Gregory brought 11 pumpkins to school for pumpkin-carving contest. Each pumpkin
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The equation which can be used to find the length of EF is option A which is EF=.
Given the length of base of right angled triangle is 12 and the hypotenuse be 24.
We are to find the length of EF and tell which equation can be used to find the length of EF.
We know that pythagoras theorem applies in a right angled triangle.
Pythagoras theorem says that in a right angled triangle the square of the hypotenuse is equal to the sum of square of base and perpendicular.
In this triangle we are only given base and hypotenuse and we have to find EF which is perpendicular.
EF=
=--1
=
=
Hence the equation which can be used to find the length of Ef is option A which is EF=.
Question is incomplete as it should include the figure also.
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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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