### Related Formula textArea of a quadrilateral with perpendicular diagonals d_1 text and d_2 = frac12 d_1 d_2$\text{Area of a quadrilateral with perpendicular diagonals } d_1 \text{ and } d_2 = \frac{1}{2} d_1 d_2$ ### Core Logic The line perpendicular to chord AB$AB$ passing through its midpoint is the diameter of the circle because the perpendicular bisector of any chord passes through the center. Thus, CD$CD$ is a diameter, making its length equal to 2R = 4$2R = 4$.

### Step 1: Find Length of Chord AB The perpendicular distance p$p$ from the center (0,0)$(0,0)$ to the line x + y - 1 = 0$x + y - 1 = 0$ is: p = frac|0 + 0 - 1|sqrt1^2 + 1^2 = frac1sqrt2$p = \frac{|0 + 0 - 1|}{\sqrt{1^2 + 1^2}} = \frac{1}{\sqrt{2}}$ Length of chord AB = 2sqrtR^2 - p^2 = 2sqrt4 - frac12 = 2sqrtfrac72 = sqrt14$AB = 2\sqrt{R^2 - p^2} = 2\sqrt{4 - \frac{1}{2}} = 2\sqrt{\frac{7}{2}} = \sqrt{14}$. ### Step 2: Area Calculation Since CD$CD$ is the perpendicular bisector of AB$AB$, the diagonals of quadrilateral ADBC$ADBC$ are perpendicular. Therefore, the area is: textArea = frac12 times AB times CD = frac12 times sqrt14 times 4 = 2sqrt14$\text{Area} = \frac{1}{2} \times AB \times CD = \frac{1}{2} \times \sqrt{14} \times 4 = 2\sqrt{14}$ ### Pattern Recognition Whenever a line passes through the midpoint of a chord and is perpendicular to it, recognize instantly that it is a diameter. The area of the quadrilateral formed is then simply the area of two triangles sharing the diameter as a base, or frac12 d_1 d_2$\frac{1}{2} d_1 d_2$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Circles Class 11 Mathematics: Straight Lines