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Answer :

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

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