+0

# Pentagon

0
85
2

The vertices of a convex pentagon are (-1,-1), (-3,4), (1,7), (6,5) and (4,-2) . What is the area of the Pentagon?

Apr 21, 2022

#1
+1

This pentagon is  irregular

See the image We can break this up into  3 triangles  AED , ADC and ABC

I'm going to  describe how to find the area of triangle  AED.....based on  this you should be able to  figure the areas of the  other two triangles  ( I'm going to express roots  as decimals to  ease the computations.....because of this your  computation will be a close approximation of the true  area )

Find the side lengths  using the distance  formula

AE  = sqrt [ (1 - -3)^2  + (7-4)^2 ]  = 5

DE =  sqrt [ (-1- -3)^2 + (4  - - 1)^2 ]  =  sqrt  [ 29 ]  ≈ 5.39

AD  = sqrt [  ( 1- -1)^2  + (7 - -1)^2 ] =  sqrt  [80 ]  ≈  8.25

Add the sides  and  divide by 2   =     [ 5 + 5.39 + 8.25]  / 2  =    9.32

Using Heron's Formula  to  calculate  the area  we have

sqrt   [s * ( s -AE)(s - DE) (s - AD)  ]   =  sqrt  [ 9.32 * (9.32  - 5) (9.32 -5.39) (9.32 - 8.25) ]  ≈   13   units^2

Follow the same  procedure for  the other two triangles .....

Then....add the ares you get to find the area of the pentagon   Apr 21, 2022
#2
+1

The entire pentagon can be enclosed in a $$9 \times 9$$ grid.

The area of the pentagon is the area of the square minus all the regions outside the pentagon, but inside the square.

We can split the area outside the pentagon into 6 distinct shapes, each of which can calculate the area as shown.

Thus, the area of the pentagon is $$81 -5-6-5-2-2.5-7 = \text{____}$$

Here is the diagram: Apr 22, 2022