# Find width and length of rectangle given diagonal and area

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.

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

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 $7$.

• 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!! – drin Jul 13 '15 at 12:30
• 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. – Zain Patel Jul 13 '15 at 14:08 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 ;) 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. $$(x-7)(x-24)=0.$$

• Although elegant, I find the method of direct substitution posted by Zain Patel to be more pedagogic and easy to understand. – Emily L. Jul 13 '15 at 12:35
• @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.:-) – Zain Patel Jul 13 '15 at 14:33