Area of Outer Triangle composed of two inner triangles

[zak100]

Homework Statement

1. [/B]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

Homework Equations

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

The Attempt at a Solution

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.[/B]

Somebody please guide me how to solve it.

Zulfi.

 

[scottdave]

Hint: They are similar triangles.

 

[scottdave]

AD = 3*CD, not CE

 

[Ray Vickson]

Homework Statement

1. [/B]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

Homework Equations

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

The Attempt at a Solution

View attachment 207329
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.[/B]

Somebody please guide me how to solve it.

Zulfi.

Use standard properties of similar triangles.

 

[scottdave]

Perhaps reading this may help you understand how the similar triangles are related. https://www.mathsisfun.com/geometry/triangles-similar.html

 

[zak100]

Hi,
Thanks for your hints. I have done following but i don't 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.

 

[scottdave]

Hi,
Thanks for your hints. I have done following but i don't 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.

You have the information there, if you rearrange it some.

Take this: AD = 3CD
and this
1/3 = DE/DG

and this: 84 = CE * DE
What is the formula (in side lengths) for the area of the big triangle? Can you substitute in things (which you can get numerical values for), to solve for a number value of the area of the big triangle?

 

[Mark44]

@zak100, here's a different, but one that is related to yours, to help you get some geometric intuition. What's the area of the large rectangle if we are given this information:
AC = 3 * BC
CE = 3 * CD
Area of small rectangle = 10

 

[zak100]

Hi,
Mr. Mark44. You have provided me a good question. I would try it once i finish my current problem.
Mr. scottdave: how you got this relationship?
1/3 = DE/DG

Can it be applied to bases also?

Zulfi.

 

[Mark44]

Mr. scottdave: how you got this relationship?
1/3 = DE/DG

This is simple algebra, and comes from the equation DG = 3 * DE, which is equivalent to DE = (1/3) * DG

Can it be applied to bases also?

Yes, the same idea can be applied.

 

[Mark44]

Mr. Mark44. You have provided me a good question. I would try it once i finish my current problem.

You should look at it now, since you're having so much difficulty with the posted problem. If you understand the relationships between similar triangles and between similar rectangles, the problem I gave can be solved by inspection (no writing needed).

 

[scottdave]

Hi,
Mr. Mark44. You have provided me a good question. I would try it once i finish my current problem.
Mr. scottdave: how you got this relationship?
1/3 = DE/DG

Can it be applied to bases also?

Zulfi.

You actually had it already, when you wrote: 1/3 = CE/AG = DE/DG
You should really try the problem which @Mark44 posted with the rectangles (note the drawing is not to scale, though). Once you understand rectangles, solving for triangles is an easy step from that.

 

[zak100]

Hi,
Thanks for your advise. If i leave it, it would be a diversion. I may concatenate from problems to problems. Actually i was not starting with a proper eq. My problem was to find the are of ADG, so i must write its eq. first:

area of ADG = 1/2 AG * DG
= 1/2 * 3CE * 3DE
= 1/2 * 9 * CE * DE
= 1/2 * 9 * 84 (Note CE * DE = 84 from post 6)
= 378
Thanks for your interest and continuous guidance.

I would now consider the rectangle problem.
Zulfi.

 

[scottdave]

area of ADG = 1/2 AG * DG
= 1/2 * 3CE * 3DE
= 1/2 * 9 * CE * DE
= 1/2 * 9 * 84 (Note CE * DE = 84 from post 6)
= 378
Thanks for your interest and continuous guidance.

I would now consider the rectangle problem.
Zulfi.

Yes you have the correct answer. Once you complete the rectangle, hopefully you will see a pattern.

 

[zak100]

Hi,
I have solved the rectangle prob:
Area of large rect = AC * CE
= 3 * BC * 3 * CD
= 9 * BC * CD
Note BC * CD = area of large rect = 10 (given)

Area of large rectangle = 9 * 10
= 90

Thanks for this prob.

Zulfi.

 

[Mark44]

Hi,
I have solved the rectangle prob:
Area of large rect = AC * CE
= 3 * BC * 3 * CD
= 9 * BC * CD
Note BC * CD = area of large rect = 10 (given)

Area of large rectangle = 9 * 10
= 90

Yes, that's correct. The idea is that since both dimensions of the small rectangle are tripled, the area of the large rectangle will be 3 * 3 = 9 times as large.
It's exactly the same idea as in your problem with the triangles. That is, the large triangle will have an area 9 times as large as the small upper triangle.