Answer:
a: [tex]x = \dfrac{9\sqrt{5}}{5}[/tex].
b: [tex]x = \sqrt{5}[/tex].
c: [tex]x = \dfrac{\sqrt{10}}{2}[/tex].
Step-by-step explanation:
Apply the Pythagorean Theorem. The square of the hypotenuse of a right triangle is the sum of the squares of the two legs.
For triangle in a:
- Hypotenuse: 9.
- First Leg: x.
- Second Leg: 2x.
Apply the Pythagorean Theorem.
[tex]9^2 = x^2 + (2x)^2\\9^2 = x^2 + 4 x^2 = 5 x^2\\x^2 = \dfrac{9^2}{5}\\x = \sqrt{\dfrac{9^2}{5}}\\\phantom{x} = \dfrac{9}{\sqrt{5}}\\\phantom{x} = \dfrac{9\sqrt{5}}{5}[/tex][tex][/tex].
As a result, [tex]x = \dfrac{9\sqrt{5}}{5}[/tex].
Try the same steps for the triangle in b. [tex]x = \sqrt{5}[/tex].
Question c includes a rectangle. The lengths of opposite sides of a rectangle are equal. As a result, for the right triangle with a right angle at the lower-right corner of the rectangle:
- Hypotenuse: 5.
- First Leg: x.
- Second Leg: 2x.
Apply the Pythagorean Theorem:
[tex]5^2 = x^2 + (3x)^2\\25 = x^2 + 9x^2\\10 x^2 = 25\\x^2 = \dfrac{5}{2}\\x = \sqrt{\dfrac{5}{2}}\\\phantom{x} = \dfrac{\sqrt{5 \times 2}}{2}\\\phantom{x} = \dfrac{\sqrt{10}}{2}[/tex]
As a result, [tex]x = \dfrac{\sqrt{10}}{2}[/tex]
