# Similar triangles!

#### Zashmar

##### New member
i need to solve for the left hand side of the tiangle, i hope the image uploaded succesfully i hav been having trouble lately though :/

Yep if any clarrification is needed please reply and i will sort it out

Thanks

#### MarkFL

Staff member
I assume that $x$ represents the entire length of the vertical leg of the largest triangle.

Can you see that there are 3 similar triangles. I suggest working with the two smaller triangles, using the equivalent ratios of the legs. I would label the horzontal leg of the smallest triangle $y$.

Can you state $y$ in terms of $x$?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
i need to solve for the left hand side of the tiangle, i hope the image uploaded succesfully i hav been having trouble lately though :/

View attachment 635

Yep if any clarrification is needed please reply and i will sort it out

Thanks
Welcome to MHB, Zashmar!

First step is to label the unknown parts so we can refer to them.
Let's call the small part on the right bottom "y".
That is, the horizontal distance from the right side of the square to the right corner.
And let's call the part of the top to the intersection "z".
That means that the part of the intersection to the right corner is (10-z).

Then you can set up 3 equations for the 3 rectangular triangles you have (Pythagoras).
Can you do that?

#### Zashmar

##### New member
Okay i think i am understanding you..

so do i use similtaneos equations?

So if i am right triangle 1 small bottom right:

(1)2 +y2 = 1+y2
so z=1+y

Does that mean for triangle number 2(top of square to very point of triangle, second biggest triangle):
(x-1)2 +1=(10-1+y)2
OKay so solve for x? and then sub back into equation?
X2 -1+1=99+y2 (i am not sure i have expanded the brackets correctly)
so x=SQR(99)+y

then 99+y2 = 99+y2

Wait i have gone wrong

Last edited:

Staff member
Give it a shot!

#### Klaas van Aarsen

##### MHB Seeker
Staff member
A new post generates a notification, but editing an old one doesn't.

Okay i think i am understanding you..

so do i use similtaneos equations?

So if i am right triangle 1 small bottom right:

(1)2 +y2 = 1+y2
so z=1+y

Almost!
With the labeling I did, the diagonal side is (10-z).
Then according to Pythagoras you get:

1+y2 = (10-z)2

Does that mean for triangle number 2(top of square to very point of triangle, second biggest triangle):
(x-1)2 +1=(10-1+y)2
Here I labeled the diagonal side as "z".
So with Pythagoras that is:

(x-1)2 +1 = z2

Btw, just now I'm deducing that with x you meant from the top to the far bottom.
Before I thought it was the part above the square.

OKay so solve for x? and then sub back into equation?
X2 -1+1=99+y2 (i am not sure i have expanded the brackets correctly)
so x=SQR(99)+y

then 99+y2 = 99+y2

Wait i have gone wrong
You need yet another equation for the big triangle.
It has sides x, (1+y), and 10.
What would Pythagoras say about their relation?