Question

in the rectangle below, ( ae = 2x + 1 ), ( bd = 3x + 12 ), and ( mangle ecb = 52^{circ} ). find ( ce ) and ( mangle eba ).

Explanation:

Step1: Use rectangle property

In a rectangle, diagonals are equal and bisect each other. So \(AE = CE\) and \(BD = AC\). Also, \(AC=2AE\). Given \(AE = 2x + 1\) and \(BD=3x + 12\), then \(3x + 12=2(2x + 1)\).

$$\begin{align*} 3x+12&=4x + 2\\ 12-2&=4x-3x\\ x&=10 \end{align*}$$

Step2: Find \(CE\)

Since \(AE = CE\) and \(AE=2x + 1\), substitute \(x = 10\) into \(AE\). Then \(AE=2\times10 + 1=21\), so \(CE = 21\).

Step3: Find \(m\angle EBA\)

We know that \(\angle ECB = 52^{\circ}\). In rectangle \(ABCD\), \(AB\parallel CD\), so \(\angle EBA=\angle EDC\) (alternate - interior angles). Also, \(BE = CE\) (diagonals bisect each other in a rectangle), so \(\angle EBC=\angle ECB = 52^{\circ}\). In \(\triangle ABC\), \(\angle ABC = 90^{\circ}\). Then \(m\angle EBA=90^{\circ}-52^{\circ}=38^{\circ}\)

Answer:

\(CE = 21\), \(m\angle EBA = 38^{\circ}\)