9
$\begingroup$

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.

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.

$\endgroup$
1
  • $\begingroup$ even without immediately realizing that there is an equilateral triangle, the trigonometry is simple: AE is simply (BC/2) / sin(30º), or simply BC. $\endgroup$
    – njzk2
    Nov 16, 2016 at 14:49

6 Answers 6

12
$\begingroup$

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

$\endgroup$
4
  • 5
    $\begingroup$ Looks like a rectangle. AB != BC. $\endgroup$
    – Holloway
    Nov 16, 2016 at 9:47
  • $\begingroup$ @Holloway, read the problem descrption. $\endgroup$
    – alexis
    Nov 16, 2016 at 12:04
  • 9
    $\begingroup$ @alexis, please point out where it says length AB = BC $\endgroup$
    – Holloway
    Nov 16, 2016 at 12:11
  • 1
    $\begingroup$ 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. $\endgroup$
    – alexis
    Nov 16, 2016 at 17:54
6
$\begingroup$

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

$\endgroup$
5
$\begingroup$

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

$\endgroup$
3
  • $\begingroup$ Why 3l ? could you specify that ? $\endgroup$ Nov 16, 2016 at 1:49
  • $\begingroup$ is the length of AE , DE , BC are equal then ? $\endgroup$ Nov 16, 2016 at 1:51
  • $\begingroup$ Because $AE=ED=BC$....the triangle $AED$ is equilateral and $AD$ (not drawn) is equal to $BC$. $\endgroup$
    – MattG88
    Nov 16, 2016 at 1:52
5
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .