Question

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

Answer 1

Join AC of the quadrilateral ABCD such that it gets divided into two triangles that are ΔACD and ΔABC.

Now, area of the triangle is given as:

$A=\frac{1}{2}(x_{1}(y_{2}-y_{3})+(x_{2}(y_{3}-y_{2})+(x_{3}(y_{1}-y_{2}))$

From ΔABC, substituting the values in the formula of the area, we have

$A=\frac{1}{2}((-4)(-5-(-2))+(-3)(-2-(-2))+(3)(-2-(-2)))$

$A=\frac{12+0+9}{2}=\frac{21}{2}sq units$

From  ΔACD, substituting the values in the formula of the area, we have

$A=\frac{1}{2}((-4)(-2-(-3))+3(3-(-2))+2(-2-(-2)))$

$A=\frac{20+15+0}{2}=\frac{35}{2}sq units$

Now, area ABCD=arΔACD +arΔABC

=$\frac{21}{2}+\frac{35}{2}=28 sq units$

Thus, the area of ABCD is 28 square units.

Answer 2

Answer:

formula - ½( x1(y2-y3) + x2(y3-y1) + x3(y1-y2)