Question

37) If A (-5,7), B(-4,5),C(-1,-6) and (4,5) and the vertices of a quadrilateral.
Find the area of the quadrilateral. L-
17​

Answer 1

Answer:

Let the vertices of the quadrilateral be A(−5,7),B(−4,−5),C(−1,−6),D(4,5).

Joining AC there are two triangles ABC,ADC

Therefore area of quadrilateral ABCD=Area of ΔABC+Area of ΔADC

Area of ΔABC= 1/2

[x1 (y2−y3 )+x 2 (y3−y1 )+x3 (y1−y 2)]

Here x1=−5,

y1=7,

x2 =−4,

y2=−5,

x3 =−1,

y3 =−6

Area of ΔABC= 1/2 [−5(−5+6)−4(−6−7)−1(7+5)]

=1/2(35) square units.

Area of ΔADC= 1/2 [x1 (y2−y3 )+x2(y3 −y1 )+x 3 (y1 −y2 )]

Here x1=−5,

y1 =7,

x2 =4,

y2=5,

x3 =−1,

y3 =−6

Area of ΔADC= 1/2 [−5(5+6)+4(−6−7)−1(7−5)]=1/2(−109)= 1/2(109) square units. (since area cannot be negative)

Therefore area of quadrilateral ABCD=Area of ΔABC+Area of ΔADC

= 1/2 (35)+1/2(109)

=1/2 (144)

=72

Therefore area of quadrilateral ABCD=72 square units.