The area of the quadrilateral with vertices \(A(1, 1)\), \(B(5, 2)\), \(C(6, 6)\) and \(D(2, 5)\), taken in this order, is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 15 square units
The diagonal AC divides the quadrilateral into the two triangles ABC and ACD, and the area is the sum of their areas by the determinant formula.
\(\displaystyle \Delta_{1} = \frac{1}{2}\begin{vmatrix}1 & 1 & 1\\ 5 & 2 & 1\\ 6 & 6 & 1\end{vmatrix}\)
\(\displaystyle \Delta_{1} = \frac{1}{2}\left[1(2 - 6) - 1(5 - 6) + 1(30 - 12)\right]\)
\(\displaystyle \Delta_{1} = \frac{1}{2}(-4 + 1 + 18) = \frac{15}{2}\)
\(\displaystyle \Delta_{2} = \frac{1}{2}\begin{vmatrix}1 & 1 & 1\\ 6 & 6 & 1\\ 2 & 5 & 1\end{vmatrix}\)
\(\displaystyle \Delta_{2} = \frac{1}{2}\left[1(6 - 5) - 1(6 - 2) + 1(30 - 12)\right]\)
\(\displaystyle \Delta_{2} = \frac{1}{2}(1 - 4 + 18) = \frac{15}{2}\)
\(\displaystyle \text{Area} = \frac{15}{2} + \frac{15}{2} = 15\)
Why the other options are incorrect
Option (1) 7.5 square units — Incorrect. This is the area of only one of the two triangles.
Option (3) 17 square units — Incorrect. The two triangles have areas 7.5 and 7.5, which add to 15.
Option (4) 30 square units — Incorrect. This adds the two determinants without the factor \(\frac{1}{2}\).