Area of Outer Triangle composed of 2 inner triangles

Thread Starter

zulfi100

Joined Jun 7, 2012
656
Hi,
In the figure below, AB=BC=CD. If the area of triangle CDE is 42, what is the area of triangle ADG.

See the attached figure in the next post. I cant upload it here.

I think we can start from area of triangle which is given by:

Area of triangle CDE = ½ * CE * DE

Or 42 = ½ * CE * DE
I can do some work to transform CE into AD or in my opinion AD = 3 * CE. But the question does not provide any relation between DE & DG. This is what I want to know.


Somebody please guide me how to solve it.

Zulfi.

 

WBahn

Joined Mar 31, 2012
29,976
If you have a square that has sides of length X and a second square that has sides of length a·X, what is the ratio of the area of the second square to that of the first?

Another way of looking at it might be to consider how you find the area in square feet of a triangle that has an area of 42 square yards.
 

WBahn

Joined Mar 31, 2012
29,976
out of curiosity - what grade is this question for?
I'm not sure when I learned about similar triangles and their properties. Certainly no later than 10th grade Geometry, which was real big on rigorous proofs. But I might have seen this stuff a bit earlier. A web search seems to indicate that, today, at least some aspects of similar triangles are commonly taught in 8th grade.
 

#12

Joined Nov 30, 2010
18,224
All polygons can be reduced to triangles. I did this for a 40 acre farm with 49 apexes.:eek:
Look up, "Hero's Formula".
 

Bordodynov

Joined May 20, 2015
3,177
I do not know if you are taught about the similarity of triangles. From the similarity of triangles it follows that the magnitude of the relations of such parties is a constant value. In this case, the ratio is 3. From this, the area is easily found. I will not write the answer. I'm afraid that I will be banned.
 

MrAl

Joined Jun 17, 2014
11,388
Hi,

Yes look up "similar" triangles. Triangles that have the same corresponding angles are similar, and similar triangles have corresponding sides that are proportional.
They are actually called "Similar Triangles".
 

Thread Starter

zulfi100

Joined Jun 7, 2012
656
Hi,
<A web search seems to indicate that, today, at least some aspects of similar triangles are commonly taught in 8th grade.>
Maybe you are right but this is related to MS/Phd.

Thanks Mr. Bordodynov. from the figure its not clear that 3CE = AG
I have done following but i dont think that its close to the solution:
AB = BC= CD
or 42 = 1/2 * CE * DE
84 = CE * DE
CD^2 = CE^2 + DE^2
CD/AD = CE/AG = DE/DG

AD = 3CD
1/3 = CE/AG = DE/DG

Some body please guide me.

Zulfi.
 

WBahn

Joined Mar 31, 2012
29,976
Hi,
<A web search seems to indicate that, today, at least some aspects of similar triangles are commonly taught in 8th grade.>
Maybe you are right but this is related to MS/Phd.
Admittedly, I have had MS and PhD students that literally have not been able to tell me the greatest common divisor of 42 and 63, but that certainly doesn't make doing so MS/PhD level work. Instead, it just means that these MS/PhD students don't have the math skills that are typically expected of someone in third or fourth grade.

Thanks Mr. Bordodynov. from the figure its not clear that 3CE = AG
How is it not clear? Each line segment with two hashes through it is the same size. The distance CE is made up of one of those lines segments while the distance AG is made of up three of them.

He drew the diagram as a series of identical (not just similar) triangles. If you know the area of one of those triangles and can count how many triangles make up the big triangle, can you determine the area of the big triangle?

I have done following but i dont think that its close to the solution:
AB = BC= CD
or 42 = 1/2 * CE * DE
84 = CE * DE
CD^2 = CE^2 + DE^2
CD/AD = CE/AG = DE/DG

AD = 3CD
1/3 = CE/AG = DE/DG

Some body please guide me.

Zulfi.
Forget about the Pythagorean Theorem! You are just throwing that equation at things because it's the only hammer you are willing to use.

The area of AGD is what you are looking for, right? So start with that.

area = 1/2 * AG * DG

You know that (ignoring units -- which I am loath to do, but I don't want to give you something else to get confused about)

42 = 1/2 * CE * DE

What is AG in terms of CE?

What is DG in terms of DE?
 

MrChips

Joined Oct 2, 2009
30,701
Forget the math.

You have three sets of parallel lines.

1. There are three parallel horizontal lines.
2. There are three parallel vertical lines.
3. There are there parallel sloping lines.

If there are three parallel lines, prove that any straight line that is intercepted by three parallel lines equally spaced will result in the line being bisected into two sections of equal length.
 

WBahn

Joined Mar 31, 2012
29,976
Forget the math.

You have three sets of parallel lines.

1. There are three parallel horizontal lines.
2. There are three parallel vertical lines.
3. There are there parallel sloping lines.

If there are three parallel lines, prove that any straight line that is intercepted by three parallel lines equally spaced will result in the line being bisected into two sections of equal length.
I think do that requires (or at least is most simply done) using the properties of similar triangles. So for someone that is having difficulty grasping how to work with similar triangles....
 

Thread Starter

zulfi100

Joined Jun 7, 2012
656
Hi,
Thanks Wbahn for the clue:
are of ADG = 1/2 * AG * DG
= 1/2 * 3CE * 3DE
= 1/2 * 9 * CE * DE
= 1/2 * 9 * 84 (From post#1)
= 1/2 * 9 * 42
= 378
Good work Mr. Bordodynov and Mr. RBR1317 for drawing additional graphics.
God bless you all.
Zulfi.
 

WBahn

Joined Mar 31, 2012
29,976
Good.

Now, let's say I have a solid shape, perhaps a pyramid of some kind, but if good be an irregular blob, and I triple it's size. By what factor does the volume increase?
 
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