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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. 

    | \
   6|  +------+
    |         |
    |         |5
    |         |

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.


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:

    | \
    |         |
   5|         |5
    |         |

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:

    | \   0.87
    |7.13     |
   5|         |5
    |         |

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 

- Doctor Peterson, The Math Forum   
Associated Topics:
High School Geometry
High School Triangles and Other Polygons

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