Math, asked by abulhusssan, 5 months ago


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

Answered by aastha28275
6

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

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