Question

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]

Answer 1

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

Area of quadrilateral $ABCD=|Areaof\Delta ABC|+|Areaof\Delta ACD|$

We have,

$\therefore$ Area of $\Delta ABC=\frac{1}{2}|(1\times -3+7\times 2+12\times 1)-(7\times 1+12\times (-3)+1\times 2)|$

$\Rightarrow \Delta ABC=\frac{1}{2}|(-3+14+12)-(7-36+2)|$

$\Rightarrow \Delta ABC=\frac{1}{2}|23+27|=25sq.units$

Also, we have

$\therefore$ Area of $\Delta ACD=\frac{1}{2}|(1\times 2+12\times 21+7\times 1)-(12\times 1+7\times 2+1\times 21)|$

$\Rightarrow \Delta ACD=\frac{1}{2}|(2+252+7)-(12+14+21)|$

$\Rightarrow \Delta ACD=\frac{1}{2}|261-47|=107sq.units$

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