The 2 equal sides of an isosceles triangle are 5.7 cm each. The base is 7 cm. What is the height?

Question

Noah Atkinson · Accepted Answer

#h=2/5sqrt2sqrt11sqrt23#

Let A, B, C be vertexes of the triangle. Normal penpendicular and median on nonequal side are the same lign segment in isosceles triangles. Then:

#color(white)(xx)3.5^2+h^2=5.7^2#

#=>h=sqrt20.24# #=>h=2/5sqrt2sqrt11sqrt23#

Caroline Aucoin · Answer

undefined To find the height of the isosceles triangle, you can use the Pythagorean theorem and the properties of right triangles. Here's how:

1. Let the two equal sides of the isosceles triangle be $ a = 5.7 $ cm each, and the base $ b = 7 $ cm.

2. Draw a perpendicular line from the vertex opposite the base to the midpoint of the base. This perpendicular line represents the height of the triangle.

3. Since the perpendicular line divides the base into two equal parts, each part will be $ \frac{7}{2} = 3.5 $ cm.

4. Now, consider one half of the base ($ 3.5 $ cm), one of the equal sides ($ 5.7 $ cm), and the height of the triangle. These form a right triangle.

5. Apply the Pythagorean theorem to find the height ($ h $) of the triangle:
$$ h^2 + (3.5)^2 = (5.7)^2 $$

6. Solve for $ h $.

$$ h^2 = (5.7)^2 - (3.5)^2 $$
$$ h^2 = 32.49 - 12.25 $$
$$ h^2 = 20.24 $$
$$ h = \sqrt{20.24} $$

7. Calculate the square root of $ 20.24 $ to find the height of the triangle.

$$ h \approx 4.49 $$

So, the height of the isosceles triangle is approximately $ 4.49 $ cm.