Question

under the QUADRATIC EQUATION(Roots)
If a triangle has an area of 56 sq. inches, with base of 9 inches longer that its altitude, find the length of its base.

Answer 1

area of triangle = 0.5 * base * length
let length of base = x
then height = x-9
so 56 = 0.5 * x(x-9)

can u solve this?

Answer 2

Let h = altitude of triangle
h+9 = base of triangle

\[h(h+9) = 56\]\[h^2+9h-56=0\]\[h=\frac{-9\pm \sqrt{36+4(56)}}{2}\]\[h=\frac{-9\pm \sqrt{260}}{2}\]

But no negative values are allowed in geometry. Therefore:
\[h=\frac{-9+ \sqrt{260}}{2}\]

Answer 3

(1/2)*b*h =56
b=h+9
=>(1/2)*(h+9)*h=56
=> h^2 + 9h -112 =0
=>h=7 , -16
as h is positive h=7
=> b = h+9 =16

Answer 4

According to Wolfram, this is the answer: 3.56225
http://www.wolframalpha.com/input/?i=%5B-9%2Bsqrt%28260%29%5D%2F2

Answer 5

its a quadratic equation
rearrange
0.5x^2 - 4.5x - 56 = 0
multiply by 2 to remove decimals
x^2 - 9x - 112 = 0
solve by factoring - if u cant use the formula

Answer 6

solution is 16 or -7 length cant be nagative so x = 16

length of base = 16 ins

Answer 7

Whoops forgot about the (1/2)*bh formula.

Answer 8

you made a slight error in ur calculation jyqft
on third line - should be 81 not 36

Answer 9

i used the formula also but i've just seen that it will factor:
x^2 - 9x - 112 = (x-16)(x+7)

Answer 10

ok with that kyra?

Answer 11

how is manila these days?