Rectangle In Triangle 5Afe12

1. **Problem statement:** We have a right-angled triangle XYZ with sides XY = 8 cm and YZ = 12 cm. Inside it, a rectangle ABCY is drawn such that B lies on XZ and AY = x cm. (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 inside \(\triangle XYZ\), and B lies on XZ, consider \(\triangle XAB\). - Notice that \(\triangle XAB\) is similar to \(\triangle XYZ\) because they share angle X and both have right angles (rectangle corner and right triangle). - Similar triangles have proportional sides: $$\frac{AB}{YZ} = \frac{AY}{XY}$$ - Substitute known lengths: $$\frac{AB}{12} = \frac{x}{8}$$ - Solve for AB: $$AB = \frac{12x}{8} = \frac{3x}{2}$$ 3. **Step (b): Find the maximum area of rectangle ABCY** - The area \(A\) of rectangle ABCY is: $$A = AB \times AY = AB \times x$$ - Substitute \(AB = \frac{3x}{2}\): $$A = \frac{3x}{2} \times x = \frac{3}{2} x^2$$ - However, \(x\) cannot be any value; it must satisfy the constraint that point B lies on XZ. - Find the equation of line XZ: - Coordinates: Let X = (0,0), Y = (8,0), Z = (8,12). - Line XZ passes through X(0,0) and Z(8,12). - Slope of XZ: $$m = \frac{12 - 0}{8 - 0} = \frac{12}{8} = \frac{3}{2}$$ - Equation of XZ: $$y = \frac{3}{2} x$$ - Point B lies on XZ and has coordinates (b_x, b_y) with: - Since B is on rectangle ABCY, its y-coordinate is \(x\) (same as AY), so: $$b_y = x$$ - Using line equation: $$x = \frac{3}{2} b_x \implies b_x = \frac{2}{3} x$$ - Length AB is horizontal distance from A to B: - A is at (0, x), B is at \(\left(\frac{2}{3} x, x\right)\) - So, \(AB = \frac{2}{3} x\) - This contradicts the earlier expression \(AB = \frac{3x}{2}\), so let's re-express carefully. - Actually, the earlier similarity ratio was: $$\frac{AB}{YZ} = \frac{AY}{XY}$$ $$\Rightarrow \frac{AB}{12} = \frac{x}{8}$$ $$\Rightarrow AB = \frac{12x}{8} = \frac{3x}{2}$$ - But from coordinate geometry, \(AB = b_x - 0 = b_x = \frac{2}{3} x\). - The discrepancy arises because the similarity ratio was misapplied. The correct approach is to use coordinate geometry. - So, the length AB is \(\frac{2}{3} x\). - Therefore, area: $$A = AB \times AY = \frac{2}{3} x \times x = \frac{2}{3} x^2$$ 4. **Find the maximum area** - The maximum value of \(x\) is limited by the side XY = 8 cm, so \(0 < x \leq 8\). - Area function: $$A(x) = \frac{2}{3} x^2$$ - Since \(A(x)\) is increasing in \(x\), maximum area occurs at \(x = 8\). - Calculate maximum area: $$A_{max} = \frac{2}{3} \times 8^2 = \frac{2}{3} \times 64 = \frac{128}{3} \approx 42.67 \text{ cm}^2$$ **Final answers:** - (a) \(AB = \frac{2}{3} x\) - (b) Maximum area of rectangle ABCY is \(\frac{128}{3} \text{ cm}^2\) or approximately 42.67 cm².