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### Triangular Garden

```
Date: 03/18/97 at 09:45:12
From: Marilee Ferguson
Subject: Triangular Garden

A garden in the shape of a right triangle has sides measuring
60 units, 80 units, and 100 units.  A fence runs from the right
angle to the hypotenuse and separates the garden into two parts
of equal perimeter.  Find the length of the fence.
```

```
Date: 04/13/97 at 02:46:37
From: Doctor Sydney
Subject: Re: Triangular Garden

Dear Marilee,

Hello!  This is a good question!  I am going to help you set up the
problem, and then I hope you'll be able to take it from there.  If
you are still stuck after you have read my response, please email us
back.  If you do email back, please indicate whether or not you have
had trigonometry, as it may come in handy in this problem!

The first thing you should do for this problem is to draw a picture
like the following one:

|\
| \
|  \ y
|   \
80 |    \
|   / \
| x/   \  100 - y
| /     \
|/_______\
60

We have our 60-80-100 triangle where the line in the middle from the
right angle to the hypotenuse is the fence whose length  you are
trying to calculate.  In the picture, I have labeled this length x.  I
have divided the hypotenuse into two segments, one of which has length
y and the other of which has length 100 - y (since the lengths of the
two segments must add up to 100).

Now we translate the word problem into algebraic equations.  Let's
look at what the problem says:

"A garden in the shape of a right triangle has sides measuring 60
units, 80 units, and 100 units.  A fence runs from the right angle to
the hypotenuse and separates the garden into two parts of equal
perimeter. Find the length of the fence."

Okay; as we said above, the length of the fence that runs from the
right angle to the hypotenuse is x, so we are looking for the value
of x.  The problem also says that the fence with length x splits the
garden into two parts of equal perimeter.  Well, we can write
equations for the perimeter of the two parts, right?  The perimeter
of the top part is just

80 + x + y.

The perimeter for the other part is

60 + (100 - y) + x.

But the problem says that the perimeters are equal, which means that

80 + x + y = 60 + (100 - y) + x.

You can then simplify this equation to find that y = 40.

Now that you know this, can  you figure out what x is?  It isn't easy,
but maybe you can figure it out now!  Good luck.  Write back if you
need more help.

-Doctor Sydney,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```
Associated Topics:
High School Geometry
High School Practical Geometry
High School Triangles and Other Polygons
High School Trigonometry

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