+363 Points $A$ and $B$ are on side $\overline{YZ}$ of rectangle $WXYZ$ such that $\overline{WA}$ and $\overline{WB}$ trisect $\angle ZWX$. If $BY = 2$ and $AZ = 4$, then what is the area of rectangle $WXYZ$?
+846 Certainly, let's find the area of rectangle WXYZ.
1. Understand the Trisection
Since WA and WB trisect ∠ZWX, we know that ∠ZWA = ∠AWB = ∠BWX.
2. Let's Define
Let's denote the length of rectangle WXYZ as 'l' (which is also equal to YZ)
Let's denote the width of rectangle WXYZ as 'w'
3. Analyze Triangles
Consider triangles ΔWZA and ΔWZB.
These triangles share a common side (WW) and have equal angles (∠ZWA = ∠BWX).
Since the rectangle has 90-degree angles, ∠WZA = ∠WBZ = 90 degrees.
Therefore, triangles ΔWZA and ΔWZB are congruent by the Angle-Side-Angle (ASA) congruence criterion.
4. Implications of Congruence
Since the triangles are congruent, their corresponding sides are equal.This means:
WA = WB
AZ = BZ
5. Calculate ZB
We are given AZ = 4.
Since AZ = BZ, we have BZ = 4.
6. Calculate YZ
YZ = BY + BZ
YZ = 2 + 4
YZ = 6
7. Determine the Length and Width
Since YZ is the length of the rectangle, l = 6.
Since WA = WB and they trisect ∠ZWX, WA and WB bisect the rectangle.
Therefore, width (w) = AZ = BZ = 4.
8. Calculate the Area
Area of rectangle WXYZ = length * width
Area = l * w
Area = 6 * 4
Area = 24
Therefore, the area of rectangle WXYZ is 24 square units.
