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ANEK [815]
3 years ago
14

When a triangle is inscribed in a semicircle where a diameter of the circle is a side of the​ triangle, the triangle formed is a

lways a right triangle. If an isosceles triangle​ (two equal​ sides) is inscribed in a semicircle with a radius of 9 ​inches, find the length of the two legs of the triangle.
Mathematics
1 answer:
sashaice [31]3 years ago
3 0

Answer:

Step-by-step explanation:

diameter = 18 inches

each leg = 18/√2 = 9√2 ≅ 12.7 inches

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In which quadrant is the point ( - 3 , -5) ?
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Answer:

3

Step-by-step explanation:

Since both coordinates are negative, the coordinate is in quadrant 3.

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A) y=(-0.49)x²+23.38x+996.03
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Use technology for part A.

Using the equation from part A, substitute 20 in place of x:
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Use calculus to find the area of the triangle with the vertices (0, 5), (2, -2), and (5, 1).
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The area of the triangle with the vertices (0, 5), (2, -2), and (5, 1) by using the calculus is  21 square unit.

We need to find the equation among all possible pairs and then integrate the equations from one co-ordinate to another co-ordinate

Equation of line passing through (0,5) and (2,-2) is

y-5 = [(-2-5)/(2-0)](x-0)

=>y-5 = (-7) /2x

=>y= (-7/2x)+5 -------(eq1)

Equation of line passing through (0,5) and (5,1) is

y-5 = [(1-5) / (5-0)](x-0)

=>y-5 = (-4/5)x

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Equation of line passing from (2,-2) and (5,1) is

y-(-2) = [[1 - (-2)] / (5-2)] / (x-2)

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Now, we use definite integration to find the area between the different equation of line.

So, area enclosed between the equations is given by the

area =\int\limits^5_2[(-4/5)x+5 - (-7/2)x + 5)dx  + \int\limits^5_1[(-4/5)x+5 -(x-4)]dx

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Using properties of integration,\int\limits x\, dx=x^{2}/2

=>area=\int\limits^5_2(27/10)x dx + \int\limits^5_1(-9/5)x+9)dx

=>area=([27/10)×[5² - 2²])/2 + [ (-9/5)×(5²-1²) ]/2 +9×(5-1)

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To know more about area of triangle, visit here:

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