Question

The area of the quadrilateral whose sides measure 9 cm, 40 cm, 28 cm and 15 cm and in which the angle between the first two sides is a right angle is

Answer 1

The area of quadrilateral is equal to sum of areas of two triangles in which it's diagonals divide it.

So, area =0.5×40×9+√s(s-a)(s-b)(s-c)

a=√{40²+9²}=41
b=28 and c=15
And s=(41+15+28)/2

Answer 2

AB= 9

BC= 40

CD= 28

DA= 15

 join A and C to form a diagonal AC in quadrilateral ABCD

using pythagorean theorem,

AB ^(2) + BC ^(2) = AC ^(2)

9*9 + 40*40 = AC ^(2)

81 + 1600 = AC ^(2)

1681 = AC^(2)

= AC

41 = AC

now, using herons formula, u can find the area of triangle ABC an d triangle ADC.

in triangle ABC

=

                                                  = 

                                                  = 

                                                  = 

                                                  = 5*3*2*3*2

                                                  = 5*36

                                                  = 180

in triangle ADC

=

                                                  =

                                                  =

                                                  =

                                                  =2*7*3*3*

                                                  = 14*9

                                                  = 126

area of ABC + area of ADC = area of ABCD

180 + 126 = 306

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