Question

x = ______6)
y = ______
z = ______
(and a right triangle with hypotenuse 14√2, one leg 14, and segments x, y, z as shown)

Explanation:

Step1: Identify the triangle type

The triangle with hypotenuse \( 14\sqrt{2} \) and the two right - angled triangles formed by the altitude \( x \) seem to be isosceles right - angled triangles. In an isosceles right - angled triangle, the legs are equal and the hypotenuse \( c \) is related to the leg \( a \) by the formula \( c = a\sqrt{2} \).

Step2: Find the value of \( x \) and \( y \)

For the upper right - angled triangle, let the leg length be \( x \) (and also \( y \) since it's isosceles right - angled). Given hypotenuse \( h=14\sqrt{2} \), from \( h = a\sqrt{2} \) (where \( a \) is the leg), if we solve for \( a \):
\[
a=\frac{h}{\sqrt{2}}=\frac{14\sqrt{2}}{\sqrt{2}} = 14
\]
So, \( x = 14 \) and \( y = 14 \) (because in the isosceles right - angled triangle, the two legs are equal and the altitude to the hypotenuse in an isosceles right - angled triangle is equal to the length of the leg of the smaller isosceles right - angled triangles formed).

Step3: Find the value of \( z \)

The triangle with side \( z \) is also an isosceles right - angled triangle (since the two legs \( x = 14 \) and \( y = 14 \)). The hypotenuse \( z \) of a right - angled triangle with legs \( a = 14 \) and \( b = 14 \) is given by the Pythagorean theorem \( z=\sqrt{a^{2}+b^{2}} \).
\[
z=\sqrt{14^{2}+14^{2}}=\sqrt{196 + 196}=\sqrt{392}=\sqrt{196\times2}=14\sqrt{2}
\]
Or, since it's an isosceles right - angled triangle, using the hypotenuse formula for isosceles right - angled triangle \( z = y\sqrt{2}=14\sqrt{2} \) (or \( z=x\sqrt{2}=14\sqrt{2} \)).

Answer:

\( x = 14 \), \( y = 14 \), \( z = 14\sqrt{2} \)