Thread: halving a rectangle

1. halving a rectangle

I'm trying to find an expression, equation.

Given rectangle x by y, what is the single value, a, by which I can reduce

each side to halve the area.

xy / 2 = (x-a)(y-a) solve for a

2. Originally Posted by gowan
I'm trying to find an expression, equation.

Given rectangle x by y, what is the single value, a, by which I can reduce

each side to halve the area.

xy / 2 = (x-a)(y-a) solve for a
Looks like you've got it. Can you not solve for a?

3. I'm brand new to this forum. I'm guessing that this is how I respond to a question.
No I can't solve for a in a way that would permit me to find 'a' given x and y.

4. Originally Posted by gowan
I'm brand new to this forum. I'm guessing that this is how I respond to a question.
No I can't solve for a in a way that would permit me to find 'a' given x and y.
$\frac{xy}{2}=xy-ax-ay+a^2$

$-\frac{xy}{2}=a^2-a(x+y)$

Note that the coeficient of the linear term is $-(x+y)$. So, can you complete the square?

5. Hello, gowan!

Given rectangle $x$ by $y$, what is the single value, $a$,
by which I can reduce each side to halve the area?

. . $(x-a)(y-a)\:=\:\frac{xy}{2}\qquad \hdots$ Solve for $a.$

We have: . $xy - ax - ay + a^2 \:=\:\frac{xy}{2} \quad\Rightarrow\quad 2xy -2ax - 2ay + 2a^2 \:=\:xy$

. . which simplifies to:. . $2a^2 - 2(x+y)a + xy \:=\:0$ . . . a quadratic in $a$

Quadratic Formula: . $a \;=\;\frac{2(x+y) \pm\sqrt{4(x+y)^2 - 8xy}}{4} \;=\;\frac{(x+y) \pm\sqrt{x^2+y^2}}{2}$

The answer is the smaller of the two values: . $a \;=\;\frac{(x+y)-\sqrt{x^2+y^2}}{2}$

6. So, once again, we have "to solve a problem, don't think at all, just memorize a formula."