Question

quadrilateral abcd is a square. find the missing values.
ad =
ac =
ec =
m∠cba =
m∠ceb =
m∠bac =
18
18√2
9√2
90°
45°

Explanation:

Step1: Find \(AD\)

In a square, all sides are equal. Given \(AB = 18\), so \(AD=18\).

Step2: Find \(\angle CBA\)

All angles in a square are \(90^{\circ}\). So \(\angle CBA = 90^{\circ}\).

Step3: Find \(AC\)

Using the Pythagorean theorem in \(\triangle ABC\) (\(AB = BC=18\)), \(AC=\sqrt{AB^{2}+BC^{2}}=\sqrt{18^{2}+18^{2}}=\sqrt{2\times18^{2}} = 18\sqrt{2}\).

Step4: Find \(\angle BAC\)

In \(\triangle ABC\), since \(AB = BC\), \(\angle BAC=\angle BCA\). And \(\angle ABC = 90^{\circ}\), so \(\angle BAC=\frac{180^{\circ}-\angle ABC}{2}=\frac{180 - 90}{2}=45^{\circ}\).

Step5: Find \(EC\)

The diagonals of a square bisect each other. So \(EC=\frac{AC}{2}\). Since \(AC = 18\sqrt{2}\), \(EC=\frac{18\sqrt{2}}{2}=9\sqrt{2}\).

Step6: Find \(\angle CEB\)

The diagonals of a square are perpendicular to each other. So \(\angle CEB = 90^{\circ}\).

Answer:

\(AD = 18\), \(\angle CBA=90^{\circ}\), \(AC = 18\sqrt{2}\), \(\angle BAC = 45^{\circ}\), \(EC=9\sqrt{2}\), \(\angle CEB=90^{\circ}\)