Re: Area help needed

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Posted by Jim on October 30, 2002 at 19:39:48:

In Reply to: Re: Area help needed posted by Brad Paul on October 30, 2002 at 12:25:01:

: : The parralel sides of the trapezium are:

: : 19.3 and 9

: : So 8.6*(19.3+9) should = 243.38

: : I hope the text book is wrong, doubt it though...anyone care to show what they get?

: Look at the equation I gave you in my 1). It gives:
: 8.6*(19.3+9)/2=121.69
: To derive this equation (which I did because I always forget this
: stuff) is to draw a trapezium and label the height h and the top and
: bottom lengths say t, and b. Now break your figure into two triangles
: and one rectangle. Find the area of all three then simplify the sum.

Thanks Brad

That did the trip, a real help. I found out the main problem with this was because I was doing it so late at night and I was forgeting to divide the following by 2:

8.6*(19.3+9)

Thanks again

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: : : The parralel sides of the trapezium are: : : : 19.3 and 9 : : : So 8.6*(19.3+9) should = 243.38 : : : I hope the text book is wrong, doubt it though...anyone care to show what they get? : : Look at the equation I gave you in my 1). It gives: : : 8.6*(19.3+9)/2=121.69 : : To derive this equation (which I did because I always forget this : : stuff) is to draw a trapezium and label the height h and the top and : : bottom lengths say t, and b. Now break your figure into two triangles : : and one rectangle. Find the area of all three then simplify the sum. : Thanks Brad : That did the trip, a real help. I found out the main problem with this was because I was doing it so late at night and I was forgeting to divide the following by 2: : 8.6*(19.3+9) : Thanks again

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