the bases of two triangles are the same length. The altitude of the first triangle is 5 cm and the area of the second triangle is 3 times the area of the first. what is the altitude of the second triangle?

area of first = (1/2)(b)(5) = 5b/2

area of second = (1/2)(b)(h) = bh/2

but we are told that bh/2 = 3(5b/2)
multiply both sides by 2

bh = 15b
divide both sides by b

h = 15

To find the altitude of the second triangle, we can use the fact that the area of the second triangle is three times the area of the first. Since the bases of the triangles have the same length, we can say that the ratio of their areas is equal to the ratio of their altitudes.

Let's denote the altitude of the second triangle as 'h'. Then, we know that the area of the first triangle is equal to (base * altitude) / 2, and the area of the second triangle is equal to (base * h) / 2.

The ratio of the areas can be expressed as:

(area of second triangle) / (area of first triangle) = (base * h) / (base * 5) = h / 5

Since we know that the area of the second triangle is three times the area of the first, we can write:

3 = h / 5

To solve for 'h', we can cross-multiply:

3 * 5 = h

Therefore, the altitude of the second triangle is 15 cm.