Question

Calculate the area of quadrilateral ABCD in which AB = 32 cm, AD = 24 cm ∠A = 90° and BC = CD = 52 cm.

Answer 1

The figure can be drawn as follows :

Here ABD is a right triangle. So the area will be :

ΔABD = `1/2(24) (32)`
          = 384
Again

BD = `sqrt(24^2 + 32^2)`

     =  `8sqrt(3^2 + 4^2)`

     =  8 ( 5 )

     = 40

Now BCD is an isosceles triangle and BP is perpendicular to BD, therefore

DP =`1/2"BD"`

     = `1/2(40)`

     = 20

From the right triangle DPC we have

PC = sqrt(52^2 - 20^2)` 

     = `4sqrt(13^2 - 5^2)`
     = 4(12)
     = 48

So
ΔDPC =`1/2 (40) (48)`
          = 960

Hence the area of the quadrilateral will be :
ΔABD + ΔDPC = 960 + 384
                        = 1344 cm²