To find the dimensions $x$ (width) and $y$ (height) of the largest rectangle inscribed in a right triangle, we need to maximize the area of the rectangle, constrained by the triangle's geometry. Here's the detailed solution:

Given:

• Base ($b$) = $35'3''$
• Height ($h$) = $118'0''$
• Hypotenuse ($c$) = $120'6''$

Convert all measurements into inches:

1. $b = 35 \times 12 + 3 = 423 \, \text{inches}$
2. $h = 118 \times 12 = 1416 \, \text{inches}$
3. $c = 120 \times 12 + 6 = 1446 \, \text{inches}$

Geometry:

A rectangle inscribed in the right triangle will have its base ($x$) along the base of the triangle and its height ($y$) along the height of the triangle. The top corner of the rectangle will lie on the hypotenuse. The relationship between $x$ and $y$ is determined by the line equation of the hypotenuse.

Hypotenuse Equation:

The hypotenuse can be described as: $\frac{x}{b} + \frac{y}{h} = 1$ Substitute $b = 423$ and $h = 1416$: $\frac{x}{423} + \frac{y}{1416} = 1$

Solve for $y$ in terms of $x$: $y = 1416 \left(1 - \frac{x}{423}\right)$

Maximize Rectangle Area:

The area of the rectangle ($A$) is: $A = x \cdot y$ Substitute $y = 1416 \left(1 - \frac{x}{423}\right)$: $A = x \cdot 1416 \left(1 - \frac{x}{423}\right)$ $A = 1416x - \frac{1416x^2}{423}$ $A = 1416x - \frac{4x^2}{1}$

Take Derivative:

To maximize $A$, set $\frac{dA}{dx} = 0$: $\frac{dA}{dx} = 1416 - \frac{8x}{1} = 0$ $1416 = 8x$ $x = \frac{1416}{8} = 177 \, \text{inches}$

Substitute $x = 177$ into the equation for $y$: $y = 1416 \left(1 - \frac{177}{423}\right)$ $y = 1416 \left(1 - 0.4187\right)$ $y = 1416 \cdot 0.5813 \approx 823.5 \, \text{inches}$

Final Dimensions:

• $x = 177 \, \text{inches}$
• $y = 823.5 \, \text{inches}$

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Would you like details on how to verify this or any further clarifications?

Related Questions:

1. What is the significance of maximizing the area of the rectangle?
2. Can this method be applied to non-right triangles?
3. How does changing the hypotenuse affect the dimensions of the rectangle?
4. What is the relationship between the area of the rectangle and the triangle's area?
5. How do these results change if the dimensions are given in different units?

Tip:

Always convert measurements to a consistent unit before calculations to avoid errors.