Answer:
The length of the base is 11 meters.
Step-by-step explanation:
The diagram of the triangle is not shown; However, the given details are enough to solve this question.
Given
Shape: Triangle
Represent the height with h and the base with b
[tex]b = 3 + 2h[/tex]
[tex]Area = 22[/tex]
Required
Find the length of the base
The area of a triangle is calculated as thus;
[tex]Area = \frac{1}{2} * b * h[/tex]
Substitute 22 for Area and 3 + 2h for b
The formula becomes
[tex]22 = \frac{1}{2} * (3 + 2h) * h[/tex]
Multiply both sides by 2
[tex]2 * 22 = 2 * \frac{1}{2} * (3 + 2h) * h[/tex]
[tex]44 = (3 + 2h) * h[/tex]
Open the bracket
[tex]44 = 3 * h + 2h * h[/tex]
[tex]44 = 3h + 2h^2[/tex]
Subtract 44 from both sides
[tex]44 - 44 = 3h + 2h^2 - 44[/tex]
[tex]0 = 3h + 2h^2 - 44[/tex]
Rearrange
[tex]0 = 2h^2 +3h - 44[/tex]
[tex]2h^2 +3h - 44 = 0[/tex]
At this point, we have a quadratic equation; which is solved as follows:
[tex]2h^2 +3h - 44 = 0[/tex]
[tex]2h^2 + 11h - 8h - 44 = 0[/tex]
[tex]h(2h + 11) - 4(2h + 11) = 0[/tex]
[tex](h - 4)(2h + 11) = 0[/tex]
Split the above
[tex](h - 4) = 0\ or\ (2h + 11) = 0[/tex]
[tex]h - 4 = 0\ or\ 2h + 11 = 0[/tex]
Solve the above linear equations separately
[tex]h - 4 = 0[/tex]
Add 4 to both sides
[tex]h - 4 + 4 = 0 + 4[/tex]
[tex]h = 0 + 4[/tex]
[tex]h = 4[/tex] ---- First value of h
[tex]2h + 11 = 0[/tex]
Subtract 11 from both sides
[tex]2h + 11 - 11 = 0 - 11[/tex]
[tex]2h = 0 - 11[/tex]
[tex]2h = -11[/tex]
Divide both sides by 2
[tex]\frac{2h}{2} = -\frac{11}{2}[/tex]
[tex]h = -\frac{11}{2}[/tex] ------ Second value of h
Since height can be negative, we'll discard [tex]h = -\frac{11}{2}[/tex]
Hence, the usable value of height is [tex]h = 4[/tex]
Recall that [tex]b = 3 + 2h[/tex]
Substitute 4 for h
[tex]b = 3 + 2(4)[/tex]
[tex]b = 3 + 8[/tex]
[tex]b = 11[/tex]
Hence, the length of the base is 11 meters
