Find the area of the quadrilateral ABCD whose vertices are respectively A (1, 1), B (7, –3) C (12, 2) and D (7,21). [4 MARKS]

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Solution

Concept : 1 Mark
Application : 1 Mark
Calculation : 2 Marks

Area of quadrilateral $A B C D = | A r e a o f Δ A B C | + | A r e a o f Δ A C D |$

We have,

$∴$ Area of $Δ A B C = \frac{1}{2} | (1 \times - 3 + 7 \times 2 + 12 \times 1) - (7 \times 1 + 12 \times (- 3) + 1 \times 2) |$

$\Rightarrow Δ A B C = \frac{1}{2} | (- 3 + 14 + 12) - (7 - 36 + 2) |$

$\Rightarrow Δ A B C = \frac{1}{2} | 23 + 27 | = 25 s q . u n i t s$

Also, we have

$∴$ Area of $Δ A C D = \frac{1}{2} | (1 \times 2 + 12 \times 21 + 7 \times 1) - (12 \times 1 + 7 \times 2 + 1 \times 21) |$

$\Rightarrow Δ A C D = \frac{1}{2} | (2 + 252 + 7) - (12 + 14 + 21) |$

$\Rightarrow Δ A C D = \frac{1}{2} | 261 - 47 | = 107 s q . u n i t s$

$∴$ Area of quadrilateral ABCD = 25 + 107 = 132 sq. units

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