Answer:
Part 1) [tex]42=\frac{1}{2}x(x-5)[/tex]
Part 2) The length of the base is [tex]x=12\ m[/tex]
Step-by-step explanation:
Part 1) Which equation would be help solve the following problem?
Let
x----> the length of the base of a triangle
y----> the height of a triangle
we know that
The area of a triangle is
[tex]A=\frac{1}{2}xy[/tex]
[tex]A=42\ m^{2}[/tex]
so
[tex]42=\frac{1}{2}xy[/tex] ------> equation A
[tex]y=x-5[/tex] -----> equation B
Substitute equation B in equation A
[tex]42=\frac{1}{2}x(x-5)[/tex]
Part 2) Find the length of the base
we have
[tex]42=\frac{1}{2}x(x-5)[/tex]
solve for x
[tex]84=x^{2}-5x\\x^{2}-5x-84=0[/tex]
Solve the quadratic equation by graphing
The solution is [tex]x=12\ m[/tex]
see the attached figure
