Question

Find the area of the quadrilateral ABCD in which AD = 24 cm, ∠BAD 90° and ∠BCD  is an equilateral triangle having each side equal to 26 cm. Also, find the perimeter of the quadrilateral.  

 

Answer 1

ΔBDC is an equilateral triangle with side `a=26` cm    

Area of ΔBDC=` sqrt3/4 a^2` 

=`sqrt3/4xx26^2` 

=`1.73/4xx676` 

=`292.37  cm^2` 

By using Pythagoras theorem in the right – angled triangle `Δ DAB, `We get

` AD^2+AB^2=BD^2` 

⇒` 24^2+AB^2=26^2` 

⇒`AB^2=26^2-24^2`  

⇒`AB^2=676-576`  

⇒ `AB^2=100` 

⇒`AB=10 cm`  

Area of the `Δ ABD=1/2 xxbxxh` 

=`1/2xx10xx24` 

=`120 cm^2`  

Area of the quadrilateral 

= Area of the quadrilateral 

                 = Are of ΔBCD+Area of  ΔABD

                 =`292.37+120` 

                =`412.37 cm^2` 

Perimeter of the quadrilateral 

                            =AB+AC+CD+AD

                           = `24+10+26+26` 

                          =86 cm