A field is in the shape of a trapezium whose parallel sides are 25m and 20m. thenon – parallel sides are 15 and 12m. Find the area of the field.
Answers
Step-by-step explanation:
Construction:
⇒Draw BE∥AD such that D−E−C
⇒Draw BM⊥DC such that D−E−M−C
□ABED is parallelogram,
⇒AD=BE=13 m
⇒AB=DE=10 m
⇒BC=14 m
⇒DC=DE+EC ....(∵D−E−C)
∴EC=25−10
∴EC=15 m
⇒In ΔBEC
⇒2S=13+14+15
⇒S=21 m
⇒Area of △BEC
⇒A=
s(s−a)(s−b)(s−c)
=
21(21−13)(21−14)(21−15)
=
21×8×7×6
=84 m
2
⇒Area of ΔBCE=
2
1
×BM×EC
⇒BM=
15
84×2
=11.2 cm
⇒Area of □ABED=11.2×10=112 m
2
⇒Area of the field □ABCD
=84+112 m
2
=196 m
2
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Answer:
Given :
A trapezium such that whose parallel sides are 25 m and 10 m. And non parallel sides are 14 m and 13 m.
Find :
Area of the field (trapezium).
Construction :
Draw a perpendicular from B on CE.
Assume :
Let ABCD a trapezium such that AD = 14 m, CD = AB = CD = 10 m, DE = 25 mand BE = 13 m respectively.
Solution :
CE = DE - DC
CE = 25 - 10 = 15 m
Semi-perimeter of ΔBCE = Sum of sides/2
s = (a + b + c)/2
Here.. a = BC, b = CE and c = EB
s = (BC + CE + EB)/2
s = (14 + 15 + 13)/2
s = 42/2
s = 21 m
From Heron's formula
Area of ΔBEC =
=
=
= 84 m² _____ (eq 1)
Area of ΔBCE
= 84 = 1/2 × b × h
= 84 = 1/2 × 15 × h
= 84 × 2 = 15h
= h = 11.2 m
(h = BF)
Area of parallelogram ABCD = b × h
=> 10 × 11.2
=> 112 m² ____ (eq 2)
Area of trapezium AEBD = (eq 1) + (eq 2)
=> 84 + 112
=> 196 m²
∴ Area of trapezium is 196 m².
