# Khan/khan-exercises

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 Area of a circle
randRange(1, 8)

Suppose the radius of a circle is \color{R_COLOR}{R}. What is its area?

Math.PI * R * R
initCircle( R ); drawRadius( R );

We know K = \pi r^2, so K = \pi \cdot \color{R_COLOR}{R}^2 = \color{K_COLOR}{R * R\pi}.

drawArea( R );
randRange(1, 8)

Suppose the area of a circle is \color{K_COLOR}{R === 1 ? "" : R * R\pi}. What is its radius?

R
initCircle( R ); drawArea( R );

We know K = \pi r^2, so r = \sqrt{K / \pi} = \sqrt{\color{K_COLOR}{R * R\pi} / \pi} = \color{R_COLOR}{R}.

randRange(1, 8)

Suppose the diameter of a circle is \color{D_COLOR}{2 * R}. What is its area?

Math.PI * R * R
initCircle( R ); drawDiameter( R );

First, find the radius: r = d/2 = \color{D_COLOR}{2 * R}/2 = \color{R_COLOR}{R}.

Now find the area: K = \pi r^2, so K = \pi \cdot \color{R_COLOR}{R}^2 = \color{K_COLOR}{R * R\pi}.

drawArea( R );
randRange(1, 8)

Suppose the area of a circle is \color{K_COLOR}{R === 1 ? "" : R * R\pi}. What is its diameter?

2 * R
initCircle( R ); drawArea( R );

First, find the radius: K = \pi r^2, so r = \sqrt{K / \pi} = \sqrt{\color{K_COLOR}{R * R\pi} / \pi} = \color{R_COLOR}{R}.

Now find the diameter: d = 2r = 2\cdot \color{R_COLOR}{R} = \color{D_COLOR}{2*R}.

drawDiameter( R );
randRange(1, 8)

Suppose the circumference of a circle is \color{C_COLOR}{2 * R\pi}. What is its area?

Math.PI * R * R
initCircle( R ); drawCircumference( R );

First, find the radius: r = c/2\pi = \color{C_COLOR}{2 * R\pi}/2\pi = \color{R_COLOR}{R}.

Now find the area: K = \pi r^2, so K = \pi \cdot \color{R_COLOR}{R}^2 = \color{K_COLOR}{R * R\pi}.

drawArea( R );
randRange(1, 8)

Suppose the area of a circle is \color{K_COLOR}{R === 1 ? "" : R * R\pi}. What is its circumference?

Math.PI * 2 * R
initCircle( R ); drawArea( R );

First, find the radius: K = \pi r^2, so r = \sqrt{K / \pi} = \sqrt{\color{K_COLOR}{R * R\pi} / \pi} = \color{R_COLOR}{R}.

Now find the circumference: c = 2\pi r = 2\pi \cdot \color{R_COLOR}{R} = \color{C_COLOR}{2*R\pi}.

drawCircumference( R );
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