The base of triangle ABC is one half the altitude. The altitude of triangle DEF is one half its base. If both triangles are equal in area, what is the ratio of the altitude of triangle ABC to the altitude of triangle DEF?

a. 1:4
b. 1:2
c. 1:1
d. 2:1
e. 4:1

please answer and explain

If we call the respective bases and altitudes a,b for ABC and d,e for DEF, then we have

a = b/2
e = d/2, so d = 2e

So, let's find the ratio b/e
Since the areas are equal,

(1/2)(ab) = (1/2)(de)
(b/2)(b) = (2e)(e)
b^2/2 = 2e^2
b^2/e^2 = 4
b/e = 2

So, (D) 2:1

To solve this problem, let's start by setting up equations for the area of each triangle.

We know that the area of a triangle is given by the formula: Area = (base * height) / 2.

Let's denote the base of triangle ABC as b and the altitude as h. According to the problem, the base of triangle ABC is one half the altitude. Thus, we can write:

b = 1/2 * h ---- (Equation 1)

Similarly, for triangle DEF, let's denote the base as d and the altitude as a. According to the problem, the altitude of triangle DEF is one-half its base. Thus, we can write:

a = 1/2 * d ---- (Equation 2)

We are also given that both triangles have equal areas. Therefore, we can equate the area of triangle ABC to the area of triangle DEF:

(b * h) / 2 = (d * a) / 2

Plugging in the values from Equations 1 and 2, we get:

((1/2 * h) * h) / 2 = (d * (1/2 * d)) / 2

Simplifying the equation, we get:

(h^2) / 4 = (d^2) / 4

Multiply both sides of the equation by 4 to eliminate the denominators:

h^2 = d^2

Take the square root of both sides of the equation:

√(h^2) = √(d^2)

h = d

From this equation, we can conclude that the altitude of triangle ABC (h) is equal to the altitude of triangle DEF (d).

Therefore, the ratio of the altitude of triangle ABC to the altitude of triangle DEF is 1:1, which corresponds to option c.