Question

type the correct answer in the box. use numerals instead of words. if necessary, use / for the fraction bar.

in \\( \triangle a b c, \overline { a b } = \overline { b c } \\) and \\( m \angle a b c = 90 ^ { \circ } \\). d and e are the midpoints of \\( \overline { a b } \\) and \\( \overline { b c } \\), respectively.

if the length of \\( \overline { c a } \\) is 9 units, the length of \\( \overline { d e } \\) is \\( \square \\) units and \\( m \angle c a b \\) is \\( \square \\)

Explanation:

Step1: Find the length of \(AB\) and \(BC\)

In right - angled isosceles triangle \(ABC\) (\(AB = BC\), \(\angle ABC=90^{\circ}\)), by Pythagoras theorem \(CA^{2}=AB^{2}+BC^{2}\). Since \(AB = BC\), let \(AB = BC=x\). Then \(CA^{2}=2x^{2}\). Given \(CA = 9\), so \(81 = 2x^{2}\), \(x^{2}=\frac{81}{2}\), \(x=\frac{9}{\sqrt{2}}\).

Step2: Use the mid - point theorem

Since \(D\) is the mid - point of \(AB\) and \(E\) is the mid - point of \(BC\), by the mid - point theorem \(DE=\frac{1}{2}AC\).

Step3: Find \(\angle CAB\)

In right - angled isosceles triangle \(ABC\) (\(AB = BC\), \(\angle ABC = 90^{\circ}\)), \(\angle CAB=\angle ACB\). Using the angle - sum property of a triangle \(\angle ABC+\angle CAB+\angle ACB = 180^{\circ}\). Substituting \(\angle ABC = 90^{\circ}\), we get \(90^{\circ}+2\angle CAB=180^{\circ}\), \(2\angle CAB = 90^{\circ}\), \(\angle CAB = 45^{\circ}\)

Answer:

The length of \(DE\) is \(4.5\) units and \(m\angle CAB\) is \(45\)