Posted by alison on Tuesday, January 4, 2011 at 7:08pm.
Let 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.
the hypotneuse is 5 meters so how could i figure the dimensions of the rectangle out?
The 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
i have the deminsions of the rectangle but i need to know how to get it , the deminsions are 8m by 3m
Are 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.
im referring to the deminsions of a rectangle with a triangle with a hypotneuse of 5m in the top right corner
Is the base of the triangle the width of the rectangle, and the height of the triangle half the length of the rectangle?
yes it certainly is
Ok. Then the perimeter of the triangle is w + l/2 + 5.
So w+l/2+5 = 12m, and 2(l+w)=22m.
thank you so very much you are awesome i wish you were my math teacher!
you 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?
Alison -- please post your last question as a separate post.
nevermind i got it thanks anyway!