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The area of a triangle is 384 sq ft. If it's base is 8 ft longer than its height. Find the length of the base and height.

Sagot :

luana
[tex]x-\ base\\ y- \ height\\\\ x-8=y\\ Area=\frac{1}{2}xy\\\\ 384=\frac{1}{2}*x(x-8)\\ 768=x^2-8x\\ x^2-8x-768=0\\(x-32)(x+24)=0\\\\ x=32\\\\y=32-8=24\\\\Length\ of\ base\ is\ 32\ and\ height\ 24.[/tex]
olemakpadu
Recall that Area of Triangle =  (1/2) *base* height.

Let the height = x.
And statement said base = 8 ft longer than height = (x+8)

Therefore:    A = (1/2)x(x+8)
                   384 = x(x+8)/2                Cross Multiplying
                    768 = x^2 + 8x     Re arrange, this is a quadratic equation.
                    x^2 + 8x = 768
                    x^2 + 8x - 768 = 0.   For quadratic factoring we think  of two numbers that would multiply to give -768 and add up to give +8.

By search, the numbers are 32 and -24.     32*-24 = -768 and (32+ -24) = 32-24 =8.

Solving:
x^2 + 8x - 768 = 0                                  Replace 8x with (32x -24x)
x^2 + 32x -24x -768 =0
x(x+32) -24(x+32)  = 0
(x+32)(x -24) = 0        Therefore (x+32) = 0  x = 0-32 =-32.
                                                (x-24) = 0  x = 0+24 =24.
x = -32 or 24.  
But our length can not be negative. Therefore x = 24. We throw away the negative answer.

Recall height = x = 24 ft.
Base = (x +8) = (24+8) = 32 ft.

Base = 32 ft, Height = 8ft. That's our answer. Hurray!
                   
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