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### Perimeter of a Strange Figure

```
Date: 03/23/2002 at 19:56:42
From: Kayleigh
Subject: Perimeter of a weird shaped figure

I don't understand how to find the area or perimeter of a shape that
looks like a rectangle with a small triangle to the left on top. The
base of the rectangle is 8 cm, the height of both is 6 cm and 5 cm,
and the one slanted side of the triangle is 7.2 cm.

+
|\7.2
| \
6|  +------+
|         |
|         |5
|         |
+---------+
8

How do I find the area and perimeter?

I don't understand line segments very well either. I know that to find
the area of a triangle the formula is 1/2b x h, and to find a
rectangle it's bxh, and a square is one side squared, and then the
perimeter for a square is the side x 4, but I don't know how to do a
triangle or shapes with multiple numbers like this. There are many
problems like this I need help on.

Thanks.
```

```
Date: 03/23/2002 at 22:45:36
From: Doctor Peterson
Subject: Re: Perimeter of a weird shaped figure

Hi, Kayleigh.

For the sake of finding the perimeter, you want to think of the shape
as just a collection of line segments placed end to end. The perimeter
is just the sum of those lengths, so all you have to do is figure out
how long each segment is.

To find the area, you want to think of the shape as made up of a
rectangle and a triangle stuck together. You have to find the area of
each one and add them. To find the area of the rectangle, you need its
length and width, and you know those. For the triangle, you want
its height and base. If it's a right triangle and you know the height
and hypotenuse, you can use the Pythagoream theorem to find the base.

First, for the perimeter, we just have to add all five sides; we know
the length of four of them. How can we find the length of the missing
side? I would continue that side to the left, dividing the shape into
a triangle and a rectangle:

+
1|\7.2
| \
+--+------+
|         |
5|         |5
|         |
+---------+
8

I replaced the 6 on the left with a 1 and a 5, because we know that
opposite sides of a rectangle are equal, and the remaining 1 cm must
be the vertical leg of the triangle.

Now we can use the Pythagorean theorem to find the length of the base
of the triangle:

1^2 + x^2 = 7.2^2

1 + x^2 = 51.84

x^2 = 51.84 - 1 = 50.84

x = sqrt(50.84) = 7.13

So we can write that in; and since this plus the side we are trying
to find add up to 8 (again, opposite sides are equal), it must be
8 - 7.13 = 0.87:

+
1|\7.2
| \   0.87
+--+------+
|7.13     |
5|         |5
|         |
+---------+
8

Now the perimeter is the sum:

P = 7.2 + 0.87 + 5 + 8 + 6 = 27.07 cm

And we have everything we need for the area, too:

A[triangle] = bh/2 = 7.13*1/2 = 3.565
A[rectangle] = bh = 8*5 = 40
A[whole] = 3.565+40 = 43.565 cm^2

Does this help? There is no formula for doing any shape like this; you
just have to know what shapes you can handle, and use your ingenuity
to find a way to break it up into those shapes and to find the missing
numbers.

- Doctor Peterson, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Geometry
High School Triangles and Other Polygons

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