Question

Please help with Question 12 at the top, i dont really want the answer. I just want how to figure it out please!

Answer 1

Answer:

a: $x = \dfrac{9\sqrt{5}}{5}$.

b: $x = \sqrt{5}$.

c: $x = \dfrac{\sqrt{10}}{2}$.

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.

$9^2 = x^2 + (2x)^2\\9^2 = x^2 + 4 x^2 = 5 x^2\\x^2 = \dfrac{9^2}{5}\\x = \sqrt{\dfrac{9^2}{5}}\\\phantom{x} = \dfrac{9}{\sqrt{5}}\\\phantom{x} = \dfrac{9\sqrt{5}}{5}$.

As a result, $x = \dfrac{9\sqrt{5}}{5}$.

Try the same steps for the triangle in b. $x = \sqrt{5}$.

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:

$5^2 = x^2 + (3x)^2\\25 = x^2 + 9x^2\\10 x^2 = 25\\x^2 = \dfrac{5}{2}\\x = \sqrt{\dfrac{5}{2}}\\\phantom{x} = \dfrac{\sqrt{5 \times 2}}{2}\\\phantom{x} = \dfrac{\sqrt{10}}{2}$

As a result, $x = \dfrac{\sqrt{10}}{2}$