles is:

A%22%22=%22%22expr%281%2F2%29%2Abase%2Aheight
A%22%22=%22%22expr%281%2F2%29%2A8%2Asqrt%283%29%2A8
A%22%22=%22%2232%2Asqrt%283%29

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%22%22=%22%22expr%281%2F2%29%2Abase%2Aheight
A%22%22=%22%22expr%281%2F2%29%2A8%2A8
A%22%22=%22%2232

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%2A%2832%2Asqrt%283%29%29%22%22%2B%22%222%2A32

64%2Asqrt%283%29%22%22%2B%22%2264cm%5E2

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

64%28sqrt%283%29+%2B+1%29cm%5E2

That's about matrix%281%2C2%2C174.8512517%2Ccm%5E2%29

Edwin

Answer by ikleyn(52781) About Me  (Show Source):
You can put this solution on YOUR website!
.


            Actually,  this problem is very simple.


Two triangles with the angles of 45 degrees form a square with the side of 8 cm,
if to attach them hypotenuse to hypotenuse.


So, the total area of these two triangles is  8%5E2 = 64 cm^2.


The great triangle on the left is the equilateral triangle with the side length a = 16 cm,
since all three its angles are 60 degrees each.


The area of such a triangle is  

    a%5E2%2A%28sqrt%283%29%2F4%29 = 16%5E2%2A%28sqrt%283%29%2F4%29 = 16%2A4%2Asqrt%283%29 = 64%2Asqrt%283%29 cm^2.


Therefore, the total area of the given kite is

    64 + 64%2Asqrt%283%29 = 64%2A%281%2Bsqrt%283%29%29 = 174.85 square centimeters approximately.    ANSWER

Solved.

------------------------

Equilateral triangles come very often in geometry problems.

Therefore,  it is useful to know this formula for their area,   a%5E2%2A%28sqrt%283%29%2F4%29,
via their side length  " a ",  in order for do not derive it every time from scratch.



Answer by mccravyedwin(407) About Me  (Show Source):
You can put this solution on YOUR website!
Here is a link to a good lesson on the special right triangles:

https://flexbooks.ck12.org/cbook/ck-12-precalculus-concepts-2.0/section/4.3/primary/lesson/special-right-triangles-pcalc/

Edwin