3. The length of the fence of a trapezium-shaped field ABCD is 130 m
and side AB is perpendicular to each of the parallel sides AD
and BC. If BC = 54 m, CD = 19 m and AD
54 m, CD = 19 m and AD = 42 m, find the area of
the field.
Answers
Answer:
The perimeter of field is 130
The perimeter of field is 130⟹AB+BC+CD+AD=130
The perimeter of field is 130⟹AB+BC+CD+AD=130AB=130−54−19−42
The perimeter of field is 130⟹AB+BC+CD+AD=130AB=130−54−19−42AB=15m
The perimeter of field is 130⟹AB+BC+CD+AD=130AB=130−54−19−42AB=15mArea is given as
The perimeter of field is 130⟹AB+BC+CD+AD=130AB=130−54−19−42AB=15mArea is given as 2
The perimeter of field is 130⟹AB+BC+CD+AD=130AB=130−54−19−42AB=15mArea is given as 21
The perimeter of field is 130⟹AB+BC+CD+AD=130AB=130−54−19−42AB=15mArea is given as 21
The perimeter of field is 130⟹AB+BC+CD+AD=130AB=130−54−19−42AB=15mArea is given as 21 AB(AD+BC)
The perimeter of field is 130⟹AB+BC+CD+AD=130AB=130−54−19−42AB=15mArea is given as 21 AB(AD+BC)2
The perimeter of field is 130⟹AB+BC+CD+AD=130AB=130−54−19−42AB=15mArea is given as 21 AB(AD+BC)21
The perimeter of field is 130⟹AB+BC+CD+AD=130AB=130−54−19−42AB=15mArea is given as 21 AB(AD+BC)21
The perimeter of field is 130⟹AB+BC+CD+AD=130AB=130−54−19−42AB=15mArea is given as 21 AB(AD+BC)21 ×15(54+42)
The perimeter of field is 130⟹AB+BC+CD+AD=130AB=130−54−19−42AB=15mArea is given as 21 AB(AD+BC)21 ×15(54+42)15×48=720m
The perimeter of field is 130⟹AB+BC+CD+AD=130AB=130−54−19−42AB=15mArea is given as 21 AB(AD+BC)21 ×15(54+42)15×48=720m 2