# Thread: Divided triangle, proportion question

1. ## Divided triangle, proportion question

Hi Forum!
Here's a question:

Nine lines parallel to the base divide the other sides into 10 equal segments and the area in 10 distinct parts. If the area of the larger of these parts is 38, then the area of the original triangle is?

Now,we can consider the remaining 9 parts of the triangle as 9/10 of it.
But, here comes the part that I don't understand.
The area can be considered as 81/100 to the entire triangle.

I'm confused with the usage of the area here. I does makes a lot of sense,
but since we are in a triangle (the remaining 9/10 will be a triangle too) the area isn't supposed to be bh1/2?

This is kind of confusing to apply. Can someone help?
Why 81/100?

2. ## Re: Divided triangle, proportion question

Originally Posted by Zellator
Hi Forum!
Here's a question:

Nine lines parallel to the base divide the other sides into 10 equal segments and the area in 10 distinct parts. If the area of the larger of these parts is 38, then the area of the original triangle is?

Now,we can consider the remaining 9 parts of the triangle as 9/10 of it.
But, here comes the part that I don't understand.
The area can be considered as 81/100 to the entire triangle.

I'm confused with the usage of the area here. I does makes a lot of sense,
but since we are in a triangle (the remaining 9/10 will be a triangle too) the area isn't supposed to be bh1/2?

This is kind of confusing to apply. Can someone help?
Why 81/100?
1. Draw a sketch!

2. By proportion you'll get:

$\dfrac bB = \dfrac9{10}~\implies~b=\frac9{10} \cdot B$

$\dfrac hH = \dfrac9{10}~\implies~h=\frac9{10} \cdot H$

3. The area of the original triangle is calculated by:

$A_o = \frac12 \cdot B \cdot H$

The area of the smaller triangle is calculated by:

$A_s = \frac12 \cdot b \cdot h = \frac12 \cdot \frac9{10} \cdot B \cdot \frac9{10} \cdot H = \frac{81}{100} \cdot A_o$

3. ## Re: Divided triangle, proportion question

Originally Posted by earboth
1. Draw a sketch!

2. By proportion you'll get:

$\dfrac bB = \dfrac9{10}~\implies~b=\frac9{10} \cdot B$

$\dfrac hH = \dfrac9{10}~\implies~h=\frac9{10} \cdot H$

3. The area of the original triangle is calculated by:

$A_o = \frac12 \cdot B \cdot H$

The area of the smaller triangle is calculated by:

$A_s = \frac12 \cdot b \cdot h = \frac12 \cdot \frac9{10} \cdot B \cdot \frac9{10} \cdot H = \frac{81}{100} \cdot A_o$
Hi earboth!
Thanks for the graph! You really exceed in Geometry!
Thanks for the easy to understand explanation, I get it now!
Using this is possible to calculate the area of the smaller triangle, and then the original triangle.
Great! Thanks again!

All the best!

4. ## Re: Divided triangle, proportion question

Hi Zellator,
Does the given area of 38 sq units refer to the larger (largest?) of the 10 parts or to the larger of two parts?

5. ## Re: Divided triangle, proportion question

Originally Posted by bjhopper
Hi Zellator,
Does the given area of 38 sq units refer to the larger (largest?) of the 10 parts or to the larger of two parts?
Hi bjhopper how are you?
The area given is of the largest portion of the divided triangle, looking at the graph of earboth, the portion underneath the red triangle.