Find the area of a quadrilateral ABCD in which AD= 24 cm, BAD = 90° and BCD forms an equilateral triangle whose each side is equal to 26 cm. (Take 
= 1.73)
Let ABCD is a quadrilateral in which AD = 24 cm and ∆BCD is an equilateral.
In right angled ∆BAD applying pythagorous theorem:
(BD)2 = (AB)2 + (AD)2
(26)2 = (AB)2 + (24)2
AB = 10 cm
Area of right angled ∆BAD = 
⇒ 
= 120 cm2
Now in equilateral ∆BCD
Let a, b and c are the sides of triangle and s is
the semi-perimeter, then its area is given by:
A = 
where 
[Heron’s Formula]
= 
= 39
A= 
A = 
= 
cm2
Hence, area of quad ABCD = 120+292.72 =412.72 cm2
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