Question

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

Answer 1

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

 

Answer 2

Answer:

Step-by-step explanation:

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