# I'm given the dimensions of a rectangle, and when they increase by 1 the area is tripled…

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.

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

• Thank you very much you explained it perfectly – Jackson Jun 12 '18 at 5:14
• @Jackson All good bro, good luck with your exams – 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.