#### Question

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 √3 = 1.73)

#### Solution

Given that, a quadrilateral ABCD in which AD = 24 cm, ∠BAD = 90°

BCD is equilateral triangle and sides BC = CD = BD = 26 cm

In ΔBAD By using Pythagoras theorem

`BA^2=BD^2-AD^2`

`⇒BA=sqrt(BD^2-AD^2)`

`=sqrt(676-576)`

`sqrt(100)=10cm`

𝐴𝑟𝑒𝑎 𝑜𝑓 Δ𝐵𝐴𝐷=`1/2`×𝐵𝐴×𝐴𝐷

=`1/2`×10×24

=`120cm^2`

𝐴𝑟𝑒𝑎 𝑜𝑓 Δ𝐵𝐶𝐷=`sqrt(a)/4xx(26)^2=292.37cm^2`

∴𝐴𝑟𝑒𝑎 𝑜𝑓 𝑞𝑢𝑎𝑑𝑟𝑖𝑙𝑎𝑡𝑒𝑟𝑎𝑙

ABCD = Area of ΔBAD + area of ΔBCD

= 120 + 292.37

= 412.37 `cm^2`

Is there an error in this question or solution?

Solution 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 √3 = 1.73) Concept: Application of Heron’S Formula in Finding Areas of Quadrilaterals.