Question

question 10 of 10
in the diagram below, \\( \triangle wxy \\) is an equilateral triangle and \\( \overline { xz } \perp \overline { wy } \\).

which statement must be true?
\\( \bigcirc \\) a. \\( wx = 2 \cdot wz \\)
\\( \bigcirc \\) b. \\( wx = 2 \cdot xz \\)
\\( \bigcirc \\) c. \\( wx = wz \\)
\\( \bigcirc \\) d. \\( wx = xz \\)

Explanation:

Step1: Recall properties of equilateral triangles

In an equilateral triangle, all sides are equal, and all angles are \(60^\circ\). Also, a perpendicular from a vertex to the opposite side bisects the opposite side and the corresponding angle. So in \(\triangle WXY\) (equilateral), \(XZ\perp WY\) implies \(WZ = ZY=\frac{1}{2}WY\), and \(\angle WXZ = 30^\circ\), \(\angle XZW = 90^\circ\).

Step2: Analyze triangle \(XZW\)

\(\triangle XZW\) is a right - triangle with \(\angle XZW = 90^\circ\) and \(\angle WXZ=30^\circ\). In a \(30 - 60 - 90\) right - triangle, the side opposite the \(30^\circ\) angle (which is \(WZ\)) is half the hypotenuse (which is \(WX\)). So, if we let \(WZ = x\), then \(WX = 2x=2\cdot WZ\).
Let's check the other options:

• Option B: In a \(30 - 60 - 90\) triangle, if the hypotenuse is \(c\), the side opposite \(60^\circ\) ( \(XZ\)) is \(\frac{\sqrt{3}}{2}c\). So \(XZ=\frac{\sqrt{3}}{2}WX\), then \(WX=\frac{2}{\sqrt{3}}XZ

eq2\cdot XZ\).

• Option C: \(WX\) is the hypotenuse of \(\triangle XZW\) and \(WZ\) is a leg, so \(WX>WZ\), \(WX

eq WZ\).

• Option D: \(WX\) is the hypotenuse of \(\triangle XZW\) and \(XZ\) is a leg, so \(WX > XZ\), \(WX

eq XZ\).

Answer:

A. \(WX = 2\cdot WZ\)