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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