Apply Pythagoras theorem, to find the lengths of x and y.

Pythagoras Theorem states that in a right-angled triangle sum of squares of perpendicular adjacent sides of a triangle is equal to the square of the third side of the triangle.

In the smaller right-angled triangle:

\(\displaystyle{x}^{{2}}+{\left({3.6}\right)}^{{2}}={6}^{{2}}\)

\(\displaystyle{x}^{{2}}={36}-{12.96}\)

\(\displaystyle{x}^{{2}}={23.04}\)

\(\displaystyle{x}={4.8}\)

In the larger right-angled triangle:

\(\displaystyle{y}^{{2}}+{6}^{{2}}={\left({6.4}+{3.6}\right)}^{{2}}\)

\(\displaystyle{y}^{{2}}+{6}^{{2}}={10}^{{2}}\)

\(\displaystyle{y}^{{2}}={100}-{36}\)

\(\displaystyle{y}^{{2}}={64}\)

\(\displaystyle{y}={8}\)

Thus, the values of x and y are 4.8 and 8 respectively.

Pythagoras Theorem states that in a right-angled triangle sum of squares of perpendicular adjacent sides of a triangle is equal to the square of the third side of the triangle.

In the smaller right-angled triangle:

\(\displaystyle{x}^{{2}}+{\left({3.6}\right)}^{{2}}={6}^{{2}}\)

\(\displaystyle{x}^{{2}}={36}-{12.96}\)

\(\displaystyle{x}^{{2}}={23.04}\)

\(\displaystyle{x}={4.8}\)

In the larger right-angled triangle:

\(\displaystyle{y}^{{2}}+{6}^{{2}}={\left({6.4}+{3.6}\right)}^{{2}}\)

\(\displaystyle{y}^{{2}}+{6}^{{2}}={10}^{{2}}\)

\(\displaystyle{y}^{{2}}={100}-{36}\)

\(\displaystyle{y}^{{2}}={64}\)

\(\displaystyle{y}={8}\)

Thus, the values of x and y are 4.8 and 8 respectively.