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Problem: What is the dimension of a rectangle that has a diagonal that measures $5/4$ units and has an area of $3/4$ units squared?

My work: Using trial and error I have come up with the dimensions $1$ and $3/4$, but I don't know how to show the solution to how I got the answer.

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Let $w$ and $h$ be the width and height. Then we have:

$wh = \dfrac34$
$w^2+h^2=\left(\dfrac54\right)^2 = \dfrac{25}{16}$

The first equation gives $w = \dfrac{3}{4h}$. Subsitute this in the second equation and you get a quadratic equation in $h^2$, which you can solve.

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  • $\begingroup$ Yep, works for me too! Thanks! Thumbs up $\endgroup$ – Ria Ann Guimary Habaña Apr 7 '15 at 1:30
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Let your rectangle have sides of length $a$ and $b$. Thus, $ab=3/4$. Also, the diagonal is $5/4$ units in length. Thus, $a^2+b^2=(5/4)^2$. Hence, you end up needing to solve $$ 16b^4-25b^2+9=0. $$ If you solve this, then you will see that you got one of the possible solutions, but there are really two (when considering the distinction between length and width to be arbitrary). However, this other possible solution involves $a$ or $b$ being negative which is clearly impossible (unless you would like for your rectangle to have negative length and width). Thus, your solution is correct.

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Define two sides of the rectangle as $ x $ and $ y $. Since the rectangle's diagonal is $ \frac{5}{4} $ units, using the Pythagorean theorem:

$$ x^{2} + y^{2} = \Big(\frac{5}{4}\Big)^{2} $$ or $$ \mathrm(1)\hspace{10mm} x^{2} + y^{2} = \frac{25}{16} $$

You're also given the rectangle's area to be $ \frac{3}{4} $ units squared. Since the area of the rectangle is the product of its sides:

$$ \mathrm(2)\hspace{10mm} xy = \frac{3}{4} $$

To solve this system of equations $(1)$ and $(2)$, begin solving for $ x $ in equation $(2)$ by dividing a $y$ to each side:

$$ x = \frac{3}{4y} $$

Substituting the value of $ x $ shown above to equation $(1)$:

$$ \Big(\frac{3}{4y}\Big)^{2} + y^{2} = \frac{25}{16} $$ or $$ \frac{9}{16y^{2}} + y^{2} = \frac{25}{16} $$

Subtract a $y^2$ to each side:

$$ \frac{9}{16y^{2}} = \frac{25}{16} - y^{2} $$

Multiply a $16y^{2}$ to each side:

$$ 9 = \Big(\frac{25}{16} - y^{2}\Big)(16y^{2}) $$

Distribute the above equation's right side:

$$ 9 = 25y^{2} - 16y^{4} $$

To make the above equation resemble a quadratic, subtract $9$ to each side:

$$ 0 = 25y^{2} - 16y^{4} - 9 $$

Now let $ w = y^{2}$. Rearranging the right side terms in descending order:

$$ 0 = -16w^{2} + 25w - 9 $$ or $$ -16w^{2} + 25w - 9 = 0 $$

The above equation is of the form $ax^{2} + bx + c = 0$, so use the quadratic formula, $$ \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} $$ to solve for $ w $, where $a = -16$, $b = 25$, and $c = -9$. Here:

$$ w = \frac{-25 \pm \sqrt{25^{2} - 4(-16)(-9)}}{2(-16)} = \frac{-25 \pm \sqrt{625 - 576}}{-32} = \frac{-25 \pm \sqrt{49}}{-32} = \frac{-25 \pm 7}{-32} $$ The solutions of $ w $ are $ w = \frac{-25 - 7}{-32} = \frac{-32}{-32} = 1 $ and $ w = \frac{-25 + 7}{-32} = \frac{-18}{-32} = 9/16 $

Since $ w = y^{2} $ from the substitution, hence $ y^2 = 1 $ or $ y^2 = 9/16 $. This rectangle problem is a real-life scenario, and thus the lengths of $ x $ and $ y $ must be positive ($ x > 0 $ and $ y > 0 $). Solving for $ y $ by taking the square root to each side:

$$ y = 1 $$ or $$ y = 3/4 $$

If $ y = 1 $, then substituting into equation $(1)$ and solving for $ x $:

$$ x^{2} + 1 = \frac{25}{16} $$ or $$ x^{2} + \frac{16}{16} = \frac{25}{16} $$ or $$ x^{2} = \frac{9}{16} $$ or $$ x = \frac{3}{4} $$

If $ y = 3/4 $, then substituting into equation $(1)$ and solving for $ x $:

$$ x^{2} + \Big(\frac{3}{4}\Big)^{2} = \frac{25}{16} $$ or $$ x^{2} + \frac{9}{16} = \frac{25}{16} $$ or $$ x^{2} = \frac{16}{16} $$ or $$ x^{2} = 1 $$ or $$ x = 1 $$

Thus, the dimensions of the rectangle are $ 1 $ unit by $ 3/4 $ unit.

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