Answer :
Area of a triangle:
A= (1/2)(bh)
In this problem we are given:
A= 178.5 [tex]in^{2} [/tex]
b= h+4 in
h= UNKNOWN ("h")
Plug in our values into the area equation and solve for "h":
[tex]A= \frac{1}{2} bh[/tex]
[tex]178.5 = \frac{1}{2} ((h+4)h)[/tex] (multiply each side by 2 to get rid of the 1/2)
[tex]357 = ((h+4)h)[/tex]
[tex]357 = h^{2} +4h[/tex]
[tex]0= h^{2} +4h -357[/tex]
(357/21=17)
[tex]0= (h+21)(h-17)[/tex] (solve both inequalities separately)
[tex]0= (h+21)[/tex]
[tex]-21= h[/tex] ( a negative height is not possible so, this answer is incorrect.)
[tex]0=(h-17)[/tex]
[tex]h=17 in[/tex]
Answer: base = (h+4)= (17+ 4) b= 21 in h= 17in
Please comment with any further questions! :)
A= (1/2)(bh)
In this problem we are given:
A= 178.5 [tex]in^{2} [/tex]
b= h+4 in
h= UNKNOWN ("h")
Plug in our values into the area equation and solve for "h":
[tex]A= \frac{1}{2} bh[/tex]
[tex]178.5 = \frac{1}{2} ((h+4)h)[/tex] (multiply each side by 2 to get rid of the 1/2)
[tex]357 = ((h+4)h)[/tex]
[tex]357 = h^{2} +4h[/tex]
[tex]0= h^{2} +4h -357[/tex]
(357/21=17)
[tex]0= (h+21)(h-17)[/tex] (solve both inequalities separately)
[tex]0= (h+21)[/tex]
[tex]-21= h[/tex] ( a negative height is not possible so, this answer is incorrect.)
[tex]0=(h-17)[/tex]
[tex]h=17 in[/tex]
Answer: base = (h+4)= (17+ 4) b= 21 in h= 17in
Please comment with any further questions! :)