Question

the roof of a house is the shape of an isosceles right triangle as shown in the diagram below. what is the height of the roof, h? 5 ft 5√2 ft 5√3 ft 5√2/2 ft

Explanation:

Step1: Analyze the isosceles right - triangle properties

In an isosceles right - triangle, the height \(h\) to the hypotenuse divides the triangle into two smaller isosceles right - triangles. Let the hypotenuse of the large isosceles right - triangle be \(c\). For an isosceles right - triangle with leg length \(a = 10\) ft, using the Pythagorean theorem \(c=\sqrt{a^{2}+a^{2}}=\sqrt{10^{2}+10^{2}}=\sqrt{200} = 10\sqrt{2}\) ft. But we can also use the property of the height in an isosceles right - triangle. Another way: consider one of the smaller right - triangles formed by the height \(h\). The hypotenuse of the smaller right - triangle is \(10\) ft (given).

Step2: Apply the Pythagorean theorem to the smaller right - triangle

Let the legs of the smaller isosceles right - triangle (since the height in an isosceles right - triangle creates two smaller isosceles right - triangles) be \(h\) (height) and the base segment. In an isosceles right - triangle with hypotenuse \(c = 10\) ft, and using the relationship \(c=\sqrt{h^{2}+h^{2}}\) (by Pythagorean theorem \(a^{2}+b^{2}=c^{2}\), here \(a = b=h\)). So \(10=\sqrt{2h^{2}}\), which simplifies to \(10 = h\sqrt{2}\).

Step3: Solve for \(h\)

From \(10 = h\sqrt{2}\), we can solve for \(h\) by \(h=\frac{10}{\sqrt{2}}\). Rationalizing the denominator: \(h=\frac{10\sqrt{2}}{2}=5\sqrt{2}\) ft.

Answer:

\(5\sqrt{2}\text{ ft}\) (the second option)