Question

A trapezium shaped field whose parallel sides are 42m and 30m and the other sides are 18m and 18m. Find the area

Answer 1

Ans =the area of trapezium is640.8m^2

Answer 2

Answer:

432√2 m²

$\rule{200}{2}$

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 =

$\frac{1}{2} \times h \times (sum \ of \ parallel \ sides)\\\\\implies \frac{1}{2} \times 12 \sqrt{2} \times (42 + 30)\\\\\implies 6\sqrt{2} \times 72\\\\\implies 432\sqrt{2}$

Hence, area of trapezium is 432√2 m²

$\rule{200}{2}$