Question

10\sqrt{2}
\sqrt{6}
45°
\sqrt{3}
\sqrt{3}y
x
find y
find x

Explanation:

Step1: Use the Pythagorean theorem for the triangle

In a right - angled triangle, by the Pythagorean theorem \(a^{2}+b^{2}=c^{2}\). Here, \(a = x\), \(b=\sqrt{6}\), \(c = \sqrt{3}y\). Also, since one of the non - right angles is \(45^{\circ}\), the triangle is an isosceles right - angled triangle (because in a right - angled triangle, if one of the acute angles is \(45^{\circ}\), the other acute angle is also \(45^{\circ}\) and the legs are equal). So \(x=\sqrt{6}\).

Step2: Use the property of a square

Since the figure with side \(10\sqrt{2}\) is a square, all sides are equal. For a square with side \(y\), using the Pythagorean theorem for the right - angled triangle formed by two adjacent sides and the diagonal (but here we can also use the fact that if the side of a square is \(y\), and we know from the triangle part (after confirming the triangle is isosceles right - angled from the \(45^{\circ}\) angle) and the square - triangle relation. In a square, if we consider the basic side - diagonal relation (but more simply, since the side of the square is \(y\) and from the triangle we first find \(x\) and then relate to the square). Since \(x = \sqrt{6}\) (from the isosceles right - angled triangle with sides \(x\) and \(\sqrt{6}\) and hypotenuse \(\sqrt{3}y\), and \(x=\sqrt{6}\) as the legs of the \(45 - 45-90\) triangle are equal). And for the square, if we assume the side \(y\) (using the fact that the side of the square is related to the triangle in a simple geometric figure where the side of the square is \(y\) and from the Pythagorean theorem for the square \(y^{2}+y^{2}=(10\sqrt{2})^{2}\). But another way: Since \(x = \sqrt{6}\) (from the \(45 - 45-90\) triangle \(x=\sqrt{6}\)), and for the square, if we consider the basic side length. Wait, no, actually, for a square of side \(y\), the diagonal \(d=\sqrt{2}y\). But here, if we assume the side of the square is \(y\) and we know from the figure (assuming the \(10\sqrt{2}\) is the side of the square). Wait, no, re - checking:
For the right - angled triangle with angles \(45^{\circ}-45^{\circ}-90^{\circ}\), the ratio of sides is \(1:1:\sqrt{2}\). Here, if the legs are \(x\) and \(\sqrt{6}\), then \(x = \sqrt{6}\) (because in \(45 - 45-90\) triangle legs are equal).
For the square, if the side is \(y\), and using the Pythagorean theorem \(y^{2}+y^{2}=(10\sqrt{2})^{2}\).

$$
LATEXBLOCK0
$$

Answer:

\(x=\sqrt{6}\), \(y = 10\)