The questions is "The length and width of a rectangle are $7$m and $5$m. When each dimension is increased by the same amount, the area is tripled. Find the dimensions of the new triangle, to the nearest tenth of a metre."`

Answer: The rectangle is $9.3$m by $11.3$m

I started with writing: $(x+7)(x+5)=105$

Then I expanded to standard form: $x^2+12x-70$

Then I completed the square to convert this to vertex form: $(x+6)^2-106$

I thought that the vertex of the parabola was going to be the dimensions of the new rectangle, but it isn't. I then checked if the roots of this equation were the answer, but they weren't. Now I'm completely lost. Any help is appreciated, my final exams are in two days.

  • 2
    $\begingroup$ Is it a rectangle, or a triangle? $\endgroup$ – Lord Shark the Unknown Jun 12 '18 at 4:31
  • $\begingroup$ Some paragraph breaks would really help with the readability of your question. Some basic MathJax formatting would also go a long way. $\endgroup$ – Xander Henderson Jun 12 '18 at 4:41

$$A=7\times5$$ $$3A=(7+x)\times(5+x)$$ These two equations show the original and new area of the rectangle $$A=35$$ $$\therefore3(35)=(7+x)\times(5+x)$$ $$105=35+12x+x^2$$ $$x^2+12x-70=0$$ From here either use graphics calculator or use quadratic formula $$x=-6+\sqrt{106}\approx4.2956$$ $$x=-6-\sqrt{106}\approx-16.29$$ It cannot be the second answer as this makes the side lengths negative

In the new rectangle, the side lengths were $$x+5=4.2956+5=9.2956$$ $$x+7=4.2956+7=11.2956$$

Therefore the new side lengths are approximately 9.3m and 11.3m

  • $\begingroup$ Thank you very much you explained it perfectly $\endgroup$ – Jackson Jun 12 '18 at 5:14
  • $\begingroup$ @Jackson All good bro, good luck with your exams $\endgroup$ – Strevo Jun 12 '18 at 7:14

A couple of things to note about your solution: you say you had $$(7+x)(5+x)=105$$ which you then expanded and rearranged to get $$x^2 + 12x - 70$$

Where did the equals sign go?

In fact, what you should have at this point is not just the expression $x^2+12x-70$, but rather the equation $$x^2+12x-70=0$$ which should make it clear that you're not looking for the vertex of the parabola $y=x^2+12x-70$, but rather its $x$-intercepts.

Essentially what happened here is that you dropped the "$=0$" from your equation, which led you to mis-read the problem as being about the vertex of a quadratic function, rather than the solution to a quadratic equation.


Given a rectangle of length $7$ and width $5$ we know that its area is $35$.
When the length and the width are increased by the same amount, which we'll call $d$, the area of the rectangle is tripled. We can write this as $(7+d)(5+d)=105$, where the new length $l=(7+d)$ and the new width $w=(5+d)$.

This is expanded to $35+12d+d^2=105$. Your error was here.
$$d^2+12d-70=0$$ $$d=\frac{-12\pm\sqrt{424}}{2}\approx 4.3,-16.3$$ We can easily eliminate $d=-16.3$, because that yields negative side lengths. Therefore, we have $l=7+4.3=11.3$ and $w=5+4.3=9.3$.

New dimensions: $11.3$, $9.3$


Assume you have a rectangle with length $ l = 7m$ and width $w = 5m$, and area $A = l*w = 35 m^2$. You want to find $l' = l + x$ and $w' = w + x$ such that $A' = 3A = 105m^2$. This reduces to $(7 +x)(5+x) = 105$, expanding with F.O.I.L. gets you $ 35 + 12x + x^2 = 105$, rearranging yields $x^2 + 12x - 70 = 0$ which can be solved with a standard quadratic formula for $x$, which you can then substitute for $l', w'$ The step you did to complete the square is a little unnecessary, as the quadratic formula is easily applied to my last equation. You should get two solutions (any quadratic equation should have 2 solutions), and you want the positive one because negative length doesn't make sense.


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.