Finding sides of rectangle

Question

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 ?

Answer 1

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, $$ a=6,\text{ if } b=6$$ $$ a=3,\text{ if } b=3$$

Answer 2

Let's denote by $a,b$ the sides of the rectangle. Since the area is equal to the perimeter we have that

$$ab=2(a+b)$$ and using the information about the diagonal it is

$$\sqrt{a^2+b^2}=3\sqrt{5}.$$ Squaring we get

$$a^2+b^2=45.$$

Thus, one has to solve the system

\begin{cases} 2a+2b &=ab\\ a^2+b^2&=45 \end{cases}

From the first equation we get $b=\dfrac{2a}{a-2}$ and subtituting in the second it is

$$a^2+\dfrac{4a^2}{(a-2)^2}=45.$$ We get

$$a^4-4a^3-37a^2+180a-180=0.$$

The integer solutions of this equation are divisors of $180.$ Since $a$ is the side of a rectangle we don't consider negative values. We see that $1$ and $2$ are not solutions but $3$ is. So we have

$$(a-3)(a^3-a^2-40a+60)=a^4-4a^3-37a^2+180a-180=0.$$ Now we work with $$a^3-a^2-40a+60$$ to find more possible solutions. We see that $6$ is a solution. And

$$a^3-a^2-40a+60=(a-6)(a^2+5a-10).$$ Thus

$$a^4-4a^3-37a^2+180a-180=(a-3)(a-6)(a^2+5a-10)=0.$$

So $$(a,b)=(3,6)\:\text{and}\: (a,b)=(6,3)$$ are integer solutions of the problem.

Now we consider the equation $a^2+5a-10=0.$ It has a positive solution which is $$\dfrac{\sqrt{65}-5}{2}.$$ Since

$$\dfrac{\sqrt{65}-5}{\dfrac{\sqrt{65}-5}{2}-2}<0$$ this doesn't give a solution to the problem.

Answer 3

From $$(3\sqrt5)^2=a^2+b^2=(a+b)^2-2ab=\frac14(ab)^2-2ab$$ $$\iff (ab)^2-8ab-180=0\iff(ab-18)(ab+10)=0$$ we find $ab=18$ or $-10$. Hence $a+b=9$. Now consider $a$ and $b$ as the solutions of $$x^2-9x+18=0\iff(x-3)(x-6)=0.$$