# How can I find the perimeter of a concave pentagon?

We know that a regular pentagon has five sides with identical lengths. But, the irregular pentagon has five sides with different angles. and moreover, the perimeter of a regular pentagon is 5a, where a is the side length of the regular pentagon. But how can I find out the formula of perimeter of an irregular concave pentagon? There is given a figure of a concave pentagon.

Here in the picture, the length of the AB and CD are Equal and the length of AB and BC are given. I have to find out the perimeter of the concave pentagon. I couldn't able to solve this anyway.

• even without immediately realizing that there is an equilateral triangle, the trigonometry is simple: AE is simply (BC/2) / sin(30º), or simply BC. – njzk2 Nov 16 '16 at 14:49

## 6 Answers

Hint: Notice how the figure looks like a rectangle with an equilateral triangle cut out.

• Looks like a rectangle. AB != BC. – Holloway Nov 16 '16 at 9:47
• @Holloway, read the problem descrption. – alexis Nov 16 '16 at 12:04
• @alexis, please point out where it says length AB = BC – Holloway Nov 16 '16 at 12:11
• Well, the problem is a rearrangement of the edges of a regular pentagon, and those edges are all equal. But I'll grant you that the final paragraph calls that into question... and I see that other readers agree with you. – alexis Nov 16 '16 at 17:54

Draw a line through $E$ parallel to $BC$. What can you say about the triangles thus formed?

The triangle $AED$ is equilateral and its base is $AD=BC=l$ so the perimeter is $3l+AB+CD$.

• Why 3l ? could you specify that ? – Salman Sourav Nov 16 '16 at 1:49
• is the length of AE , DE , BC are equal then ? – Salman Sourav Nov 16 '16 at 1:51
• Because $AE=ED=BC$....the triangle $AED$ is equilateral and $AD$ (not drawn) is equal to $BC$. – MattG88 Nov 16 '16 at 1:52

We know the triangle $AED$ is an equilateral triangle. Therefore $AD=AE=DE=BC$. Thus the perimeter is:$$P=3BC+2AB$$

Perimeter of the retangle: 2AB + 2BC

Perimeter of the equilateral triangle: 3BC

Add the two together and remove the two BC lines (left edge of the retangle, left edge of the triangle): 2AB + 3BC

The perimeter formula is 5a. Since it's a square with an inscribed equilateral triangle, the other two (triangle) sides are equal to the 3rd -- and the remaining 3 sides of the square.

3 sides of a square + 2 of a triangle