A field is in the form of a Trapezium whose parallel sides are 42 metre and 30 metre if its non parallel sides are 18 and 18 metre find its area
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432√2 m²
Step-by-step explanation:
Refer the attachment for figure
In figure, ABCD is the trapezium with height h, AB = 30 m and CD = 42 m.
Consider triangle ADE and BFC. They both are congruent as :-
∠ AED = ∠BFC (90° each)
AD = BC (18 m given)
AE = BF (both length is h)
By, RHS ΔADE ≅ ΔBFC
Hence, By CPCT, DE = FC
let, DE = FC be x
and, EF = 30 m( since, ABEF is a rectangle, opposite sides are equal)
So, we get now
DE + EF + FC = DC
⇒ x + 30 + x = 42
⇒ 2x = 42 - 30
⇒ 2x = 12
⇒x = 12/2
⇒ x = 6 m
⇒ED = FC = x = 6 m
Now, apply Pythagoras Theroem in any triangle to get 'h'
By Pythagoras Theroem,
h² + ED² = AD²
⇒ h² = AD² - ED²
⇒ h² = 18² - 6²
⇒ h² = 324 - 36
⇒ h² = 288
⇒ h = √288
⇒ h = 12√2 m
So, area of trapezium =
Hence, area of trapezium is 432√2 m²
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