The base of a triangle is 6 cm longer than its height. If the area of the triangle is 140cm^2, what is the length of the base?
Im just confused on how the equation should be formed. Thanks.
B = H + 6
Area = HB/2
Substitute H+6 for B in the second equation and solve for H. Insert that value into the first equation to solve for B. Check by putting both values into the second equation.
140 = H(H+6)/2
70 = H^2 + 6H
To solve this problem, we can use the formula for the area of a triangle: Area = (base * height) / 2.
Let's denote the length of the height of the triangle as "h" (in cm). According to the problem, the base of the triangle is 6 cm longer than the height. So, the length of the base can be expressed as "h + 6" (in cm).
We are given that the area of the triangle is 140 cm². Therefore, we can set up the equation:
140 = (h + 6) * h / 2.
To solve this equation, we can multiply both sides by 2 to eliminate the fraction:
280 = h^2 + 6h.
Next, we rearrange the equation in the standard quadratic form:
h^2 + 6h - 280 = 0.
Now, we need to solve this quadratic equation for "h". We can do this by factoring, completing the square, or using the quadratic formula. Let's use factoring:
(h - 14)(h + 20) = 0.
This equation is satisfied when either (h - 14) = 0 or (h + 20) = 0.
If (h - 14) = 0, then h = 14. But since the height cannot be negative, we discard this solution.
If (h + 20) = 0, then h = -20. However, the height of a triangle cannot be negative, so we discard this solution as well.
Therefore, there are no valid solutions for "h" in this problem.
This implies that there is no triangle with an area of 140 cm², given that the base is 6 cm longer than the height.