a park in the shape of a quadrilateral ABCD has angle C is equal to 90 degree a b is equal to 9 M BC is equal to 12 CD is equal to 5 M and 80 is equal to 8 metre how much area does it occupy
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By Pythagoras Theorem (In triangle ABC)
AC^2=AB^2+BC^2
= 9^+12^
= 81+144
= 225
AC^2=225
Therefore..... AC= 15m
Area of quadrilateral ABCD = Area of triangle ABC + Area of triangle ADC
Area of triangle ABC = 1/2*AB*BC
= 1/2*9*12
= 54m^2.......... equation (1)
Area of triangle ADC = 1/2*AD*DC
= 1/2*8*5
= 20m^2.......... equation (2)
Adding (1) and (2),
54m^2+20m^2=74m^2
Hence, quadrilateral ABCD occupies 74m^2 of area.
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