Question

en los problemas 6 a 9, indica si un triángulo con las longitudes de lado dadas es un triángulo rectángulo.

1. 5, 12, 13

$5^2 + 12^2 = 13^2$
$25 + 144 = 13^2$
$169 = 13^2$ (true)

1. 6, 8, 11

$6^2 + 8^2 = 11^2$
$36 + 64 = 11^2$
$100 \
eq 11^2$ (false)

1. 7, 7, 10

$7^2 + 7^2 = 10^2$
$49 + 49 = 10^2$
$98 \
eq 10^2$ (false)

1. $\frac{9}{2}$, 6, $\frac{15}{2}$

Explanation:

Step1: Identify the largest side

The sides are $\frac{9}{2}$, $6$, and $\frac{15}{2}$. First, convert $6$ to halves: $6 = \frac{12}{2}$. Now, the sides are $\frac{9}{2}$, $\frac{12}{2}$, $\frac{15}{2}$. The largest side is $\frac{15}{2}$.

Step2: Apply Pythagorean theorem

Check if $(\frac{9}{2})^2 + (\frac{12}{2})^2 = (\frac{15}{2})^2$.
Calculate each term:
$(\frac{9}{2})^2 = \frac{81}{4}$, $(\frac{12}{2})^2 = \frac{144}{4}$, $(\frac{15}{2})^2 = \frac{225}{4}$.
Sum the two smaller squares: $\frac{81}{4} + \frac{144}{4} = \frac{225}{4}$.

Step3: Compare

Since $\frac{81}{4} + \frac{144}{4} = \frac{225}{4}$, the equation holds.

Answer:

True (the triangle with sides $\frac{9}{2}$, $6$, $\frac{15}{2}$ is a right triangle)