Math Help - largest possible perimeter

1. largest possible perimeter

Three non-overlapping regular plane polygons, at least two of which are congruent, all have side length 1. The polygons meet at a point A in such a way that the sum of the three interior angles at A is 360. Thus the three polygons form a new polygon with A as an interior point. What is the largest possible perimeter that this polygon can have?

a) 12 b) 14 c) 18 d ) 21 e)24

The answer is d)21 but I think I can make all of them, if not more.
Could someone please take a look at my attached work and
show me what I am not understanding?

Thanks.

Vicky.

2. Originally Posted by Vicky1997
Three non-overlapping regular plane polygons, at least two of which are congruent, all have side length 1. The polygons meet at a point A in such a way that the sum of the three interior angles at A is 360. Thus the three polygons form a new polygon with A as an interior point. What is the largest possible perimeter that this polygon can have?

a) 12 b) 14 c) 18 d ) 21 e)24

The answer is d)21 but I think I can make all of them, if not more.
Could someone please take a look at my attached work and
show me what I am not understanding?

Thanks.

Vicky.
1. The formula you used determines the sum of all interior angles of a polygon. The value of one single interior angle is calculated by

$|\alpha| = 180^\circ \cdot \left(\frac{n-2}{n}\right)$

where n is the number of vertices of the polygon.

2. Let n denote the number of vertices of the congruent polygons and k the number of vertices of the other one then you'll get:

$2 \cdot 180^\circ \cdot \left(\frac{n-2}{n}\right) + 180^\circ \cdot \left(\frac{k-2}{k}\right) = 360^\circ$

Simplify this equation and you'll get

$n = \frac{4k}{k-2}$

3. The perimeter of the 3 polygons is

$p = 2n+k - 6$

4. I'll leave the rest for you.

3. Originally Posted by earboth
1. The formula you used determines the sum of all interior angles of a polygon. The value of one single interior angle is calculated by

$|\alpha| = 180^\circ \cdot \left(\frac{n-2}{n}\right)$

where n is the number of vertices of the polygon.

2. Let n denote the number of vertices of the congruent polygons and k the number of vertices of the other one then you'll get:

$2 \cdot 180^\circ \cdot \left(\frac{n-2}{n}\right) + 180^\circ \cdot \left(\frac{k-2}{k}\right) = 360^\circ$

Simplify this equation and you'll get

$n = \frac{4k}{k-2}$

3. The perimeter of the 3 polygons is

$p = 2n+k - 6$

4. I'll leave the rest for you.

Thank you!! Thankyou!!

I think I got the answer.
Could you please take a look at my work

Vicky.

4. Originally Posted by Vicky1997
Thank you!! Thankyou!!

I think I got the answer.
Could you please take a look at my work

Vicky.
$p=2(n-2)+(k-2)$