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?
+130071 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
+2668 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:
