Given that the number inside each region of the triangle is the area of that region, find the area of
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D B D C = A A D B A A D C = A O D B A O D C
4 0 + 3 0 + 3 5 8 4 + a + y = 3 5 a ⟹ 3 5 ( 8 4 + a + y ) = 1 0 5 a ⟹ 2 9 4 0 + 3 5 y = 7 0 a ( 1 )
F B A F = A C F B A C F A = A O F B A O F A
3 0 + 3 5 + a 8 4 + y + 4 0 = 3 0 4 0 ⟹ 3 0 ( 1 2 4 + y ) = 4 0 ( 6 5 + a ) ⟹ 1 1 2 0 + 3 0 y = 4 0 a ( 2 )
Solve for a in terms of y in ( 2 ) then substitute in ( 1 ) , we have
a = 4 0 1 1 2 0 + 3 0 y
2 9 4 0 + 3 5 y = 7 0 ( 4 0 1 1 2 0 + 3 0 y ) = 1 . 7 5 ( 1 1 2 0 + 3 0 y ) = 1 9 6 0 + 5 2 . 5 y
2 9 4 0 − 1 9 6 0 = 5 2 . 5 y − 3 5 y
9 8 0 = 1 7 . 5 y
y = 5 6