# Area Problem #16

A regular pentagon and a square both share a side with a regular decagon. The green shape is made from the vertices of the square and pentagon. If the side length of the decagon is 2cm, what is the area of the green shape?

## 8 thoughts on “Area Problem #16”

1. The black shape is a decagon. Should it show an octagon, or is the text wrong?

2. it’s decagon all the way. I accidentally wrote octagon when I published! Rewrote straight away but realise that the email version probably says octagon. I blame baby brain.

3. EN says:

Fingers crossed π
Area = 1/2 (2sqrt(5+2sqrt5) + (3+sqrt5) sqrt(5+2sqrt5) – 2(3+sqrt5)) = 5.899095667

• I’m afraid not. It should work out to very, very close to 4cm^2. So close in fact, that I convinced myself I’d found a new ‘thing’ and that it was perfectly equal to the area of the square. It’s not perfectly equal to the square, but it is extremely close in area.

4. Ben says:

We got 3.9969826…. cm^2

5. EN says:

Of course it is! Mine was the area of the trapezium and I forgot to subtract the triangle which was sqrt(10+2sqrt5)/2 so I am getting now 3.996982634 cm^2. Thanks.