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

Hi!

Cheers!

• 89

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...:)

• -4

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

• 0

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.......

• 9

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

23/2 + 33/2

( 23 + 33 )/2

56 / 2

28

therefore totall area = 28 unit square

thumbs plzzzzzzzzzzzz

• 33

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........

• 10

thumbs up plzzzzzzzz

• 6

o yeah...its by mistake...

• -4
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