Find 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 right angle .

Explanation

Let ABCD be the given quadrilateral in which AB= 9 cm , BC=40 cm ,CD=28 cm DA= 15 cm and `angle ABC = 90^(@)`.
By pythagoras' theorem ,
`AC=sqrt(AB^(2)+BC^(2)) "units"=sqrt((9)^(2)+(40)^(2)) cm `
`= sqrt( 1681) cm = 41 cm.`
` "Area of " Delta ABC =((1)/(2) xxABxxBC)`
`=((1)/(2) xx9xx40 ) cm^(2) = 180 cm^(2).`
`"In" Delta ACD ,`
Let `a=AC=41 cm , b= CD = 28 cm and c=DA =15 cm.`
`therefore s= (1)/(2) (41+28 +15) cm = 42 cm.`
` therefore (s-a) =1cm ,(s-b) = 14cm and (s-c) = 27 cm.`
Area of `Delta ACD =sqrt(s(s-a)(s-a)(s-c))"sq units"`
` = sqrt( 42xx1xx14xx27)cm^(2) `
`=(14xx3xx3)cm^(2)=126 cm^(2)`.
Area of the quadrilateral ABCD
`="area of " Delta ABC +"Area of "Delta ACD`
`=(180 +126) cm ^(2) = 306 cm^(2) `.