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April 30, 2016
Posted by **alison** on Tuesday, January 4, 2011 at 7:08pm.

- math -
**Marth**, Tuesday, January 4, 2011 at 7:12pmLet l be the length, and w be the width, of the rectangle.

The perimeter of the rectangle is 2(l+w).

The perimeter of the triangle is l+w+sqrt(l^2+w^2). sqrt(l^2+w^2) is the hypotenuse of the triangle.

There are two equations and two variables. Find l and w. - math -
**alison**, Tuesday, January 4, 2011 at 7:16pmthe hypotneuse is 5 meters so how could i figure the dimensions of the rectangle out?

- math -
**Marth**, Tuesday, January 4, 2011 at 7:19pmThe perimeter of the rectangle is 22m. So 2(l+w) = 22.

l+w = 11

l = 11-w

Also, l^2+w^2 = 5^2 by Pythagorean Theorem.

Substitute in l = 11-w: (11-w)^2+w^2=5^2 - math -
**alison**, Tuesday, January 4, 2011 at 7:24pmi have the deminsions of the rectangle but i need to know how to get it , the deminsions are 8m by 3m

- math -
**Marth**, Tuesday, January 4, 2011 at 7:26pmAre you referring to the triangle formed by the diagonal of the rectangle and two of its sides? If the dimensions are 8m by 3m, the hypotenuse of said triangle cannot be 5m.

- math -
**alison**, Tuesday, January 4, 2011 at 7:29pmim referring to the deminsions of a rectangle with a triangle with a hypotneuse of 5m in the top right corner

- math -
**Marth**, Tuesday, January 4, 2011 at 7:33pmIs the base of the triangle the width of the rectangle, and the height of the triangle half the length of the rectangle?

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**alison**, Tuesday, January 4, 2011 at 7:35pmyes it certainly is

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**Marth**, Tuesday, January 4, 2011 at 7:37pmOk. Then the perimeter of the triangle is w + l/2 + 5.

So w+l/2+5 = 12m, and 2(l+w)=22m. - math -
**alison**, Tuesday, January 4, 2011 at 7:41pmthank you so very much you are awesome i wish you were my math teacher!

- math -
**alison**, Tuesday, January 4, 2011 at 7:45pmyou are going to end up hating me i am really bad at dimensions so please bear with me.

a dorm room your sister is considering seems small.the room is rectangular and has a perimeter of 42 feet. the room is 4 feet longer than its width. what are the dimensions of the room? - math -
**Ms. Sue**, Tuesday, January 4, 2011 at 7:54pmAlison -- please post your last question as a separate post.

- math -
**alison**, Tuesday, January 4, 2011 at 7:54pmnevermind i got it thanks anyway!