Question

Find the perimeter and area of the quadrilateral ABCD in which AB = 17 cm, AD = 9 cm, CD = 12 cm, ∠ACB = 90° and AC = 15 cm. 

   

Answer 1

In the right angled : ΔACB 

`AB^2=BC^2+AC^2` 

⇒ `17^2=BC^2+15^2` 

⇒ `17^2-15^2=BC ^2` 

⇒ `64=BC^2` 

⇒`BC=8 cm` 

Perimeter=`AB+BC+CD+AD` 

=`17+8+12+9` 

= `46 cm^` 

Area of `ΔABC=1/2(bxxh`)  

=`1/2(8xx15)` 

=`60 cm^2` 

In : ΔADC 

`AC^2=AD^2+CD^2` 

So, ΔADC is a right- angled triangle at D 

Area of `ΔADC=1/2xxbxxh` 

                       =`1/2xx9xx12` 

                      =`54 cm^2` 

∴ Area of the quadrilateral  =` Area of ΔABC+Area of ΔADC` 

=`60+54` 

=`114 cm^2`