SOLUTION: (1) The perimeter of a rectangle is 100 and its diagonal has length "x". The area of this rectangle,is (A) 625-x^2 (b) 625-(x^2)/2 (c) 1250-x^2 (d) 1250-(x^2)/2

Question 77435: (1) The perimeter of a rectangle is 100 and its diagonal has length "x". The area of this rectangle,is
(A) 625-x^2 (b) 625-(x^2)/2 (c) 1250-x^2 (d) 1250-(x^2)/2
Answer by ptaylor(2198) (Show Source): You can put this solution on YOUR website!
Perimeter(P)=two times length (L) plus two times width (W) or
P=2L+2W
2L+2W=100 divide both sides by 2
L+W=50-----------------------------------eq1
Area of rectangle (A)=length(L)*width(W) or
A=L*W-------------------------------------eq2
When we draw the diagonal in a rectangle we, in effect, divide the rectangle into two identical right triangles. Here, we can apply the Pythagorean Theorem:
L^2+W^2=x^2--------------------------------eq3
Clearly, we have enough information to solve these non-linear simultaneous equations----BUT THAT'S NOT RECOMMENDED! Instead, lets square both sides of eq1 and we get:
(L+W)^2=50^2
L^2+2L*W+W^2=2500 lets rewrite this:
L^2+W^2+2L*W=2500 but
L^2+W^2=x^2 and
L*W=A
substitute these into the equation and we get:
x^2+2A=2500 subtract x^2 from both sides
x^2-x^2+2A=2500-x^2 collect like terms
2A=2500-x^2 divide both sides by 2
A=(2500-x^2)/2
A=1250-(x^2)/2--------------------------------ans
Frequently, when a problem appears much too difficult, there will be an easy way to solve it

Hope this helps---ptaylor

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