Question

a trapezium shaped field whose parallel sides are 42m and 30m and other sides are 18m and18m find its area​

Answer 1

A Trapezium shaped field whose parallel sides are 42 m, 30m. Other sides are 18, 18 m. Finding the area,

We know that,

Area of trapezium is given by

$\bold{A = \frac{1}{2} h(a +b)}$

Now,

$\textbf{The Half of Difference } \\ \textbf{of the parallel sides,} \: \: \: \: \: \\ \textbf{Other sides and Height } \\ \textbf{ of \: the \: trapezium form} \\ \textbf{ a right angled triangle. }$

So, 42 - 30 = 12

Half of twelve = 6.

Now, Using Pythagoras theorem,

6² + h² = 18²

h² = 18² - 6² = 324 - 36= 288

h = √288 = 12 √2

Now,

$Area = \frac{1}{2} h ( a + b) \\ \\ A= \frac{1}{2} (12 \sqrt{2} )(42 + 30) \\ \: \: \: \: \: = 6 \sqrt{2} (72) = 432 \sqrt{2}$

Therefore, The area of this trapezium is $432 \sqrt{2}$ 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