0
$\begingroup$

The diagonal of rectangle is 25, its area is 168, find width and length. I tried solving this problem using trigonometry since diagonal and two sides forms a right triangle, from area i got that a=168/b, considering that area is a*b=168, I reached to a quadratic equation but i got a negative root, does anyone have any idea? Besides the solutions are a=7 and b=24.

$\endgroup$
  • $\begingroup$ $a*b=168$ and $a^2+b^2=25^2$ $\endgroup$ – Hexacoordinate-C Jul 13 '15 at 12:15
  • $\begingroup$ i did that but what after? $\endgroup$ – drin Jul 13 '15 at 12:17
1
$\begingroup$

You have two equations $$ab = 168 \implies a = \frac{168}{b}$$ and $$a^2 + b^2 = 25^2$$

Substituting the first into the second and multiplying throughout by $b^2$ yields $$\frac{168^2}{b^2} + b^2 = 25^2 \implies 168^2 + b^4 = 25^2b^2$$ This is a quadratic in $b^2$ that gives us solutions $$b^2 = 49 \quad \text{ or } \quad 576$$

Hence $b = \pm 7$ and $b = \pm 24$. We neglect the negative solutions to get $$a = 24, b=7 \quad \text{or} \quad a=7, b= 24$$

Which is just symmetric. So you can simply say that one side is $24$ and the other is $7$.

$\endgroup$
  • $\begingroup$ i don't understand what to do after 168^2+b^4=25^2*b^2, if i try to solve this as a quadratic equation i get b^2-25*b+168=0, which leads me to negative root!! $\endgroup$ – drin Jul 13 '15 at 12:30
  • $\begingroup$ Let $u = b^2$ to get 168^2 + u^2 = 24^2 u$, now solve this as a quadratic in $u$ and then back-substitute. $\endgroup$ – Zain Patel Jul 13 '15 at 14:08
0
$\begingroup$

It normal to find negative solution, they do mathematics. In real world of course lengh is positive !

Just look $$7^2+24^2=25^2=(-7)^2+24^2=7^2+(-24)^2$$

Hope it helps you ;)

$\endgroup$
0
$\begingroup$

We have $$ab=168\tag 1$$ $$a^2+b^2=25^2\tag2$$ So, from $(1)(2)$, we have $$(a+b)^2=a^2+b^2+2ab=25^2+2\cdot 168=961\Rightarrow a+b=31.\tag3$$

Now from $(1)(3)$, we know that $a,b$ are the solutions of $$x^2-31x+168=0,$$ i.e. $$(x-7)(x-24)=0.$$

$\endgroup$
  • $\begingroup$ Although elegant, I find the method of direct substitution posted by Zain Patel to be more pedagogic and easy to understand. $\endgroup$ – Emily L. Jul 13 '15 at 12:35
  • 1
    $\begingroup$ @EmilyL. I was considering going with this method, but then I re-evaluated based on the level of experience demonstrated by the OP and went with the (less elegant) method.:-) $\endgroup$ – Zain Patel Jul 13 '15 at 14:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.