# Find the Area of this shape:

• Jun 24th 2008, 05:10 PM
E=MChuh?
Find the Area of this shape:
The question posed is..

Find the Area of this shape: (Forgive my crappy art job)

http://img57.imageshack.us/img57/4121/mathqin6.jpg

Can someone help me work this problem out? I thought it was basically LxW but according to my book my answer is wrong. How exactly do you work this problem? Why? (any notes/info is helpful thank you.)
• Jun 24th 2008, 05:22 PM
This diagram might help. You are right that it is basically LXW.

Attachment 6953

If you post the answer you and the book got, someone here can tell you if you are right.
• Jun 24th 2008, 06:10 PM
E=MChuh?
Quote:

This diagram might help. You are right that it is basically LXW.

Attachment 6953

If you post the answer you and the book got, someone here can tell you if you are right.

Okay, I did 5mi x 7mi + 4mi x 2mi = 43mi area.

even Navigating through the books 'answer guide' is hard.. but unless i'm mistaken the book says the answer is 29mi^2
• Jun 24th 2008, 06:13 PM
Jhevon
Quote:

Originally Posted by E=MChuh?
Okay, I did 5mi x 7mi + 4mi x 2mi = 43mi area.

even Navigating through the books 'answer guide' is hard.. but unless i'm mistaken the book says the answer is 29mi^2

cut it into two rectangles and find the area of each

draw a horizontal line to complete the top of the larger rectangle. now we have a 4 by 2 rectangle and a 7 by 3 rectangle

so the area is (4*2) + (7*3) = 8 + 21 = 29 square units

• Jun 24th 2008, 06:52 PM
Mathnasium
Imagine it was a complete rectangle, i.e., without the upper right corner cut out. It would measure 5 x 7 and thus have an area of 35 square miles.

Now, we have to SUBTRACT the area of the upper right, the part that is removed. That is a rectangle with dimensions 2 x 3 (the 2 is given; the 3 is 7-4, as pointed out elsewhere). The rectangle that is removed, thus, has an area of 6 square miles.

Take the total area of the "complete" rectangle and subtract the removed part:

35 - 6 = 29 square miles.
• Jun 25th 2008, 02:18 PM
E=MChuh?
Thanks
ahhh I see how it works now, thanks everyone :).
• Jun 25th 2008, 04:30 PM