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=42 m, find the area of the field.

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Solution

The perimeter of field is $130$
$⟹ A B + B C + C D + A D = 130 A B = 130 - 54 - 19 - 42 A B = 15 m$
Area is given as $\frac{1}{2} A B (A D + B C) \frac{1}{2} \times 15 (54 + 42) 15 \times 48 = 720 m^{2}$

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Q. Question 3
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Length of the fence of a trapezium-shaped field ABCD is 120 m. If BC = 48 m, CD = 17 m and AD = 40 m, find the area of this field, given that side AB is perpendicular to the parallel sides AD and BC.

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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=42 m, find the area of the field.

The perimeter of field is 130⟹AB+BC+CD+AD=130AB=130−54−19−42AB=15mArea is given as 12AB(AD+BC)12×15(54+42)15×48=720m2

The perimeter of field is $130$
$⟹ A B + B C + C D + A D = 130 A B = 130 - 54 - 19 - 42 A B = 15 m$
Area is given as $\frac{1}{2} A B (A D + B C) \frac{1}{2} \times 15 (54 + 42) 15 \times 48 = 720 m^{2}$