Answer:
A
Step-by-step explanation:
The segment from the vertex to the base , bisects the base at right angles
Then there are 2 right triangles with legs x and 4 and hypotenuse [tex]\sqrt{52}[/tex]
Using Pythagoras' identity in either of the 2 right angles
x² + 4² = ([tex]\sqrt{52}[/tex] )²
x² + 16= 52 ( subtract 16 from both sides )
x² = 36 ( take the square root of both sides )
x = [tex]\sqrt{36}[/tex] = 6 → A
