A=πr√(h^2-r^2) make "r" the
subject of the formula
show workings
#thanks
(A/pi)^2 = r^2 (h^2-r^2)
=-r^4+h^2r^2
r^4 - h^2 r^2 + (A/pi)^2 = 0
r^2 = [h^2 +/-sqrt(h^4-4A^2/pi^2)]/2
r = sqrt { [h^2 +/-sqrt(h^4-4A^2/pi^2)]/2 }
A=πr√(h^2-r^2)
A/π=r√(h^2-r^2)
(A/π)^2=r(h^2-r^2)
are my step right?, please help me from here. Thanks
To make "r" the subject of the formula A=πr√(h^2-r^2), we need to isolate and solve for "r". Here are the steps to achieve that:
Step 1: Start with the given formula: A = πr√(h^2 - r^2).
Step 2: Divide both sides of the equation by the constant term π: A/π = r√(h^2 - r^2).
Step 3: Square both sides of the equation to eliminate the square root: (A/π)^2 = r^2(h^2 - r^2).
Step 4: Expand the equation: A^2/π^2 = r^2(h^2 - r^2).
Step 5: Distribute r^2 to the terms inside the brackets: A^2/π^2 = r^2h^2 - r^4.
Step 6: Rearrange the equation to get all terms on one side: r^4 + r^2h^2 - A^2/π^2 = 0.
Step 7: Now, we have a quadratic equation in terms of r^2. We can solve this equation by applying the quadratic formula: r^2 = (-b ± √(b^2 - 4ac)) / 2a.
In our case, a = 1, b = h^2, and c = -A^2/π^2.
Step 8: Apply the quadratic formula: r^2 = (-(h^2) ± √((h^2)^2 - 4(1)(-A^2/π^2))) / (2 * 1)
Simplifying further will give us:
r^2 = (-h^2 ± √(h^4 + 4A^2/π^2)) / 2
Step 9: Take the square root of both sides to determine the positive and negative values of r: r = ± √((-h^2 ± √(h^4 + 4A^2/π^2)) / 2)
Both positive and negative values of r satisfy the original equation.
Therefore, we have obtained the solution for "r" as r = ± √((-h^2 ± √(h^4 + 4A^2/π^2)) / 2).
Please note that the plus/minus sign indicates we will have two solutions for "r."