Question

Find the area of the quadrilateral abcd, whose vertices are(-3,-1),B(-2,-4),C(4,-1) ANDD(3,4)

Answer 1

Area of quadrilateral ABCD can be easily found by finding the areas of Δ ABC and Δ ACD .

Another method exists and we can use that method also. We can plot the points in graph and then find the area of the triangles .

But I will do in the hardest way possible .

Area of Δ ABC

$Area\:of\:triangle=\frac{1}{2}[x1(y2-y3) -x2(y3-y1) -x3(y1-y2)]\\\\\implies \frac{1}{2}\times [(-3)(-4+1)+(-2)(-1+1)+(4)(-1+4)]\\\\\implies \frac{1}{2}\times [-3\times -3+ (-2\times 0)+4\times 3]\\\\\implies \frac{1}{2}\times [9+0+12]\\\\\implies \frac{1}{2}\times 21\\\\\implies \frac{21}{2}$

Area of Δ ACD

$Area\:of\:triangle=\frac{1}{2}[x1(y2-y3) -x2(y3-y1) -x3(y1-y2)]\\\\\implies \frac{1}{2}[(-3)(4+1)+3(-1+1)+4(-1-4)]\\\\\implies \frac{1}{2}[-3\times 5+3\times 0 +4\times(-5)]\\\\\implies \frac{1}{2}\times [-15+0-20]\\\\\implies \frac{1}{2}\times 35\\\\\implies \frac{35}{2}$

Area of ABCD

Area = area of Δ ABC + area of Δ ACD

$\implies \frac{21}{2}+\frac{35}{2}\\\\\implies \frac{21+35}{2}\\\\\implies \frac{56}{2}\\\\\implies 28$

$\mathfrak{\underline{ANSWER}}\\\\\texttt{The area of the quadrilateral ABCD is}\:\mathsf{28\:units^2}$