We are given a square $ABCD$, and points $P$ and $Q$ on sides $BC$ and $CD$, respectively, such that $BP = CQ = \frac{1}{3} AB$. We are also told that the area of triangle $BRP$ is 10 cm², and we need to find the area of square $ABCD$.

Step-by-Step Solution:

1. Understanding the geometry:

   • Let the side length of square $ABCD$ be $s$.
   • Since $BP = \frac{1}{3} AB$, we know that $BP = \frac{1}{3} s$, and similarly, $CQ = \frac{1}{3} s$.

   Now, we place the square on a coordinate plane with the following coordinates:

   • $A(0, 0)$
   • $B(s, 0)$
   • $C(s, s)$
   • $D(0, s)$

2. Coordinates of points $P$ and $Q$:

   • Point $P$ lies on side $BC$, and $BP = \frac{1}{3} s$, so the coordinates of $P$ are $(s, \frac{s}{3})$.
   • Point $Q$ lies on side $CD$, and $CQ = \frac{1}{3} s$, so the coordinates of $Q$ are $\left( \frac{s}{3}, s \right)$.

3. Equation of line $AP$: The line $AP$ passes through points $A(0, 0)$ and $P(s, \frac{s}{3})$. The slope of $AP$ is: $\text{slope of } AP = \frac{\frac{s}{3} - 0}{s - 0} = \frac{1}{3}$ The equation of line $AP$ is: $y = \frac{1}{3}x$

4. Equation of line $PQ$: The line $PQ$ passes through points $P(s, \frac{s}{3})$ and $Q\left( \frac{s}{3}, s \right)$. The slope of $PQ$ is: $\text{slope of } PQ = \frac{s - \frac{s}{3}}{\frac{s}{3} - s} = \frac{\frac{2s}{3}}{-\frac{2s}{3}} = -1$ The equation of line $PQ$ is: $y - \frac{s}{3} = -1(x - s)$ Simplifying this: $y = -x + \frac{4s}{3}$

5. Finding the coordinates of point $R$: Point $R$ is the intersection of lines $AP$ and $PQ$. We solve the system of equations: $y = \frac{1}{3}x \quad \text{and} \quad y = -x + \frac{4s}{3}$ Setting the two equations equal: $\frac{1}{3}x = -x + \frac{4s}{3}$ Solving for $x$: $\frac{1}{3}x + x = \frac{4s}{3}$ $\frac{4}{3}x = \frac{4s}{3}$ $x = s$ Substituting $x = s$ into $y = \frac{1}{3}x$: $y = \frac{1}{3}s$ So, the coordinates of $R$ are $(s, \frac{s}{3})$.

6. Area of triangle $BRP$: The area of triangle $BRP$ can be calculated using the formula for the area of a triangle with vertices at $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$: $\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|$ For triangle $BRP$, the coordinates of $B$, $R$, and $P$ are:

   • $B(s, 0)$
   • $R(s, \frac{s}{3})$
   • $P(s, \frac{s}{3})$

   Substituting these into the area formula: $\text{Area} = \frac{1}{2} \left| s\left(\frac{s}{3} - \frac{s}{3}\right) + s\left(\frac{s}{3} - 0\right) + s\left(0 - \frac{s}{3}\right) \right|$ Simplifying: $\text{Area} = \frac{1}{2} \left| s(0) + s\left(\frac{s}{3}\right) + s\left(-\frac{s}{3}\right) \right|$ $\text{Area} = \frac{1}{2} \left| s\left(\frac{s}{3} - \frac{s}{3}\right) \right|$ $\text{Area} = \frac{1}{2} \left| 0 \right|$ Thus, the area is