Find the altitude of the isosceles triangle shown in the figure.Round your answer to two decimal places.? $$\theta = 45 ^ { \circ } , b = 11$$ ?
A)2.00
B)11.00
C)5.50
D)22.00?
E)7.50

Question

Find the altitude of the isosceles triangle shown in the figure.Round your answer to two decimal places.?  $$\theta = 45 ^ { \circ } , b = 11$$ ?
A)2.00 
B)11.00 
C)5.50 
D)22.00? 
E)7.50

Quizplus · Accepted Answer

Since the given triangle is isosceles, we know that the altitude will bisect the base. Therefore, we can use the Pythagorean Theorem to find the altitude. 
Let the altitude be h and the length of each congruent side be x. Then we have: 
$(\frac{x}{2})^2 + h^2 = x^2$
Simplifying, we get: 
$\frac{x^2}{4} + h^2 = x^2$
$h^2 = \frac{3}{4}x^2$
$h = \frac{x}{2}\sqrt{3}$
We are given that each side of the triangle has length 11, so we can substitute x=11 into the formula above to get: 
$h = \frac{11}{2}\sqrt{3}\approx 5.50$
Therefore, the altitude is approximately 5.50 units. The correct answer is (C).

Find the Altitude of the Isosceles Triangle Shown in the Figure.Round