Given ABCD is a quadrilateral having sides AB=6cm, BC=8cm, CD=12cm and DA=14 cm. Now. Join AC. We have, ABC is a right angled triangle at B. Now, AC2=AB2+BC2 [by Pythagoras theorem]Now, AC2=AB2+BC2 [by Pythagoras theorem] =62+82=36+64=100=62+82=36+64=100 ⇒ AC=10cm [taking positive square root]⇒ AC=10cm [taking positive square root] ∴Area of quadrilateral ABCD=Area of △ABC+Area of △ACD∴Area of quadrilateral ABCD=Area of △ABC+Area of △ACD Now, area of △ABC=12×AB×BC [∵area of triangle=12(base×height)]Now, area of △ABC=12×AB×BC [∵area of triangle=12(base×height)] =12×6×8=24cm2=12×6×8=24cm2 In △ACD, AC=a=10cm,CD=b=12cmIn △ACD, AC=a=10cm,CD=b=12cm and DA=c=14cmand DA=c=14cm Now,semi-perimeter of △ACD,s=a+b+c2=10+12+142=362=18cmNow,semi-perimeter of △ACD,s=a+b+c2=10+12+142=362=18cm Area of △ACD=s(s−a)(s−b)(s−c)−−−−−−−−−−−−−−−−−√ [by Heron's formula]Area of △ACD=s(s-a)(s-b)(s-c) [by Heron's formula] =18(18−10)(18−12)(18−14)−−−−−−−−−−−−−−−−−−−−−−−−√=18(18-10)(18-12)(18-14) =18×8×6×4−−−−−−−−−−−−√=(3)2×2×4×2×3×2×4−−−−−−−−−−−−−−−−−−−−−−−√=18×8×6×4=(3)2×2×4×2×3×2×4 =3×4×23–√×2=246–√cm2=3×4×23×2=246cm2 From Eq. (i) Area of quadrilateralABCD=Area of △ABC+Area of △ACDArea of quadrilateralABCD=Area of △ABC+Area of △ACD =24+246–√=24(1+6–√)cm2=24+246=24(1+6)cm2 Hence, the area of quadrilateral is 241+6–√−−−−−−√cm2241+6cm2 .