# Khan/khan-exercises

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 Special right triangles
randRange(2, 7)

In the right triangle shown, AC = BC = AC.

What is AB?

betterTriangle( 1, 1, "A", "B", "C", AC, AC, "x" );
AC * AC * 2

We know the length of each leg, and want to find the length of the hypotenuse. What mathematical relationship is there between a right triangle's leg and its hypotenuse?

\qquad AC^2 + BC^2 = x^2

\qquad AC^2 + AC^2 = x^2

\qquad x^2 = 2 \cdot AC^2

\qquad x = \sqrt{2 \cdot AC^2}

\qquad x = \sqrt{2} \cdot \sqrt{AC^2}

2 * randRange(2, 6)

In the right triangle shown, AC = BC and AB = AB.

How long are each of the legs?

betterTriangle( 1, 1, "A", "B", "C", "x", "x", AB );
AB * AB / 2

We know the length of the hypotenuse, and want to find the length of each leg. What mathematical relationship is there between a right triangle's legs and its hypotenuse?

\qquad x^2 + x^2 = AB^2

\qquad 2 \cdot x^2 = AB^2

\qquad x^2 = AB * AB / 2

\qquad x = \sqrt{AB * AB / 2}

\qquad x = \sqrt{AB * AB / 4 \cdot 2}

\qquad x = \sqrt{AB * AB / 4} \cdot \sqrt{2}

2 * randRange(2, 6)

In the right triangle shown, AC = BC and AB = AB\sqrt{2}.

How long are each of the legs?

betterTriangle( 1, 1, "A", "B", "C", "x", "x", AB + "\\sqrt{2}" );
AB * AB

We know the length of the hypotenuse, and want to find the length of each leg. What mathematical relationship is there between a right triangle's legs and its hypotenuse?

\qquad x^2 + x^2 = AB^2

\qquad 2 \cdot x^2 = (AB\sqrt{2})^2

\qquad 2 \cdot x^2 = AB^2 \cdot (\sqrt{2})^2

\qquad 2 \cdot x^2 = AB * AB \cdot 2

\qquad x^2 = AB * AB

randRange(2, 6) randFromArray([[1, ""], [3, "\\sqrt{3}"]]) BC + BCrs rand(2) ["\\angle A = 30^\\circ", "\\angle B = 60^\\circ"][ANGLE]

In the right triangle shown, mAB and BC = BC + BCrs.

How long is AB?

betterTriangle(1, sqrt(3), "A", "B", "C", BC + BCrs, "", "x"); if (ANGLE === 0) { arc([0, 5 * sqrt(3) / 2], 0.8, 270, 300); label([-0.1, (5 * sqrt(3) / 2) - 1], "{30}^{\\circ}", "below right"); } else { arc([5/2,0], 0.5, 120, 180); label([5/2-0.2, 0], "{60}^{\\circ}", "above left"); }
4 * BC * BC * BCr

We know the length of a leg, and want to find the length of the hypotenuse. What is the relationship between the sides of a 30 - 60 - 90 triangle?

This is a 30-60-90 triangle with a hypotenuse of length 1.

betterTriangle(1, sqrt(3), "A", "B", "C", "\\dfrac{1}{2}", "\\dfrac{\\sqrt{3}}{2}", 1); arc([0, 5 * sqrt(3) / 2], 0.8, 270, 300); arc([5/2,0], 0.5, 120, 180); label([-0.1, (5 * sqrt(3) / 2) - 1], "{30}^{\\circ}", "below right"); label([5/2-0.2, 0], "{60}^{\\circ}", "above left");

The ratio of AB : BC is 1 : \dfrac{1}{2}.

Therefore, \dfrac{x}{BC + BCrs} = \dfrac{1}{\frac{1}{2}} = 2.

x = 2 \cdot BC + BCrs

x = BC * 2 + BCrs

3 * randRange( 2, 6 ) randFromArray([ [1, "", (AC * 2 / 3) + "\\sqrt{3}", AC * AC * 4 / 3], [3, "\\sqrt{3}", (AC * 2), AC * AC * 4] ]) AC + ACrs rand(2) ["\\angle A = 30^\\circ", "\\angle B = 60^\\circ"][ANGLE]

In the right triangle shown, mAB and AC = AC + ACrs.

How long is AB?

betterTriangle(1, sqrt(3), "A", "B", "C", "", AC + ACrs, "x"); if (ANGLE === 0) { arc([0, 5 * sqrt(3) / 2], 0.8, 270, 300); label([-0.1, (5 * sqrt(3) / 2) - 1], "{30}^{\\circ}", "below right"); } else { arc([5/2,0], 0.5, 120, 180); label([5/2-0.2, 0], "{60}^{\\circ}", "above left"); }
AB

We know the length of a leg, and want to find the length of the hypotenuse. What is the relationship between the sides of a 30 - 60 - 90 triangle?

This is a 30-60-90 triangle with a hypotenuse of length 1.

betterTriangle(1, sqrt(3), "A", "B", "C", "\\dfrac{1}{2}", "\\dfrac{\\sqrt{3}}{2}", 1); arc([0, 5 * sqrt(3) / 2], 0.8, 270, 300); arc([5 / 2,0], 0.5, 120, 180); label([-0.1, (5 * sqrt(3) / 2) - 1], "{30}^{\\circ}", "below right"); label([5 / 2 - 0.2, 0], "{60}^{\\circ}", "above left");

The ratio of AB : AC is 1 : \dfrac{\sqrt{3}}{2}.

Therefore, \dfrac{x}{AC + ACrs} = \dfrac{1}{\frac{\sqrt{3}}{2}} = \dfrac{2}{\sqrt{3}}.

x = \dfrac{2}{\sqrt{3}} \cdot AC + ACrs

x = \dfrac{2 * AC}{\sqrt{3}}

x = \dfrac{2 * AC}{\sqrt{3}} \cdot \dfrac{\sqrt{3}}{\sqrt{3}}

x = \dfrac{2 * AC \cdot \sqrt{3}}{3}

So, x = ABs.

randRange( 2, 6 ) randFromArray([ [1, ""], [3, "\\sqrt{3}"] ]) 2*BC + BCrs rand(2) ["\\angle A = 30^\\circ", "\\angle B = 60^\\circ"][ANGLE]

In the right triangle shown, mAB and AB = 2 * BC + BCrs.

How long is BC?

betterTriangle(1, sqrt(3), "A", "B", "C", "x", "", 2 * BC + BCrs); if (ANGLE === 0) { arc([0, 5 * sqrt(3) / 2], 0.8, 270, 300); label([-0.1, (5 * sqrt(3) / 2) - 1], "{30}^{\\circ}", "below right"); } else { arc([5/2,0], 0.5, 120, 180); label([5/2-0.2, 0], "{60}^{\\circ}", "above left"); }
BC * BC * BCr

We know the length of the hypotenuse, and want to find the length of the shortest side. What is the relationship between the sides of a 30 - 60 - 90 triangle?

This is a 30-60-90 triangle with a hypotenuse of length 1.

betterTriangle(1, sqrt(3), "A", "B", "C", "\\dfrac{1}{2}", "\\dfrac{\\sqrt{3}}{2}", 1); arc([0, 5 * sqrt(3) / 2], 0.8, 270, 300); arc([5/2,0], 0.5, 120, 180); label([5/2-0.2, 0], "{60}^{\\circ}", "above left"); label([-0.1, (5 * sqrt(3) / 2) - 1], "{30}^{\\circ}", "below right");

The ratio of BC : AB is \dfrac{1}{2} : 1.

Therefore, \dfrac{x}{ABdisp} = \dfrac{1}{2}.

x = BC + BCrs

3 * randRange( 2, 6 ) randFromArray([ [1, "", (AC * 2 / 3) + "\\sqrt{3}", AC * AC * 4 / 3], [3, "\\sqrt{3}", (AC * 2), AC * AC * 4] ]) rand(2) ["\\angle A = 30^\\circ", "\\angle B = 60^\\circ"][ANGLE]

In the right triangle shown, mAB and AB = ABs.

How long is AC?

betterTriangle(1, sqrt(3), "A", "B", "C", "", "x", ABs); if (ANGLE === 0) { arc([0, 5 * sqrt(3) / 2], 0.8, 270, 300); label([-0.1, (5 * sqrt(3) / 2) - 1], "{30}^{\\circ}", "below right"); } else { arc([5/2,0], 0.5, 120, 180); label([5/2-0.2, 0], "{60}^{\\circ}", "above left"); }
AC * AC * ACr

We know the length of the hypotenuse, and want to find the length of the longest side. What is the relationship between the sides of a 30 - 60 - 90 triangle?

This is a 30-60-90 triangle with a hypotenuse of length 1.

betterTriangle(1, sqrt(3), "A", "B", "C", "\\dfrac{1}{2}", "\\dfrac{\\sqrt{3}}{2}", 1); arc([0, 5 * sqrt(3) / 2], 0.8, 270, 300); arc([5 / 2,0], 0.5, 120, 180); label([-0.1, (5 * sqrt(3) / 2) - 1], "{30}^{\\circ}", "below right"); label([5 / 2 - 0.2, 0], "{60}^{\\circ}", "above left");

The ratio of AC : AB is \dfrac{\sqrt{3}}{2} : 1.

Therefore, \dfrac{x}{ABs} = \dfrac{\sqrt{3}}{2}.

x = \dfrac{\sqrt{3}}{2} \cdot ABs

x = AC / 3 \cdot \sqrt{3} \cdot \sqrt{3}

So, x = AC + ACrs.