Question

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

Answer 1

Given that :-

AB = 30 M

DC = 42M

AD=BD=18M

$\rule{200}{2}$

Construction - AE and BF perpendicular to DC

$\rule{200}{2}$

Now,

AB = EF =30M

In Triangle ADE and BCF :-

☸️ AD = BC (GIVEN)

☸️ $\angle E$ = $\angle F$ [90 degrees]

☸️ AE = BF [distance between parallel lines are equal]

$\boxed{ADE\:congruent\:to\:BCF}$

Reason - RHS congruency criteria

So,

DE = CF (C. P. C. T)

$\rule{200}{3}$

Now,

DE+CF+30 = 42

DE+CF = 42-30 =12

DE = CF = 6

$\rule{200}{2}$

Now,

In right triangle ADE

AD = 18 M

DE=6M

So,

AE² = AD² - DE² [Pythagoras Theorem]

So,

$ae = \sqrt{ {18}^{2} - {6}^{2}} \\ \\ \\ = \sqrt{324 - 36} \\ \\ \\ = \sqrt{288} \\ \\ \\ = 16.97(approx \cdot)$

Now,

Area of Trapezium :-

$\boxed{ \sf{ \dfrac{1}{2} \times(sum \: of \: \parallel \: sides) \times h}}$

So,

Area of field will be :-

$\frac{1}{2} \times (42 + 30) \times 16.97 \\ \\ \\ = \frac{1}{2} \times 72 \times 16.97 \\ \\ \\ = 36 \times 16.97 \\ \\ \\ = 610.92 {m}^{2}$

$\rule{200}{4}$

So,

Area of field is equal to 610.92 M² approximately.

On rounding off we can say that

Area of field = 611 m²

Answer 2

$<b><i>Answer</b>$: 432√2 m²

$<b>Step-by-step explanation</b>$:

Draw BE || AD, and we get ||gm ADEB and ΔBCE.

Area of ΔBCE = $<b>√[s(s - a)(s - b)(s - c)]</b>$

Where, $<b>s = ( a + b + c )/2</b>$

Here, s = [ 18 + 18 + ( 42 - 30 )] / 2

s = ( 18 + 18 + 12 ) / 2

s = 24 m

Area = √[ 24(24 - 18)(24 - 18)(24 - 12)]

= √( 2 × 12 × 6 × 6 × 12 )

= 12 × 6 √2

= 72√3

$<b>Area = 1/2 × base × height</b>$

72√2 = 1/2× 12 × h

72√2 × 2/12 = h

12√2 = $<b>Height of paralleogram and triangle</b>$

Area of trapezium = 1/2( a + b )h

Area = 1/2( 42 + 30 ) × 12√2

$<b>$Area = 432√2 m²