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
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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
=>y = (-4/5)x+5-------(eq2)
Equation of line passing from (2,-2) and (5,1) is
y-(-2) = [[1 - (-2)] / (5-2)] / (x-2)
=>y+2=(3/3)(x-2)
=>y=x-4--------(eq3)
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 =[(-4/5)x+5 - (-7/2)x + 5)dx +
[(-4/5)x+5 -(x-4)]dx
=>area=(7/2-4/5)x dx +
((-9/5)x+9)dx
Using properties of integration,
=>area=(27/10)x dx +
(-9/5)x+9)dx
=>area=([27/10)×[5² - 2²])/2 + [ (-9/5)×(5²-1²) ]/2 +9×(5-1)
=>area=(27/20)×(25-4) + (-9/5)×24+9×4
=>area = (27×21)/20 + (-216)/5+ 36
=>area=(567/20) - (216/5) + 36
=>area= [(567-261×4)+(36×20)]/20
=>area=[(567-864)+720]/20
=>area=423/20
=>area=21 square unit.
Hence, area of triangle is 21 square unit.
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