M and N are the midpoints of AB and CD, respectively, of a square ABCD whose side is 12 cm. Take a point P on MN and let AP = r cm and PC = s cm. The area of the triangle whose sides are r, s, 12 cm is

Correct option is A

Given:
ABCD is a square with side = 12 cm
M and N are the midpoints of sides AB and CD respectively
Point P lies on line MN
AP = r cm, PC = s cm
AC = 12 cm (diagonal of the square)
We are to find the area of triangle APC, whose sides are r, s, and 12 cm

Formula Used:

$$\\[1em]\textbf{} \\\text{Area of triangle with sides } a, b, c \text{ is:} \\\text{Area} = \frac{1}{4} \sqrt{(a + b + c)(a + b - c)(a + c - b)(b + c - a)}\\[1em]\textbf{Solution:} \\\text{Let } a = r, \quad b = s, \quad c = 12 \\\text{Area} = \frac{1}{4} \sqrt{(r + s + 12)(r + s - 12)(r + 12 - s)(s + 12 - r)}$$$$\\[1em]\textbf{} \\\text{Area of triangle with sides } a, b, c \text{ is:} \\\text{Area} = \frac{1}{4} \sqrt{(a + b + c)(a + b - c)(a + c - b)(b + c - a)}\\[1em]\textbf{Solution:} \\\text{Let } a = r, \quad b = s, \quad c = 12 \\\text{Area} = \frac{1}{4} \sqrt{(r + s + 12)(r + s - 12)(r + 12 - s)(s + 12 - r)}$$

​$$\text{Let the points be:} \\\quad \bullet \; A = (0, 0) \\\quad \bullet \; C = (12, 12) \\\quad \bullet \; P = (6, y)\\[1em]\text{Use coordinate-based area formula:} \\\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|\\[1em]\text{Plug in the values:} \\\text{Area} = \frac{1}{2} |0(12 - y) + 12(y - 0) + 6(0 - 12)| \\\ \\= \frac{1}{2} |12y - 72| = \frac{1}{2}(12y - 72) \quad \text{(since } y \geq 6\text{)}\\[1em]\text{Try } y = 12: \\\text{Area} = \frac{1}{2} (144 - 72)\\\ \\ = \frac{1}{2} \times 72 = 36 \; \text{cm}^2\\[1em]\text{So, the \textbf{maximum possible area} of triangle APC is } \boxed{36 \; \text{cm}^2}, \text{ and this happens when \textbf{P is at point N}.}$$​

Final Answer: (A) 36 cm²