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I am in 6th grade, and I was given this extension for extra practice. The question is:

The ratio of one base to the height to the other base of a trapezoid is $2:3:4$. The area of the trapezoid is 117 square cm. Find the length of the bases and the height.

So far I did $A = \frac{a+b}{2}h \Rightarrow 117=\frac{a+2x}{2}3x$ which gets me to this equation $2x^2+ax-78=0$. Now I am stuck as I don't know how to solve for $a$ or $x$, but can someone nudge me in the right direction?

Adam Karlson
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Harry Iguana
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1 Answers1

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I edited your question a little, so that it is more readable now.

As you are given the ratios, you should let the common ratio to be the same.

We will denote the length of bases by $b$ and $a$ and the height by $h$. If you let $b = 2x$, then you get $h = 3x$ and $a = 4x$. You have made a slight error of not expressing $a$ in terms of $x$.

Now you should be able to solve for $x$ (which should come out to be $\sqrt{13}$, I believe)

Adam Karlson
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  • Ah so I actually referred "a" as the upper base of trapezoid 1. Could you explain why 4x is used to substitute a? I am still a little bit confused. – Harry Iguana May 07 '21 at 02:28
  • You are given three lengths corresponding to the trapezoid. I will give you another way to do it. Suppose that the upper base is smaller than the lower base. Then, you are given $$\text{lower base: height: upper base} = 2:3:4$$ Always look at such ratios pairwise. What this tells you is that the ratio of lower base to height is 2:3, that is, the height is 3/2 times that of lower base. Similarly, the upper base is 4/3 of the height. Take your height as unknown, $h$ and solve for it. This is another way.

    Earlier, I assumed that my lower base is of size 2x (e.g if it was 7cm, then x = 3.5cm)

     – Adam Karlson May 07 '21 at 10:18
  • Yeah my original equation I also made 2x as lower base. It is a little more clear now thanks. – Harry Iguana May 07 '21 at 11:46