Question

In the given figure, ABCD is a field in the form of a quadrilateral whose sides are
indicated in the figure. If angle DAB = 90°,
find the area of the field.​

Answer 1

=ar△ABD+ar△BDC

$=ar\triangle ABD+ar\triangle BDC$

$AB×AD= </p><p>2</p><p>1</p><p> </p><p> ×40×9$

=180㎡

$Again in △ABD,BD= </p><p>(AB) </p><p>2</p><p> +(AD) </p><p>2</p><p> </p><p> </p><p> </p><p>$

(Using Pythagoras theorem)

$= </p><p>(40) </p><p>2</p><p> +(9) </p><p>2</p><p> </p><p> </p><p> = </p><p>1681</p><p> </p><p> =41㎝$

$Now, ar△BDC= </p><p>s(s−a)(s−b)(s−c)</p><p> </p><p> ,s=$

$2</p><p>a+b+c</p><p> </p><p> = </p><p>2</p><p>41+15+28</p><p> </p><p> =42㎝$

$∴ar△BDC=$

$42(42−4)(42−15)(42−28)</p><p> </p><p> =$

$42×1×27×14</p><p> </p><p> </p><p>=$

$14×3×9×3×14</p><p> </p><p> =3×3×14=126㎡$

∴Area of quadrilateral ABCD=180+126=306㎡