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The number of diagonals of two different regular...

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The number of diagonals of two different regular...

by BTGmoderatorLU » Sat Oct 21, 2017 6:27 pm
The number of diagonals of two different regular polygons each consists of two identical digits.
The square root of the sum of the sides of both polygons equals the number of sides of a

A. nonagon
B. hendecagon
C. heptagon
D. pentagon
E. triskaidecagon

The OA is D.

Experts, can you help me with this PS question please? I don't understand it.
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by EconomistGMATTutor » Sat Oct 28, 2017 2:25 am
Dear LUANDATO.

I understand you because this is a hard question. But I will try to give you a good explanation.

First, the number of diagonals of a polygon is $$D=\frac{\left(n-3\right)\cdot n}{2}$$ where n is the number of sides.

Now, we have that the number of diagonals of each polygon is {11, 22, 33, 44, 55, 66, 77, 88, 99}.

So, we can set the equations $$\frac{\left(n-3\right)\cdot n}{2}=11,\ 22,\ 33,\ .\ .\ .\ ,\ 99.$$

The only solutions we are interested in are those where n is a positive integer.

The unique numbers of diagonals that satisfy this are 44 and 77. Let's prove it:

$$ \frac{\left(n-3\right)\cdot n}{2}=44\ <=>\ n^2-3n=88\ <=>\ n^2-3n-88=0,$$

The solutions of the quadratic equation above are n=11 and n=-8. We have to select n=11.

$$ \frac{\left(n-3\right)\cdot n}{2}=77\ <=>\ n^2-3n=154\ <=>\ n^2-3n-154=0,$$

The solutions of the quadratic equation above are n=14 and n=-11. We have to select n=14.

So, we have two regular polygons, one has 11 sides and the other has 14 sides.

Finally, the square root of the sum of the sides of both polygons is $$\sqrt{11+14}=\sqrt{25}=5.$$

And this is the number of sides of a Petangon.

So, the answer is option D.

I hope this can help you.

I'm available if you'd like any follow up.

Regards.
GMAT Prep From The Economist
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