The problem states that a triangular warning sign has an area of 340 square inches. The base of the sign is 14 inches longer than the altitude. We need to find the altitude of the triangle.

1. Problem Description

The problem states that a triangular warning sign has an area of 340 square inches. The base of the sign is 14 inches longer than the altitude. We need to find the altitude of the triangle.

2. Solution Steps

Let $h$ be the altitude (height) of the triangle.

Let $b$ be the base of the triangle.

We are given that the base is 14 inches longer than the altitude, so we can write $b = h + 14$.

The area of a triangle is given by the formula:

$Area = \frac{1}{2} \cdot base \cdot height$

We are given that the area is 340 square inches. Substituting the given information into the formula:

$340 = \frac{1}{2} \cdot b \cdot h$

Substitute $b = h + 14$ into the area equation:

$340 = \frac{1}{2} \cdot (h+14) \cdot h$

Multiply both sides by 2:

$680 = (h+14) \cdot h$

$680 = h^2 + 14h$

Rearrange the equation into a quadratic equation:

$h^2 + 14h - 680 = 0$

Now we need to solve this quadratic equation for $h$. We can use the quadratic formula:

$h = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

In our equation, $a = 1$, $b = 14$, and $c = -680$.

$h = \frac{-14 \pm \sqrt{14^2 - 4(1)(-680)}}{2(1)}$

$h = \frac{-14 \pm \sqrt{196 + 2720}}{2}$

$h = \frac{-14 \pm \sqrt{2916}}{2}$

$h = \frac{-14 \pm 54}{2}$

We have two possible solutions for $h$:

$h = \frac{-14 + 54}{2} = \frac{40}{2} = 20$

$h = \frac{-14 - 54}{2} = \frac{-68}{2} = -34$

Since the altitude cannot be negative, we take the positive solution: $h = 20$.

The altitude is 20 inches. Then the base is $b = h + 14 = 20 + 14 = 34$ inches.

Let's verify the area: $\frac{1}{2} \cdot 34 \cdot 20 = 17 \cdot 20 = 340$. The area is correct.

3. Final Answer

The altitude is 20 inches.