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

Draw a line from B to D

Now we have,
Area of quad. ABCD = | Area of ∆ DAB | + | Area of ∆ BCD |

So,
Area of ∆ DAB = $$| \ \frac{1}{2} [2(-2+5) \ - \ 4(-5-3) \ - \ 3(3+2)] \ |$$ = $$\ | \frac{1}{2} [6 + 32 - 15 ] \ |$$ = $$\frac{23}{2}$$ sq. units

Similarly, Area of ∆ BCD = $$| \ \frac{1}{2} [-3(-2-3) \ + \ 3(3+5) \ + \ 2(-5+2)] \ | \ = \ | \ \frac{1}{2} [15 + 24 - 6 ] \ |$$ = $$\frac{33}{2}$$ sq. units

$$\therefore$$ Area of quad. ABCD = $$\frac{23}{2} \ + \ \frac{33}{2}$$ = $$\frac{56}{2} \ = \ 28$$ sq. units