Question

Find the area of quadrilateral ABCD in which AB=9cm, BC=40cm,CD= 28cm ,DA=15cm and angle ABC=90°

Answer 1

area (triangle abc)=half*base*height
                             =1%2*9*40
                             =180
area of triangle adc=half*base*height
                               =1%2*28*15
                               21o
area of quadrilateral ABCD=area of triangle abc +area of triangle adc
                                           =180+210
                                           =390

Answer 2

Thank you for asking this question. Here is your answer:

DC² = 28² – 9²

= 703

= 26.51 (DC)

Now we will find the area of ADC

s = ½(26.51+ 28 + 9) = 31.76

Area = √[ 31.76(31.76-26.51)( 31.76-28)( 31.76-9)]

= 119.45

Now we will find the area of ABC

s = ½(28+40+15) = 41.5

Area = √[ 41.5(41.5-28)( 41.5-15)( 41.5-40)]

= 149.23

So the total area is = 119.45 + 149.23

= 268.68 cm²

If there is any confusion please leave a comment below.