Find the area of the quadrilateral whose vertices taken in order are (-4, -2), (-3, -5 ), (3, -2), and (2,3)........

Let A (-4,-2) B(-3,-5) C(3,-2) and D (2,3)

join BD. then,

area tri. ABD + area tri. BCD = area quad. ABCD.

so, area tri. ABD = 1/2 {-4(-5-3) -3(3+2) +2(-4+3)}

  = 1/2{ 32 -15 -2}

  = 1/2(15) = 15/2  ...(1)

area tri. BCD = 1/2 {-3(-2-3) +3(3+5) +2(-5+2)}

  =1/2 { 15 +24 -6}

  = 1/2 (33) = 33/2  .....(2)

adding (1) and (2) we get,

  area quad. ABCD = 15/2 + 33/2

  = (15 + 33)/2

  = 48/2 = 24..sq units...ans...:)

my teacher said ans is 28......... also if we take diagonal AC then ans is coming as 13.........

the answer wiil come the same, the only thing which you have to keep in mind is that take the coordinates as  area tri.DAC and area tri. ABC.......

 let A =( -4 , -2 )

      B = (-3 , -5 )

      C = ( 3 , -2 )

       D = ( 2 , 3 )

now in triangle ABD

area = 1/2 { 2*(-2 + 5 ) - 4( -5 -3 ) -3( 3 + 2 ) }

            1/2 { 2*3  -  4* (-8)  - 3 * 5 )

            1/2{ 6 + 32 - 15}

            23/2

therefore area of triangle ABD = 23/2 unit square ------------1

now area of triangle BCD

area = 1/2{ -3( -2 -3 ) + 3( 3 + 5 ) + 2( -5 + 2 ) }

            1/2 { -3*-5 + 3*8 + 2*(-3)}

             1/2 { 15 + 24 - 6 }

             33/2

therefore area of triangle BCD = 33/2 unit square ------------- 2

adding 1 & 2

23/2 + 33/2

( 23 + 33 )/2

56 / 2

28

therefore totall area = 28 unit square

thumbs plzzzzzzzzzzzz

 hey vartika 

ur answer is wrong because in area of triangle ABD by mistake u have written 2(-4 + 3) correct point is 2( -2 + 5).......

plzzzz check it........

 thumbs up plzzzzzzzz