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I have a question: If I have a square that is $50\ ft$ by $50\ ft$ it equals to $2500\ ft^2$. Simple enough.

Now, if I divide $2500\ ft^2$ area in half, that equals to $1250\ ft^2$. How do I know what the length and width of this area are?

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    $\begingroup$ Is "divide in half"="shrink the width"? Otherwise there are a lot of solutions. $\endgroup$
    – Emil
    Feb 6 '18 at 17:51
  • $\begingroup$ 25ft by 25ft does not equal to 1250sqft area. $\endgroup$
    – ben
    Feb 6 '18 at 17:52
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    $\begingroup$ @ben Draw a picture. There are many many ways to divide a square into two sectors of equal area. It all depends on how you divide. $\endgroup$ Feb 6 '18 at 17:55
  • $\begingroup$ What I should have clarified, is that the I'm dividing the area of the square proportionally. $\endgroup$
    – ben
    Feb 6 '18 at 22:19
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The easiest thing to do is cut the square in half to make a rectangle. Then it has a length of 50 feet and width of 25 feet. But I assume you're not asking that, since it's so obvious. I assume you're asking about a square that has half the area.

So how would you find the length and width of a square that has half the area of the original square? Draw the diagonals of the square, cutting it into four right triangles. Take two of those triangles, (half the area), and put them together making a new square. The length and width of this new square is what you want, and you see from how you constructed it, that the edge of the square is equal to half the diagonal of the original square.

So we just need to calculate the length of the diagonal of the original square. The diagonal is the hypotenuse of a right triangle, with the edges of the original square its legs, so the diagonal is the square root of 5000, or 50 times the square root of two.

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Of course length and width of $1250$ $ft^2$ area change depending on how you divide the square into two. I will only give some examples since there are infinitely many such divisions of a square:

enter image description here

For example, in part $a)$, we have $50 \times 25$ rectangle, whose area is $1250\ ft^2$. That's probably what you were looking for.

But in part $b)$, we don't even have a rectangle, instead, we have a isosceles right angle triangle whose sides are $50\ ft$. Still the area is $\frac{50 \cdot 50}{2} = 1250\ ft^2$.

Part $c)$ is the reason why I said there are infinitely many such divisions. You may have a trapezoid, whose short base is $a\ ft$, long base is $50-a\ ft$, height is $50\ ft$. Then area of each trapezoid is $\frac{[a+(50-a)]\cdot 50}{2} = 1250\ ft^2$ independent from the value of $a$. So if we assume $a \in \mathbb{R}^+$ with $0 < a < 25$, there are infinitely many such real numbers.

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  • $\begingroup$ dividing into right triangles made sense and it worked giving me 35.355ft by 35.355ft $\endgroup$
    – ben
    Feb 6 '18 at 18:05
  • $\begingroup$ But is there an easy online calculator that I cause for future use so that I don't have to take the long road? $\endgroup$
    – ben
    Feb 6 '18 at 18:06
  • $\begingroup$ Oh, I think you found it by $\sqrt{1250}$. But area of an isosceles right angle triangle with sides $a$ is not $a \cdot a$ so $a \cdot a = 1250$ is not correct. Area should be equal to $\frac{a \cdot a}{2}$ so we should have $\frac{a \cdot a}{2} = 1250 \implies a^2 = 2500 \implies a = 50\ ft$ as shown in part $b)$. $\endgroup$
    – ArsenBerk
    Feb 6 '18 at 18:12
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There are many ways to divide a square area into two parts.A straight line or any other anti-symmetric curve ( odd function $ y=f(x)\, eg., \, y=\tanh x )$ with respect to the square's center as origin bifurcates total area of the square.

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