Question

using the provided figure below, what is the length of \\( \overline{dm} \\)?

Explanation:

Step1: Identify triangle properties

Since \(\angle L = 45^{\circ}\) and \(\angle M=90^{\circ}\), \(\triangle LMN\) is a \(45 - 45-90\) triangle. In a \(45 - 45 - 90\) triangle, the sides are in the ratio \(1:1:\sqrt{2}\). Also, \(\triangle LDM\) is a \(45 - 45-90\) triangle (because \(\angle L = 45^{\circ}\) and \(\angle LDM = 90^{\circ}\)).

Step2: Use the property of \(45 - 45-90\) triangle

In a \(45 - 45-90\) triangle, if the hypotenuse is \(h\), then the legs \(a=b=\frac{h}{\sqrt{2}}\). Here, the hypotenuse \(LN = 22\) cm. In \(\triangle LDM\), let \(DM=x\). Since \(\triangle LDM\) is a \(45 - 45-90\) triangle and using the trigonometric relation \(\sin45^{\circ}=\frac{DM}{LM}\) (but also from the side - ratio property of \(45 - 45-90\) triangle in a more straightforward way, if we consider the fact that the projection from the right - angled vertex of a right - angled isosceles triangle to the hypotenuse divides the hypotenuse into two equal parts). The length of \(DM\) (using the formula for the length of the altitude to the hypotenuse of a right - angled isosceles triangle \(h=\frac{c}{2}\) where \(c\) is the hypotenuse of the large right - angled isosceles triangle).

$$DM=\frac{22}{\sqrt{2}}\times\frac{\sqrt{2}}{\sqrt{2}}=\frac{22\sqrt{2}}{2} = 11\sqrt{2}\approx15.56$$

Answer:

\(11\sqrt{2}\text{ cm}\approx15.6\text{ cm}\)