solving rectangle

Question

Diagonal of a rectangle is $13$ cm. If we extend the length of the rectangle for $4$ cm and width for $7$ cm, then diagonal will be longer for $7$ cm as well. Find sides (length and width) of the rectangle. So $d^2=a^2+b^2 \implies b^2=d^2-a^2 \implies b^2=13^2-a^2 \implies b^2=169-a^2$ and $a^2+b^2=169$ \begin{align*} d'^2 & =a'^2+b'^2\\ (d+7)^2 & =(a+4)^2+(b+7)^2\\ 20^2 & = a^2+8a+16+b^2+14b+49\\ 400 & = 8a+16+14b+49+169\\ \frac{166}{2} & = \frac{8a+14b}{2}\\ 83 & =4a+7b \end{align*} ... That's what I've done. What should I do next?

Answer 1

You can substitute in $a=\sqrt{169-b^2}$ and do the algebra. I don't think the numbers will be very pretty.

Answer 2

You have $$ a^2+b^2=169 $$

$$ 4a+7b=83 \Longleftrightarrow a=\frac{83-7b}{4} $$

After substitution:

$$ \frac{(83-7b)^2}{16}+b^2=169 $$

$$ \frac{49b^2}{16}-\frac{1162b}{16}+\frac{6889}{16}-169=0 $$

Finally, use 'abc-formula' with a = 49/16, b = 1162/16 and c = 6889/16-169.

Answer 3

Let us rework the problem; this might make the answer easier to see. Let $a$ and $b$ be the side lengths of the rectangle, in which case $13 = \sqrt{a^{2} + b^{2}}$ is the diagonal. As you have already noted, we have $169 = a^{2} + b^{2}.$

When we enlarge the rectangle, we can use Pythagoras once again to get the following relation: $$20 = \sqrt{(a + 4)^{2} + (b + 7)^{2}}.$$

This expands to $$400 = (a + 4)^{2} + (b + 7)^{2},$$ which you have discovered to simplify into $4a + 7b = 83.$

Now, we begin the messy work. Rearranging the previous equation, $a = \frac{83 - 7b}{4}.$ We substitute this into $169 = a^{2} + b^{2}$ and solve as follows: $$\left(\frac{83 - 7b}{4}\right)^{2} + b^{2} = 169$$ $$\frac{6889 - 1162b + 49b^{2}}{4} + b^{2} = 169,$$ which yields the quadratic form $$\frac{53}{4}b^{2} - \frac{581}{2}b + \frac{6889}{4} = 0.$$

But the discriminant of the equation is less than $0,$ signaling no real solutions. Looking back at the problem, this makes sense. If you expand the width by $7$ units and the length by a non-zero amount, the diagonal should expand by more than $7$ units.