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    http://files.physicsandmathstutor.co...ce%20Paper.pdf

    In Q2(a), how do we find the area of the triangle? I'm confused with the solution that's given here in the solution bank.
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    (Original post by sabahshahed294)
    http://files.physicsandmathstutor.co...ce%20Paper.pdf

    In Q2(a), how do we find the area of the triangle? I'm confused with the solution that's given here in the solution bank.
    POQ is a right-angle triangle. P is on the y-axis so that's when the curve has x=0. That will be the height of the triangle. Q is found by forming an equation which is tangent to P then seeing where it crosses the x-axis. This will be the length of the triangle.

    1/2 * Height * length = your area.

    Red is the curve, blue is the tangent line, and green is the area you need:
    Name:  SDSAD.PNG
Views: 40
Size:  12.9 KB
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    (Original post by sabahshahed294)
    http://files.physicsandmathstutor.co...ce%20Paper.pdf

    In Q2(a), how do we find the area of the triangle? I'm confused with the solution that's given here in the solution bank.
    The area of a triangle is  \frac{1}{2}bh . P is where the curve meets the y axis so the y coordinate of P is the height of the triangle. Where the tangent to the curve meets the x axis (Q) is the other corner, and since the triangle is right angled the x coordinate of Q must be the base. So you need to find the coordinates of P and Q. Start with P, which is where the curve intersects the y axis.
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    (Original post by RDKGames)
    POQ is a right-angle triangle. P is on the y-axis so that's when the curve has x=0. That will be the height of the triangle. Q is found by forming an equation which is tangent to P then seeing where it crosses the x-axis. This will be the length of the triangle.

    1/2 * Height * length = your area.

    Red is the curve, blue is the tangent line, and green is the area you need:
    Name:  SDSAD.PNG
Views: 40
Size:  12.9 KB
    Your diagram helped me out, thanks!

    (Original post by sindyscape62)
    The area of a triangle is  \frac{1}{2}bh . P is where the curve meets the y axis so the y coordinate of P is the height of the triangle. Where the tangent to the curve meets the x axis (Q) is the other corner, and since the triangle is right angled the x coordinate of Q must be the base. So you need to find the coordinates of P and Q. Start with P, which is where the curve intersects the y axis.
    Yes, I got what to do now. Thanks for your time!
 
 
 
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