Question

if quadrilateral abcd is a square and ac = 6√2, find the missing values. ad = m∠aeb = ab = m∠adc = ed = :: 6 :: 3√2 :: 90° :: 45° :: 9 :: 6√2 :: 30°

Explanation:

Step1: Use Pythagorean theorem in square

In a square \(ABCD\) with side - length \(a\) and diagonal \(d\), by the Pythagorean theorem \(d^{2}=a^{2}+a^{2}=2a^{2}\). Given \(d = AC=6\sqrt{2}\), then \((6\sqrt{2})^{2}=2a^{2}\), \(72 = 2a^{2}\), \(a^{2}=36\), \(a = 6\). So \(AD = AB=6\).

Step2: Property of square's diagonals

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

Step3: Angle in a square

Each interior angle of a square is \(90^{\circ}\), so \(\angle ADC=90^{\circ}\).

Step4: Diagonal - bisecting property

The diagonals of a square bisect each other. Since \(AC = 6\sqrt{2}\), then \(ED=\frac{1}{2}BD\) and \(AC = BD\), so \(ED=\frac{1}{2}\times6\sqrt{2}=3\sqrt{2}\).

Answer:

\(AD = 6\)
\(m\angle AEB=90^{\circ}\)
\(AB = 6\)
\(m\angle ADC = 90^{\circ}\)
\(ED = 3\sqrt{2}\)