Question

solve the missing value:
ab = 18
bc=
ac = 18\sqrt{2}
ao=
m\angle abo=
m\angle aob=

Explanation:

Step1: Find \(BC\)

Since \(ABCD\) is a square, all sides are equal. Given \(AB = 18\), so \(BC=AB = 18\).

Step2: Find \(AO\)

The diagonals of a square are equal and bisect each other. Given \(AC = 18\sqrt{2}\), and \(AO=\frac{1}{2}AC\). Then \(AO=\frac{1}{2}\times18\sqrt{2}=9\sqrt{2}\).

Step3: Find \(m\angle ABO\)

In square \(ABCD\), \(\triangle ABO\) is an isosceles right - triangle (\(AO = BO\) as diagonals of a square bisect each other and \(AB\) is a side). So \(m\angle ABO = 45^{\circ}\).

Step4: Find \(m\angle AOB\)

In \(\triangle ABO\), using the angle - sum property of a triangle (\(\angle OAB+\angle ABO+\angle AOB = 180^{\circ}\)). Since \(\angle OAB=\angle ABO = 45^{\circ}\), then \(m\angle AOB=180-(45 + 45)=90^{\circ}\).

Answer:

\(BC = 18\), \(AO=9\sqrt{2}\), \(m\angle ABO = 45^{\circ}\), \(m\angle AOB = 90^{\circ}\)