1. **Problem statement:** We have a right-angled triangle XYZ with legs XY = 8 cm and YZ = 12 cm. Inside it, a rectangle ABCY is drawn such that AY = x cm and B lies on the hypotenuse XZ. (a) We need to find a triangle similar to $\triangle XAB$ and then express the length AB in terms of $x$. (b) We want to find the maximum area of rectangle ABCY. 2. **Step (a): Find a similar triangle and express AB in terms of x** - Since ABCY is a rectangle, $\angle AYB = 90^\circ$. - $\triangle XAB$ shares angle $\angle XAB$ with $\triangle XYZ$. - Both triangles have a right angle: $\angle XAB = 90^\circ$ and $\angle XYZ = 90^\circ$. Therefore, $\triangle XAB \sim \triangle XYZ$ by AA similarity (two angles equal). 3. **Using similarity ratios:** - Corresponding sides are proportional: $$\frac{AB}{YZ} = \frac{AX}{XY} = \frac{XB}{XZ}$$ - We know: - $XY = 8$ cm - $YZ = 12$ cm - $AY = x$ cm - Since ABCY is a rectangle, $AY$ is along XY, so: $$AX = XY - AY = 8 - x$$ - Using the ratio: $$\frac{AB}{12} = \frac{8 - x}{8} \implies AB = 12 \times \frac{8 - x}{8} = 12 \times \left(1 - \frac{x}{8}\right) = 12 - \frac{12x}{8} = 12 - \frac{3x}{2}$$ 4. **Step (b): Find the maximum area of rectangle ABCY** - The area $A$ of rectangle ABCY is: $$A = AY \times AB = x \times \left(12 - \frac{3x}{2}\right) = 12x - \frac{3x^2}{2}$$ - To find the maximum area, differentiate $A$ with respect to $x$ and set to zero: $$\frac{dA}{dx} = 12 - 3x = 0 \implies 3x = 12 \implies x = 4$$ - Check the second derivative: $$\frac{d^2A}{dx^2} = -3 < 0$$ which confirms a maximum at $x = 4$. - Substitute $x = 4$ back to find maximum area: $$A_{max} = 12(4) - \frac{3(4)^2}{2} = 48 - \frac{3 \times 16}{2} = 48 - 24 = 24$$ **Final answers:** (a) $AB = 12 - \frac{3x}{2}$ cm (b) Maximum area of rectangle ABCY is $24$ cm$^2$.