Question

drag the tiles to the boxes to form correct pairs. not all tiles will be used.
determine each length in right triangle abc.

Explanation:

Step1: Find the length of \(AB\)

Since \(\triangle ABC\) is a right - isosceles triangle (\(\angle A = 45^{\circ},\angle C=45^{\circ},\angle B = 90^{\circ}\)), and \(AC=AD + DC=8 + 8=16\).
By the property of a \(45 - 45-90\) triangle (\(AB = BC\) and \(AC=\sqrt{2}AB\)).
We know that \(AC = 16\), then \(AB=\frac{AC}{\sqrt{2}}=\frac{16}{\sqrt{2}}=8\sqrt{2}\) (rationalizing the denominator: \(\frac{16}{\sqrt{2}}=\frac{16\sqrt{2}}{2}\)).

Step2: Find the length of \(BD\)

First, find the area of \(\triangle ABC\) in two ways.
The area of \(\triangle ABC\) using \(AB\) and \(BC\): \(S=\frac{1}{2}\times AB\times BC\). Since \(AB = BC = 8\sqrt{2}\), \(S=\frac{1}{2}\times8\sqrt{2}\times8\sqrt{2}=\frac{1}{2}\times128 = 64\).
The area of \(\triangle ABC\) using \(AC\) and \(BD\): \(S=\frac{1}{2}\times AC\times BD\). We know \(AC = 16\) and \(S = 64\).
Substituting into the formula \(\frac{1}{2}\times16\times BD=64\).
Solving for \(BD\): \(8BD=64\), so \(BD = 8\).

Answer:

\(AB\to8\sqrt{2}\), \(BD\to8\)