The sides of a quadrilateral ABCD are 6cm,8cm,12cm and 14cm (taken in order).
i.e. ∠ABC=90o
Now, let us join the points A & C.
So, we get two triangles namely △ABC (a right angled triangle) and △ACD.
Applying Pythagoras theorem in △ABC, we get
AC=√AB2+BC2=√62+82=√36+64=√100=10cm
So, the area of △ABC=12×base×height
=12×AB×BC
=12×6×8=24
Now, in △ACD, we have
AC=10cm,CD=12cm,AD=14cm.
According to Heron's formula the area of triangle (A)=√[s(s−a)(s−b)(s−c)]
where, 2s=(a+b+c).
Here, a=10cm,b=12cm,c=14cm
s=(10+12+14)2=362=18
Area of △ACD=√[18×(18−10)(18−12)(18−14)]
=√(18×8×6×4)
=√(2×3×3×2×2×2×2×3×2×2)
=√[(2×2×2×2×2×2×3×3)×2×3]
=2×2×2×3×√6
=24√6
So, total area of quadrilateral ABCD =△ABC+△ACD
=24+24√6
=24(√6+1)
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