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Find the area of the hexagen whose Vertices taken in order

are (5.0), (4,2), (1,3), (-2,2), (-3,-1) and (o,-47)​

1 Answer

7 votes

Answer:

Explanation:

A(5,0),B(4,2), C(1,3),D(-2,2),E(-3,-1),F(0-4)

we divide the hexagon in triangles and trapeziums

area Δ BCD (with coordinates B(4,2),C(1,3),D(-2,2)) =1/2(2+4)×2=6

area Δ DEG (with coordinates D(-2,2),E(-3,-1),G(-2.-2))=1/2(2+2)×2=4

area ΔAOF(with coordinates A(5,0),O(0,0),F(0,-4))=1/2×5×4=10

area trapezium ABDH (with coordinates A(5,0),B(4,2),D(-2,2),H(-2,0))=1/2[(2+5)+(2+4)]×2=13

area trapezium OHGF (with coordinates O(0,0),H(-2,0),G(-2,-2),F(0,-4))=1/2[4+2]×2=6

Total area=6+4+10+13+6=39 sq units.

User Sharikov Vladislav
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