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).

Answer 1

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

We have,

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

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

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

Also, we have

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

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

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

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