Question

Find the value of x in the isosceles triangle shown below​

Answer 1

Answer:

x = $\sqrt{119\\}$

Step-by-step explanation:

formula for this is $\sqrt{a^{2}-(b/2)^{2}}$ where a is the altitude and b is the base

altitude = 12

base = 10

$\sqrt{12^{2}-(10/2)^{2} }$

$\sqrt{144-5^{2} }$

$\sqrt{144-25}$

$\sqrt{119}$ ≈ 10.9

so I would say x = $\sqrt{119}$ or 10.9, but seeing as thats not an option, then I'm not sure