the base of a triangle is 4dm longer than its altitude, and its area is16dm^2. Find the lengths of the base and its altitude.

A = 1/2 bh

h = altitude
h + 4 = base

16 = 1/2 (h + 4)h
Solve for h

Nothing

please jelp me

To find the lengths of the base and altitude of a triangle, given the area, we can use the formula for the area of a triangle:

Area = (1/2) * base * altitude

Let's assign variables to the base and altitude: let the altitude be 'x' dm, and the base be 'x + 4' dm (since the base is 4 dm longer than the altitude).

Now we can rewrite the area formula with the known values:

16 dm^2 = (1/2) * (x + 4) dm * x dm

To solve for x, let's simplify the equation:

16 dm^2 = (1/2) * (x^2 + 4x) dm^2

Multiplying both sides by 2:

32 = x^2 + 4x

Rearranging the equation:

x^2 + 4x - 32 = 0

Now we have a quadratic equation. To solve it, we can factor it or use the quadratic formula. Let's use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

For our equation, a = 1, b = 4, and c = -32. Substituting these values into the quadratic formula:

x = (-4 ± √(4^2 - 4 * 1 * -32)) / (2 * 1)

Simplifying further:

x = (-4 ± √(16 + 128)) / 2

x = (-4 ± √144) / 2

Using the square root of 144:

x = (-4 ± 12) / 2

This gives us two possible solutions:
1. x = (-4 + 12) / 2 = 8 / 2 = 4 dm
2. x = (-4 - 12) / 2 = -16 / 2 = -8 dm

Since lengths cannot be negative, we discard the negative solution, so the altitude of the triangle is 4 dm.

Now we can find the base length by substituting the altitude value into the equation: base = altitude + 4.

The base length = 4 dm + 4 dm = 8 dm.

Therefore, the lengths of the base and altitude of the triangle are 8 dm and 4 dm, respectively.