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What is the area in square inches of the pentagon shown?

 

 Dec 3, 2020
 #1
avatar+114592 
+1

Probably an easier way to do this....but.....

 

Let the bottom left vertex =  (0,0)  = A

Let the bottom right vertex =  (8,0)   = B

Let the  top left vertex  =  (0,10) = E

Let the top right vertex  ( 8,12) = C

Let the apex vertex =  D

 

CE =   sqrt ( 8^2 + 2^2)  = sqrt  ( 68)  = 2sqrt (17)

 

Semi-perimeter  of triangle   CED  =  ( 4 + 6 + 2sqrt (17) ) / 2 =    5 + sqrt (17)

 

Area of triangle   CED  =

   

sqrt  [ (5 + sqrt (17) )  ( 5 + sqrt (17) - 2sqrt (17)) (5 + sqrt (17) - 6)  ( 5 + sqrt (17) - 4 )  ]  =

 

sqrt [ (5 + sqrt (17) ) (  5 -sqrt (17)) ( sqrt (17) - 1) ( sqrt (17) + 1)  ]   = 

 

sqrt [ (5 + sqrt (17) ) (5-sqrt (17) ) (sqrt (17) -1) (sqrt (17 + 1)  ]   =

 

sqrt  [ (25 -17)  ( 17 - 1)  ]  =

 

sqrt [  8 * 16 ]  =  

 

sqrt ( 64 * 2)  =    8sqrt (2)

 

And  ABCE  is a trapezoid  with  bases of  10 and 12  and a height of 8

 

[ ABCE ] =    (1/2) [ 8] [ 10 + 12 ]=    4 * 22   =  88

 

So...area of the pentagon  = [  88  +  8 sqrt (2) ] in^2  =    8 (11 + sqrt (2))  in^2

 

 

cool cool cool

 Dec 3, 2020
 #2
avatar+902 
+2

8 * 10 = 80 u2

 

(2 * 8) / 2 = 8 u2

 

Using Heron's formula you can find the area of ΔABC.    [ABC] = 11.314 u2

 

 Dec 3, 2020
edited by jugoslav  Dec 3, 2020

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