Question

Find the area of the quadrilateral whose vertices are(2,4),(3,5),(-3,4),&(-4,-3).​

Answer 1

Answer:

28 unit square

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

therefore totall area = 28 unit square

Answer 2

Answer:

Let A(–4,–2), B(–3,–5), C(3,–2) and D(2,3) be the vertices of the quadrilateral ABCD.

Area of a quadrilateral ABCD= Area of △ABC+ Area of △ACD

By using a formula for the area of a triangle =

2

1

∣x

1

(y

2

−y

3

)+x

2

(y

3

−y

2

)+x

3

(y

1

−y

2

)∣

Area of △ABC

=

2

1

[−4(−5+2)+−3(−2+2)+3(−2+5)]

=

2

1

[12+9]

=

2

21

sq.units

Area of △ACD=

2

1

[−4(3+2)+−2(−2+2)+3(−2−3)]

=

2

1

[−20−15]

=

2

35

sq.units

∴Area of quadilateral=

2

21

+

2

35

=

2

56

=28 sq.unit