Answer:
Fence of the trapezium shaped field \[ABCD=120\text{ }m\] \[\Rightarrow \] \[AB+BC+CD+DA=120\] \[\Rightarrow \] \[AB+48+17+40=120\] \[\Rightarrow \] \[AB+105=120\] \[\Rightarrow \] \[AB=120105\] \[\Rightarrow \] \[AB=15\text{ }m\] \[\therefore \] Area of the field \[=\frac{(BC+AD)\times AB}{2}\] \[=\frac{(48+40)\times 15}{2}\] \[=660\,{{m}^{2}}\].
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