It is given that area of rectangle = its perimeter and the diagonal is 3√5. We are to find it's area by use of quadratic equations. The problem is that I'm getting a 4 degree equation when I substitute for a in the second equation. How do I find the sides of rectangle ?
Algebra Precalculus – How to Find Sides of a Rectangle
algebra-precalculusareaquadratics
Related Solutions
$(x-y)^2=(x+y)^2-4xy=(24)^2-4.135=576-540=36$
Hence, $(x-y)=6$
Can you take it from here?
Quadrilaterals are not rigid: you can fix a side and move one of the other vertices freely while respecting all side lengths. The area changes when you move the vertex.
(image from http://www.mathsisfun.com/definitions/rigid.html)
Best Answer
Let, length$=a$ and weidth $=b$ So, $$ab=2(a+b)..............(1)$$ $$\text{and,}$$ $$a^2+b^2=(3 \sqrt(5))^2...............(2)$$ From (2) we get,
\begin{array}{ll} & a^2+b^2 &=(3 \sqrt(5))^2\\ & \implies a^2+b^2 +2ab &=(3 \sqrt(5))^2 +2ab \\ & \implies (a+b)^2 &=45+4(a+b) \\ & \implies (a+b)^2 -4(a+b)-45 &=0 \\ &Let,&\\ & & a+b=p\\ & \implies p^2 -4p-45 &=0 \\ & \implies (p-9)(p+5)&=0 \\ & p=9,-5 &[\text{Here, p=-5 is not possible, because the sum of two lengths can't be negative}]\\ &Hence,&\\ & a+b &=9 \\ & \implies a &=9-b..........(3) \end{array}
From equation (1) we get now,
\begin{array}{ll} & (9-b)b &=2(9-b+b)\\ & \implies b^2-9b+18 &=0 \\ & \implies (b-6)(b-3) &=0 \\ & \implies b &=3 \text{ or, } 6 \end{array}
From (3) we get, $$ a=6,\text{ if } b=6$$ $$ a=3,\text{ if } b=3$$