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## Comment on

Polygons## Hi,

Can we assume a polygon to be regular if nothing is mentioned?

## No. If the question doesn't

No. If the question doesn't stated that the polygon is regular, we can't assume that it's regular.

## Thanks

## https://gre.myprepclub.com/forum

I have a potential dumb question here: How can we assume that all the sides of convex polygon are equal?

## Question link: https:/

Question link: https://gre.myprepclub.com/forum/length-of-a-side-of-regular-pentagon-wi...

Good question.

The word "regular" tells us that all sides are equal.

Without the word "regular," we wouldn't be able to make any conclusions about the lengths of the sides.

Cheers,

Brent

## I think in the last class you

## That's correct; the 4 angles

That's correct; the 4 angles in a quadrilateral will ALWAYS add to 360°.

So, to answer your second question, the angles in a quadrilateral CANNOT have a sum less than 180°.

Where do I say that the angles in a quadrilateral have a sum less than 180°?

Cheers,

Brent

## please you mentioned that to

## Question link: https:/

Question link: https://gre.myprepclub.com/forum/given-figure-is-that-of-a-regular-hexag...

The 6 angles in a hexagon will add to 720 degrees.

In the my solution to the above question, I divide the regular hexagon into 6 equilateral triangles.

Notice that each equilateral triangle has one of its angles at the CENTER of the hexagon.

These six angles at the CENTER are not part of the angles in a hexagon. So, we can disregard them.

When we disregard them, we see that the angles add to 720 degrees.

Does that help?

Cheers,

Brent

## Dear Brent,

https://gre.myprepclub.com/forum/a-regular-polygon-of-24-sides-is-inscribed-in-a-circle-12630.html

From your solution to this question, can I infer that perimeter of a polygon inscribed in a circle will always be smaller than the circumference of the inscribing circle cause distance from one point to another along a straight line will be smaller than the distance along a curve?

## Absolutely! The perimeter of

Absolutely! The perimeter of a polygon inscribed in a circle will always be less than the circumference of the inscribing circle.

## Hi Brent,

https://gre.myprepclub.com/forum/in-a-particular-seven-sided-polygon-the-sum-of-four-equal-12768.html

For this I had approached in this way

As sum of the interioir angles=900

the 4 equal angles will be 900/4 = 225

so Qa = 225

Is this the right approach

Thanks

## Question link: https:/

Question link: https://gre.myprepclub.com/forum/in-a-particular-seven-sided-polygon-the...

Your approach yielded the correct answer, but it's not valid.

If each of the four angles were 225, then the other three angles would be 0 degrees (which is impossible)

Cheers,

Brent

## I solved this problem by

https://gre.myprepclub.com/forum/what-is-the-area-of-the-figure-below-11775.html

## Question link: https:/

Question link: https://gre.myprepclub.com/forum/what-is-the-area-of-the-figure-below-11...

That's a perfectly valid approach. Nice work!

## Hi Brent, regarding this

https://gre.myprepclub.com/forum/a-rectangle-is-inscribed-in-a-hexagon-that-has-all-sides-of-equal-leng-25565.html

I have a question.

(If answering it requires some horrible proof, please ignore this or give me the first step and I will go from there.)

In my solution (see below, step 3), I assumed that the area of each shaded part equaled 1/2 of an equilateral triangle.

Question: But how do we *know* that each of the shaded portions has an area equal to 1/2 of an equilateral triangle?

Is the answer connected to 30-60-90 triangles?

The short version of this theory: take your region A and the UNshaded right triangle below it.

Two adjacent 30-60-90 triangles consist of side-30° angle-side, where sides are equal. The two triangles are thus congruent.

An equilateral triangle consists of two congruent 30-60-90 triangles, so region A (and its UNshaded counterpart) must each have area equal to 1/2 of area of equilateral triangle.

I could be way off base. I am too close to this question at the moment.

I solved in a way similar to the way you did. I just assigned a value for area x.

1. Let area x of hexagon =36

2. Divide the hexagon into 6 congruent equilateral triangles.

Then each of the six equilateral triangles has area 36/6 = 6

3. [I assumed this part and only thought about it later.]

The area of each shaded triangle =

1/2 of an equilateral triangle with area of 6.

So the area of each shaded region equals 6/2 = 3

4. Total area of shaded parts = 3+3+3+3= 12

5. If X = 36, and shaded parts' total = 12, then

the shaded parts are 12 of 36 total parts

= 12/36

= 1/3 of total area, which

= 1/3x = x/3

I might be missing something really obvious about the "shaded = 1/2 of equilateral triangle" issue.

## It's a solid solution. Nice

It's a solid solution. Nice work!

Here's how we know about the shaded parts representing half on an equilateral triangle: https://imgur.com/u3gmHmP

All of this is based on the fact that all regular hexagons are comprised of 6 equilateral triangles.

## That image is PERFECT (and