1. **Problem statement:** We have a square divided into 4 parts: triangles A and B, and trapeziums C and D. Given: - Area ratio of A to B is $13:1$. - Area ratio of C to D is $3:1$. We need to find the ratio of the area of C to the area of the whole square in simplest form. 2. **Understanding the problem:** - The square's total area is the sum of areas of A, B, C, and D. - Let areas be $A$, $B$, $C$, and $D$ respectively. 3. **Expressing areas using ratios:** - Since $A:B=13:1$, let $A=13x$ and $B=x$ for some $x>0$. - Since $C:D=3:1$, let $C=3y$ and $D=y$ for some $y>0$. 4. **Total area of the square:** $$\text{Square area} = A + B + C + D = 13x + x + 3y + y = 14x + 4y$$ 5. **Relating areas of triangles and trapeziums:** - The square is divided into two halves vertically or horizontally (implied by the problem). - The left half contains triangles A and B, so their combined area is $14x$. - The right half contains trapeziums C and D, so their combined area is $4y$. 6. **Since the square is divided into two halves, the sum of areas on each side must be equal:** $$14x = 4y$$ 7. **Solve for $y$ in terms of $x$:** $$y = \frac{14x}{4} = \frac{7x}{2}$$ 8. **Find the ratio of $C$ to the square:** $$\frac{C}{\text{Square}} = \frac{3y}{14x + 4y}$$ Substitute $y = \frac{7x}{2}$: $$\frac{C}{\text{Square}} = \frac{3 \times \frac{7x}{2}}{14x + 4 \times \frac{7x}{2}} = \frac{\frac{21x}{2}}{14x + 14x} = \frac{\frac{21x}{2}}{28x}$$ 9. **Simplify the fraction:** $$\frac{21x/2}{28x} = \frac{21}{2} \times \frac{1}{28} = \frac{21}{56} = \frac{3}{8}$$ 10. **Final answer:** The ratio of the area of trapezium C to the area of the square is $\boxed{3:8}$ in simplest form.