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Answer:
4.24
Step-by-step explanation:
First, let's set up the equation: [tex]\frac{2+x^2}{2}=10[/tex]
Next, clear the fraction by multiplying both sides by 2:[tex]2*(\frac{2+x^2}{2}=10)=2+x^2=20[/tex]
Then, subtract 2 from both sides: [tex]x^2=18[/tex]
Finally, take a square root of both sides and round: [tex]\sqrt{x^2=18} =[/tex](Rounded to the nearest hundredth) 4.24
We will see that the base of the triangle measures 4 units.
For a triangle of base b and height h, the area is:
A = b*h/2
In this case, we know that:
A = 10 square units.
h = b + 2
Then we can write:
10 = b*h/2
If we replace the second equation into the above one, we get:
10 = b*(b + 2)/2
Now we can solve this for b:
[tex]20 = b^2 + b[/tex]
Then we need to solve the quadratic equation:
[tex]b^2 + b - 20 = 0[/tex]
The solutions are given by Bhaskara's formula.
[tex]b = \frac{-1 \pm \sqrt{1^2 - 4*1*(-20)} }{2} \\\\b = \frac{-1 \pm 9 }{2}[/tex]
The solution that we care for is the positive one:
b = (-1 + 9)/2 = 4
The base measures 4 units.
If you want to learn more about triangles:
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