Question

Solve for x using the figure to the right.
x=

Answer 1

Finding the hypotenuse of the small triangle:

• 
  `a^(2) = x^(2) + 5^(2)`
• =>
  `\sqrt{a^(2)} = \sqrt{x^(2) + 5^(2)}`
• =>
  `a = \sqrt{x^(2) + 25}`

Finding the hypotenuse of the big triangle:

• 
  `b^(2) = x^(2) + 45`
• 
  `\sqrt{b^(2) } = \sqrt{x^(2) + 45^(2) }`
• 
  `b = \sqrt{x^(2) + 2025 }`

Finding the value of x.

• 
  `(45 + 5)^(2) = (\sqrt{x^(2) + 25} )^(2) + (\sqrt{x^(2) + 2025} )^(2)`
• 
  `(50)^(2) = (\sqrt{x^(2) + 25} )(\sqrt{x^(2) + 25} ) + (\sqrt{x^(2) + 2025} )(\sqrt{x^(2) + 2025} )`
• 
  `2500 = x^(2) + 25 + {x^(2) + 2025}`
• 
  `2500 = 2x^(2) + 2050`
• 
  `450 = 2x^(2)`
• 
  `225 = x^(2)`
• 
  `x = \±15`

Answer 2

x = 15

Step-by-step explanation:

we have to find both the hypotenuse of the triangles to solve this.

for the smaller triangle:

x² + 5² = h²

h = √x² + 5²

for the bigger triangle:

x² + 45² = h²

h = √x² + 45²

now that if you look, we found the sides - adjacent and leg side of the Δ

solving using Pythagoras theorem:

a² + b² = c²

(√x² + 45²)² + (√x² + 5²)² = 50²

x² + 2025 + x² + 25 = 2500

2x² = 450

x² = 225

x = ± 15

x = 15