Question

The sides of a triangle have lengths of x, x + 5, and 25. If the longest side is 25, which of the following values of x would make the triangle right? A. 10 B. 12 C. 15 D. 17

Answer 1

To find which value of x makes the triangle a right triangle with the longest side being 25, apply the Pythagorean theorem. The only suitable value for x is 15, so option C (x = 15) is correct.

Step-by-step explanation:

The sides of a triangle have lengths of x, x + 5, and 25. To determine which value of x makes the triangle a right triangle with the longest side being 25, you can use the Pythagorean theorem.

According to the theorem, for a right triangle, the sum of the squares of the two shorter sides must equal the square of the hypotenuse. In this case, the hypotenuse (the longest side) is 25, so we look for an x such that:

x^2 + (x + 5)^2 = 25^2

Simplifying this equation, we get:

x^2 + x^2 + 10x + 25 = 625
2x^2 + 10x + 25 = 625
2x^2 + 10x - 600 = 0
x^2 + 5x - 300 = 0

This is a quadratic equation, and we solve for x. Factoring the equation, we get:

(x - 15)(x + 20) = 0

Setting each factor equal to zero gives us the possible values for x:

x - 15 = 0 or x + 20 = 0
x = 15 or x = -20

Since a side length cannot be negative, the only suitable value for x is 15. Therefore, option C (x = 15) makes the triangle a right triangle.

Answer 2

In a right angle, the sum of the squares of the two shorter sides is equal to the square of the longest side.
Here, the longest side is 25, the shorter sides are x and x+5.


`x^2+(x+5)^2=25^2 \\ x^2+x^2+10x+25=625 \\ 2x^2+10x+25=625 \ \ \ |-625 \\ 2x^2+10x-600=0 \ \ \ |/ 2 \\ x^2+5x-300=0 \\ x^2+20x-15x-300=0 \\ x(x+20)-15(x+20)=0 \\ (x-15)(x+20)=0 \\ x-15=0 \ \lor \ x+20=0 \\ x=15 \ \lor \ x=-20`

The length of a side must be a positive number, so x=15.
The answer is C.