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Sagot :
1>
Let the base be "x"
According to the question,
h = x - 13
A = 24 in sq.
Now,
[tex] Area of triangle = \frac{1}{2}~b~h [/tex]
[tex]24= \frac{1}{2}(x)(x-13) [/tex]
[tex]24 *2= (x)(x-13) [/tex]
[tex]48= x^{2} - 13x[/tex]
[tex]0= x^{2} - 13x -48[/tex]
Factorizing the equation, x² - 13x -48, we get,
[tex]0= x^{2}+3x - 16x -48[/tex]
[tex]0= x(x+3) - 16(x+3)[/tex]
[tex]0= (x + 3)(x-16)[/tex]
NOW, using zero product property, we get,
Either,
x + 3 = 0
x = -3
Or,
x - 16 = 0
x = 16
Since, distance can't be negative, we have, x = 16 in.
So, the length of the base is 16 inches.
2>
umm...i don't know how to do it...can you post a picture of the same kind of (solved) question....i mean like an example
Let the base be "x"
According to the question,
h = x - 13
A = 24 in sq.
Now,
[tex] Area of triangle = \frac{1}{2}~b~h [/tex]
[tex]24= \frac{1}{2}(x)(x-13) [/tex]
[tex]24 *2= (x)(x-13) [/tex]
[tex]48= x^{2} - 13x[/tex]
[tex]0= x^{2} - 13x -48[/tex]
Factorizing the equation, x² - 13x -48, we get,
[tex]0= x^{2}+3x - 16x -48[/tex]
[tex]0= x(x+3) - 16(x+3)[/tex]
[tex]0= (x + 3)(x-16)[/tex]
NOW, using zero product property, we get,
Either,
x + 3 = 0
x = -3
Or,
x - 16 = 0
x = 16
Since, distance can't be negative, we have, x = 16 in.
So, the length of the base is 16 inches.
2>
umm...i don't know how to do it...can you post a picture of the same kind of (solved) question....i mean like an example
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