Answer:
Heavenly, we need the forumula for the area of a triangle, do you know it?
Step-by-step explanation:
Area of a triangle is (1/2 ) * B*H
where B = base
H = height
then
A = 1/2* B * H
A is given to be 20
B = b
H = b+4
plug all those into the formula,
20 = (1/2)*b*(b+4)
use your algebra skillz :P
20 = 1/2 *( [tex]b^{2}[/tex] + 4b)
20*2 = [tex]b^{2}[/tex] + 4b
40 = [tex]b^{2}[/tex] + 4b
0 = [tex]b^{2}[/tex] + 4b-40
rewrite the above so it's in x format
[tex]x^{2}[/tex] + 4x -40 = 0
use the quadratic formula to find x , and for our problem, remember that x represents the base, or b, but don't mix it up with the b in the quadatic fomula, with the b of the triangle, which is the base, they are differnt "b"s :P
X = (b±[tex]\sqrt{b^{2}-4*a*c }[/tex] ) / 2*a
a = 1
b = 4
c = -40
x = (4±[tex]\sqrt{4^{2}-4*1*(-40) }[/tex] ) /2*1
x = ( 4±[tex]\sqrt{16+160}[/tex] ) / 2
x = (4±[tex]\sqrt{176}[/tex] ) / 2
x = (4±13.266499)/2
x= 2±6.633249
x = 2 + 6.633249 = 8.633
b= 8.633 cm
