Question 262938
If the perimeter of a rectangle is 26 meters, and the area is 30 square meters,
what is the diagonal?

2.  A right triangle has area 30 square meters and perimeter 30 meters. What is
its hypotenuse?

I understand how to do #1 (via quadratic equation), but I don't see how to set-up #2, with the triangle perimeter info.  The solution must be under my nose, but I just don't see it (as of yet).

Thank you.

1. Perimeter of triangle = 2*(l+w)

26=  2*(l+w)
13= l+w
LB =30

Let length be x
Width = 13-x

X*(13-x)= 30

13x-x^2=30

X^2-13x+30=0

X^2-10x-3x+30=0
X(x-10)-3(x-10)=0
(x-10)(x-3) =0

X=3 or 10  The length and breadth can be said to be 10 and 3  meters


2.  Area of right triangle =30 square meters
     
•a + b + h = 30  a  and b  are the sides and h the hypo

(1 / 2) a b = 30   or    a b = 60 

a 2 + b 2 = h 2 
•Rewrite the equation a + b + h = 60 as follows 

a + b = 30 - h 
•Square both sides 

(a + b)2 = (30 - h)2 
•Expand both sides 

a2 + b2 + 2 a b = 302 + h2 - 60 h 
•Combine the equation a 2 + b 2 = h 2 with the above equation to obtain 

2 a b = 302 - 60 h 
•a b is known to be equal to 60, hence the above equation becomes 

120 = 302 - 60 h 
•Solve for h to obtain 

h = 13 units 
•Substitute h by 13 in the equation a + b + h = 60 to obtain 

a + b +13=30  =17
•Since a b = 60, then b = 60 / a which is substituted in the equation a + b = 60 to obtain 

a + 60 / a - 17 = 0 
•Multiply all terms by a to obtain a quadratic equation of the form 

a 2 + 60 - 17 a = 0 
•Solve the above equation to obtain two solutions 
a^2-17a+60=0
Use the equation a b = 60 to obtain 

when a 12 , b = 5 and when a = 5 , b =12 
•The two sides of the right triangle and the hypotenuse are, respectively, given by 

5 units, 12 units and 13 units. 

mananth@hotmail.com