SOLUTION: The perimeter of a rectangle is 34cm, and the diagonal is 13cm. Determine the lengths of the sides of the rectangle.
I need to set this up in a quadratic equation then solv
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Question 464769: The perimeter of a rectangle is 34cm, and the diagonal is 13cm. Determine the lengths of the sides of the rectangle.
I need to set this up in a quadratic equation then solve. I have several problems like this and knowing how to set them up would be a big help. Thanks.
Found 2 solutions by richard1234, lwsshak3:
Answer by richard1234(7193) (Show Source): You can put this solution on YOUR website!
If the 5-12-13 triangle sticks out at you, and you recognize 5+12 = 17, times 2 is 34, you're done.
To set it as a quadratic, we can let L and 17-L be the length and width (17-L instead of 34-L because the perimeter is 2(L+W) = 2(L + 17-L) = 34). The triangle with sides L, 17-L, and the diagonal of length 13 must satisfy the Pythagorean theorem, i.e.
^2 = 169)
(L-12) = 0 \Rightarrow L = 5, 12)
This implies the dimensions are 5 cm by 12 cm (note that both dimensions are under L because we can choose either length to be L).
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
The perimeter of a rectangle is 34cm, and the diagonal is 13cm. Determine the lengths of the sides of the rectangle.
...
let x=length of rectangle
let y=width of rectangle
..
2x+2y=perimeter=34
divide by 2
x+y=17
..
x^2+y^2=13^2=169 (Pythagorean Theorem)
y^2=169-x^2
y=ħsqrt(169-x^2)
since y>0, use the positive sqrt
x+y=17
x+sqrt(169-x^2)=17
sqrt(169-x^2)=17-x
square both sides
169-x^2=289-34x+x^2
2x^2-34x+120=0
x^2-17x+60=0
(x-5)(x-12)=0
x=5
y=17-x=12
or
x=12
y=17-x=5
..
ans:
length of rectangle=5
Width of rectangle=12
or
length of rectangle=12
Width of rectangle=5
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