Please help- Similar tringles

[eureka]

In triangle ABC, D and E are the points on AB and AC, DE//BC. Area of ADE : area of DBCE = 1 : 2. Find the AD : DB.
Having difficulty connecting the ratio of areas into the ratio of corresponding sides.

 

[Deleted member 4993]

In triangle ABC, D and E are the points on AB and AC, DE//BC.
Area of ADE : area of DBCE = 1 : 2. Find the AD : DB.
Having difficulty connecting the ratio of areas into the ratio of corresponding sides.

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

https://www.freemathhelp.com/forum/threads/read-before-posting.109846/#post-486520​

Please share your work/thoughts about this problem.

 

[Dr.Peterson]

Are you aware that the areas of similar figures are proportional to the squares of their sides?

 

[eureka]

Are you aware that the areas of similar figures are proportional to the squares of their sides?

Yes,

 

[Dr.Peterson]

The figures you worked with are not similar; in fact, they are not both triangles!

Try working with ADE and ABC instead.

 

[eureka]

Let area of ADE 1k and DECB 2k that leads the area of ABC is 3k.
area ADE : area of ABC= 1 : 3. Is that correct?

 

[Deleted member 4993]

Let area of ADE 1k and DECB 2k that leads the area of ABC is 3k.
area ADE : area of ABC= 1 : 3. Is that correct?

Yes,,,

 

[eureka]

So, as I showed in my previous work ( in the image I uploaded), area of ADE/ area of DECB = 1/2 =(1/sqrt2)^2 doesn't apply? Is it because DECB not being a triangle?
Don't we square the ratio [ (1/3)^2]. I am kind of confused.

 

[Deleted member 4993]

So, as I showed in my previous work ( in the image I uploaded), area of ADE/ area of DECB = 1/2 =(1/sqrt2)^2 doesn't apply? Is it because DECB not being a triangle?
Don't we square the ratio [ (1/3)^2]. I am kind of confused.

You need to use:

area ADE : area of ABC= 1 : 3. (your response #6) ....................................This will be helpful ratio.​

Then use (response #3)

the areas of similar figures are proportional to the squares of their sides​

​

Then:

AE/AC = AD/AB = DE/BC = ?​

continue.....

 

[eureka]

AE/AC=AD/AB=DE/BC= 1k/3k= 1/3
So AD/AB = 1/3

If we were asked to find area triangle ADE : area of triangle ABC = (1/3)^2 = 1/9. Am I right?

 

[Deleted member 4993]

AE/AC=AD/AB=DE/BC= 1k/3k= 1/3
So AD/AB = 1/3

If we were asked to find area triangle ADE : area of triangle ABC = (1/3)^2 = 1/9. Am I right? No!

You had already calculated:

area triangle ADE : area of triangle ABC = 1/3 .............................. your response #6. Now you have to:

Find the AD : DB

 

[Dr.Peterson]

AE/AC=AD/AB=DE/BC= 1k/3k= 1/3
So AD/AB = 1/3

If we were asked to find area triangle ADE : area of triangle ABC = (1/3)^2 = 1/9. Am I right?

You're going the wrong direction.

You know that area ADE : area of ABC= 1 : 3 . The ratio of areas is the square root of the ratio of sides; so the ratio of sides is the square root of the ratio of areas! That's what you did with the wrong ratio in #4. Do it now to find the ratio AD:AB; then use that to get AD: DB.

 

[eureka]

You're going the wrong direction.

You know that area ADE : area of ABC= 1 : 3 . The ratio of areas is the square root of the ratio of sides; so the ratio of sides is the square root of the ratio of areas! That's what you did with the wrong ratio in #4. Do it now to find the ratio AD:AB; then use that to get ADB.

So AD:AB = 1:3. AND ADB=1:2 ?

 

[Deleted member 4993]

So AD:AB = 1:3. .... INCORRECT AND ADB=1:2 ?

You need to use:

area of ADE : area of ABC= 1 : 3. (your response #6) ....................................This will be helpful ratio.
Then use (response #3)

the areas of similar figures are proportional to the squares of their sides. Then:

AE/AC = AD/AB = DE/BC = ?

(AE/AC)² = (AD/AB)² = (DE/BC)² = \(\displaystyle \frac{area ADE }{ area \ of \ ABC}\)

AE/AC = AD/AB = DE/BC = \(\displaystyle \sqrt{\frac{area \ of \ ADE}{area \ of \ ABC}}\)

 

[joshuaa]

you will be surprised if i told you that

[MATH]AD > DB[/MATH]

 

[Dr.Peterson]

So AD:AB = 1:3. AND ADB=1:2 ?

I think you need to slow down and write out each small step of your thinking, so that you can check each detail as you do it. Don't make leaps.

 

[eureka]

(AE/AC)² = (AD/AB)² = (DE/BC)² = \(\displaystyle \frac{area ADE }{ area \ of \ ABC}\)

AE/AC = AD/AB = DE/BC = \(\displaystyle \sqrt{\frac{area \ of \ ADE}{area \ of \ ABC}}\)

Thank you Mr Khan.

From my previous work, Area of ADE : area of ABC= 1:3 Which means AD/AB = 1/sqrt 3.

So, BD= AB-AD= sqrt 3 - 1. Therefore AD : DB= 1: sqrt 3 -1.
I am sorry, I am just trying but messed up with this problem.