# Special triangles and ratio of areas help?

• Feb 11th 2011, 04:18 PM
picklepie159
Special triangles and ratio of areas help?
Hi Mathforums!
I've been lurking around here for a round a week and found you guys are really helpful. Just today, I've been hit with two hard problems (to me, at least. The class geniuses had no problems whatsoever with these) I was sent home and was told to bring in the answer on Monday.
So I've attached the problems, and there are some scribbles on them. Please take a look at them and see if you can help.

So for the first one, since the diagonals add up to 28, and I discovered it consisted of 4 30:60:90 triangles, I took one of the 4 small triangles. The side facing 30 became x, and the side facing 60 is x(sqrt 3).
So then x + x (sqrt 3) = 14

Now, I use pythagorean;s theorem
x^2 + ( 3 )x^2 = c^2, where C is an outer side.

Now I get 2 equations - 4x^2= c^2
and x (1 + (sqrt 3)) = 14
so I get x = 14/ (1 + (sqrt 3))

I plug that in, I get c^2= 4x^2
c^2 = (4 x 196) / (4 + (sqrt 3))
c^2 = 784 / (4 + (sqrt 3))

Now I'm stuck. Can anyone help me?
• Feb 11th 2011, 04:24 PM
picklepie159
And also, for the second question, my mind is blown.(Angry)
Here's all I have.
The two upper triangle's areas add up to equal the triangle on the bottoms' area.

The ratio of the areas between the triangles are : 16/25, 9/16, and 9/25.
Therefore, the ratios of the corresponding sides are their square roots:
4/5, 3/4, and 3/5 are the ratios of proportionality ( I think thats what its called?)

• Feb 11th 2011, 06:20 PM
Soroban
Hello, picklepie159!

Quote:

1, The sum of the diagonals of rhombus $\displaystyle ABCD$ is 28
. . and $\displaystyle \angle CBA = 2(\angle DAB).$

Find the perimeter of the rhombus.

Code:

             A             *           /|\           / | \         / y|  \         /  |  \       /    |    \     B * - - + - - * D       \ x  |    /         \  |  /         \  |  /           \ | /           \|/             *             C

You are correct; we have 30-60 right triangles.

And we have: .$\displaystyle 2x + 2y \:=\:28 \quad\Rightarrow\quad x + y \:=\:14$ .[1]

We also have: .$\displaystyle y \,=\,\sqrt{3}\,x$ .[2]

Substitute [2] into [1]: .$\displaystyle x + \sqrt{3}\,x \:=\:14 \quad\Rightarrow\quad x(1 + \sqrt{3}) \:=\:14$

. . $\displaystyle x \:=\:\dfrac{14}{1+\sqrt{3}} \quad\Rightarrow\quad x\:=\: 7(\sqrt{3}-1)$

Substitute into [2]: .$\displaystyle y \:=\:\sqrt{3}\cdot7(\sqrt{3}-1) \quad\Rightarrow\quad y \:=\:7(3-\sqrt{3})$

Pythagorus says: .$\displaystyle AB^2 \:=\:x^2+y^2$

. . $\displaystyle AB^2 \;=\;\left[7(\sqrt{3}-1)\right]^2 + \left[7(3-\sqrt{3})\right]^2$

. . . . . . $\displaystyle =\;49(3-2\sqrt{3}+1) + 49(9-6\sqrt{3}+3)$

. . . . . . $\displaystyle =\;49(4 - 2\sqrt{3} + 12 - 6\sqrt{3})$

. . . . . . $\displaystyle =\; 49(16-8\sqrt{3}) \;=\;392(2-\sqrt{3})$

. . .$\displaystyle AB \;=\;\sqrt{392(2-\sqrt{3})} \;=\;14\sqrt{4-2\sqrt{3}}$

$\displaystyle \text{Therefore: \,Perimeter}\;=\;4(AB) \;=\;56\sqrt{4-2\sqrt{3}} \;\approx\;41$

• Feb 12th 2011, 11:51 AM
picklepie159
Thanks a lot! I see what I did wrong. I was hoping someone could nudge me along in the right direction for the 2nd question? Thanks!
• Feb 12th 2011, 02:40 PM
Wilmer
1st one is quite simple; seems Soroban took the long way...

we have x = 14 / (1 + sqrt(3))

outer side = 2x (property of 30-60-90 triangle)

so perimeter = 4(2x) = 8x = 112 / (1 + sqrt(3)) = ~41
• Feb 18th 2011, 03:27 AM
Kaloda
the answer in number 2 is 144!
• Feb 18th 2011, 11:29 AM
Wilmer
Agree with 144; easy enough IF XYZ is equilateral triangle:
diagram sure looks so; but not stated.

So assuming 4 equilateral triangles (3 inside lines parallel to sides):
side of XYZ = sum of sides of the other 3;
Formula: SIDE = 2(AREA) / 3^(1/4)

k=3^(1/4), a= areaXYZ side, b=area25 side, c=area16 side, d=area9 side

a = b + c + d : works out to :
a = 2sqrt(25)/k + 2sqrt(16)/k + 2sqrt(9)/k = ~18.236;
Makes area = 144.