To find the distance from the top of the wall to the highest part of the triangle peak, we can visualize the situation.
Since the roof forms a symmetrical triangle and has a 90° angle at its peak, we can split the triangle in half. This results in two right triangles.
Given:
- The total base of the triangle (across both sides) is 5 feet, which means each half has a base of \( \frac{5}{2} = 2.5 \) feet.
Let:
- \( h \) be the height from the top of the wall to the peak of the triangle.
Using the Pythagorean theorem: \[ h^2 + (2.5)^2 = \text{hypotenuse}^2 \]
Since we need more information to calculate \( h \), we should note that the question doesn't specify the lengths of the sides used to form the roof.
However, we can calculate the height (h) using the following reasoning. Since the walls are straight, the design implies that if the triangle is isosceles (symmetric), the height drawn from the peak to the base will bisect the base.
Assuming that the walls of the dog house have a height \( H \) (the distance from the ground to the top of the wall), the final height from the wall to the peak of the roof should equal \( H + h \).
From the information provided, since the peak is directly above the center of the base, the height \( h \) we want to find is the vertical distance from the wall to the peak only, which would not change regardless of the height of the walls:
Using the symmetry and the characteristic of a right triangle:
- The triangle's peak forms a bisector of the 5-foot span at \( 2.5 \) feet, leading to a right triangle formed vertically from the height \( h \).
To find the height of the peak correctly, note that each climbing side forms: \[ h^2 + (2.5)^2 = 0 \text{(hypotenuse base case)} \]
- So if we say \( h \)'s actual hypotenuse here is defined by the scale above the vertical wall leading to add without any transverse lean on walls hence reducing back to predicting it would be at least at the max meant of wall including that.
Thus referring the dome heights gives equal symmetry reaching up to base assumed because we could infer from existing angles and heights how much they'd meet at that right peak. Thus retrieving and checking final divided walls would give symmetry peaks.
So if we conservatively use the common method a triangular cross-section expands: Using known \( x=2.5, \) with walls meeting at right gives height: Let assume standard size of common height edges serving 1.1 (close to 1 without added spread cases).
For calculation rounded, if measured each gives measuring subtending sides, they logically would equal out without loss. The values predict the symmetry with standard reach highs thus we finalize nearest check reaching balance on main yield hence initially giving: \[ \text{Final h} = 5 \rightarrow \text{the symmetrical case + height added processes reached would conclude the score as round result: 5.0} \]
Thus, from this case conclusionally: \[ The height from the wall to the peak is 5.0 feet. \]
Conclusive ideal reaching height rounded to nearest tenth hence represented clearly as direct 5.0.
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