Question

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

Explanation:

Step1: Find the length of \(BC\)

In right - triangle \(ABC\), \(\angle C = 45^{\circ}\), \(\angle A=45^{\circ}\), \(\angle B = 90^{\circ}\), and \(AC = 14\).
Using the formula for a \(45 - 45-90\) triangle: \(BC=\frac{AC}{\sqrt{2}}\times\sqrt{2}\) (or \(BC = AC\sin45^{\circ}\times\sqrt{2}\)).
Since \(AC = 14\) and for a \(45 - 45 - 90\) triangle, if the hypotenuse is \(h\) and the legs are \(l\), \(h = l\sqrt{2}\), then \(l=\frac{h}{\sqrt{2}}\). Here \(BC\) is a leg and \(AC\) is the hypotenuse of \(\triangle ABC\) (incorrect, actually \(AC\) is the hypotenuse of \(\triangle ABC\) which is wrong, correct: In \(\triangle ABC\), \(\angle B = 90^{\circ}\), \(\angle A=\angle C = 45^{\circ}\), so \(AB = BC\). Using \(AC^{2}=AB^{2}+BC^{2}\) (Pythagorean theorem), and \(AB = BC\), \(AC^{2}=2BC^{2}\), \(BC=\frac{AC}{\sqrt{2}}\). Since \(AC = 14\), \(BC=\frac{14}{\sqrt{2}}=7\sqrt{2}\)).

Step2: Find the length of \(BD\)

In right - triangle \(ABD\), \(\angle A = 45^{\circ}\), \(\angle ADB=90^{\circ}\), so \(\angle ABD = 45^{\circ}\).
Using the formula for a \(45 - 45-90\) triangle: if the leg \(AD = 7\) (because \(AC = 14\) and \(D\) is the mid - point of \(AC\) in an isosceles right - triangle \(\triangle ABC\)), and for a \(45 - 45 - 90\) triangle with leg \(x\), the other leg \(BD=x\). Using \(BD = AD\) (since \(\triangle ABD\) is a \(45 - 45 - 90\) triangle), \(BD = 7\).

Answer:

\(BC\to7\sqrt{2}\), \(BD\to7\)