Question

find the area of a quadrilateral ABCD in which AB=3cm,BC=4cm,CD=4cm,DA=5cm and AC=5cm.​

Answer 1

2S=5+4+3 ⇒ S=6

Area of ΔABC

=  

s(s−a)(s−b)(s−c)

​  

 

=  

15(15−9)(15−8)(15−13)

​  

=  

6(1)(2)(3)

​  

=  

36

​  

=6

⇒For area of ΔADC,  

⇒2S=5+4+5  

⇒ S=7cm

⇒Area of ΔADC

=  

7(7−5)(7−5)(7−4)

​  

=  

7×2×2×3

​  

 

=2  

21

​  

=9.16m  

2

 

⇒Area of quad ABCD= Area of (△ABC+△ADC)

=6+9.16

=15.16  cm  

2

≈15.2  cm  

2