Points $A$ and $B$ are on side $\overline{YZ}$ of rectangle $WXYZ$ such that $\overline{WA}$ and $\overline{WB}$ trisect $\angle ZWX$. If $WX = 2$ and $XY = 3$, then what is the area of rectangle $WXYZ$?

Akhain1 Apr 18, 2024

#1**0 **

Since WA and WB trisect ∠ZWX, we can divide ∠ZWX into three congruent angles of measure 3180∘=60∘. Since ∠WXY and ∠ZWX are right angles, then ∠XYW=90∘−60∘=30∘.

[asy] pair W,X,Y,Z,A,B; W = (0,0); X = (2,0); Y = (2,3); Z = extension(W,W+(0,3),X,X+(2,0),false); A = (W + X)/3; B = (2*X + W)/3; draw(W--X--Y--Z--cycle); draw(W--A); draw(W--B); label("W",W,SW); label("X",X,SE); label("Y",Y,NE); label("Z",Z,NW); label("A",A,S); label("B",B,S); [/asy]

pen_spark

We are given that WX=2 and XY=3. Since △WXY is a 30-60-90 triangle, then WY=WX⋅3=2⋅3. The area of rectangle WXYZ is the product of the length and width, which is WX⋅WY=2⋅(23)=43 square units.

Boseo Apr 18, 2024