# Khan/khan-exercises

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Update colors, wording, and angle markers

commit 8c1c334b3d1bcac66cc98db1c6af16fa0c9a2bdd 1 parent cbb135b
beneater authored
Showing with 120 additions and 141 deletions.
1. +120 −141 exercises/solving_similar_triangles_2.html
261 exercises/solving_similar_triangles_2.html
 @@ -8,7 +8,7 @@
- randomTriangleAngles.triangle() + randomTriangleAngles.triangle() [[0, 1], []] [[1], [2]] @@ -61,9 +61,6 @@
- What is the length of the side AC? -
-
var aBounding = TR_A.boundingRange(1); var bBounding = TR_D.boundingRange(1); @@ -72,40 +69,36 @@ var minY = Math.min(aBounding[1][0], bBounding[1][0]); var maxY = Math.max(aBounding[1][1], bBounding[1][1]); - init({ - range: [ [minX, maxX ], [ minY, maxY ] ] - }) + init({ range: [[minX, maxX], [minY, maxY]] }) // draw an arc representing congruent angles BAD, CAD - style({ - stroke: "green", - }); - - // draw the arc demonstrating congruent angles farther out if the angle is small + style({ stroke: GREEN }); var angle_mod = ((180 - TR_A.angles[0]) / 135) + 0.2; + arc(TR_A.points[0], angle_mod, 0, TR_A.angles[0]); + line([(angle_mod - 0.1) * cos(3 * TR_A.angles[0] / 4 * PI / 180) + TR_A.points[0][0], + (angle_mod - 0.1) * sin(3 * TR_A.angles[0] / 4 * PI / 180) + TR_A.points[0][1]], [ + (angle_mod + 0.1) * cos(3 * TR_A.angles[0] / 4 * PI / 180) + TR_A.points[0][0], + (angle_mod + 0.1) * sin(3 * TR_A.angles[0] / 4 * PI / 180) + TR_A.points[0][1]]); + line([(angle_mod - 0.1) * cos(TR_A.angles[0] / 4 * PI / 180) + TR_A.points[0][0], + (angle_mod - 0.1) * sin(TR_A.angles[0] / 4 * PI / 180) + TR_A.points[0][1]], [ + (angle_mod + 0.1) * cos(TR_A.angles[0] / 4 * PI / 180) + TR_A.points[0][0], + (angle_mod + 0.1) * sin(TR_A.angles[0] / 4 * PI / 180) + TR_A.points[0][1]]); - arc( TR_A.points[0], angle_mod, 0, TR_A.angles[0] ); - - style({ - stroke: "black", - }); + style({ stroke: "black" }); TR_B.draw(); TR_B.drawLabels(); - style({ - stroke: "black", - }); TR_C.draw(); TR_C.drawLabels(); - // draw a red line over the line segment in question - line( QUESTION_POINT_Q, QUESTION_POINT_R, { stroke: "red" } ); + // draw a pink line over the line segment in question + line(QUESTION_POINT_Q, QUESTION_POINT_R, { stroke: PINK, strokeWidth: 3 });
- roundTo( 1, TEMP_AB * TEMP_CD / TEMP_BD ) + roundTo(1, TEMP_AB * TEMP_CD / TEMP_BD) TR_A.points[0] TR_A.points[2] @@ -116,82 +109,79 @@ return pointE; }() - function(){ - var trD = new Triangle([0,0], [], 3, {}, [POINT_D, TR_A.points[1], POINT_E]); + var trD = new Triangle([0, 0], [], 3, {}, [POINT_D, TR_A.points[1], POINT_E]); trD.labels = { "points": ["", "", "E"] }; return trD; }() - -
-
- What is the length of the side AC? (Round to 1 decimal place).
+

+ What is the length of side AC? (Round to 1 decimal place) +

AC

There aren't any similar triangles in the problem figure, but we can make - a new traingle with a few lines.
+ a new triangle with a few lines.
1. Draw a line parallel to the line - \color{red}{AC}, through the point B.
+ \pink{AC}, through the point B.
2. Extend the line AD out to meet it at a new point E.

- style({ - stroke: "purple", - }); + style({ stroke: "purple" }); TR_D.draw(); TR_D.drawLabels();

- Now we have a very useful triangle - \triangle BDE which is similar to \triangle ACD. + Now we have a very useful triangle \triangle BDE + which is similar to \triangle ACD.

- Since \color{red}{AC} and BE are parallel, we - know that \angle CAE - is equal to \angle AEB. + Since \pink{AC} and BE are parallel, we + know that \angle CAE + is congruent to \angle AEB.

- style({ - stroke: "green", - }); - // draw the arc demonstrating congruent angles farther out if the angle is small + style({ stroke: GREEN }); var angle_mod_d = ((180 - TR_D.angles[2]) / 180) + 0.2; - arc( TR_D.points[2], angle_mod_d, 180 + TR_D.angles[2], 180 + TR_D.angles[2]*2 ); + arc(TR_D.points[2], angle_mod_d, 180 + TR_D.angles[2], 180 + TR_D.angles[2] * 2); + line([(angle_mod_d - 0.1) * cos((180 + TR_D.angles[2] * 1.5) * PI/180) + TR_D.points[2][0], + (angle_mod_d - 0.1) * sin((180 + TR_D.angles[2] * 1.5) * PI/180) + TR_D.points[2][1]], + [(angle_mod_d + 0.1) * cos((180 + TR_D.angles[2] * 1.5) * PI/180) + TR_D.points[2][0], + (angle_mod_d + 0.1) * sin((180 + TR_D.angles[2] * 1.5) * PI/180) + TR_D.points[2][1]]);

- It's now clear that \angle BAE - is equal to \angle AEB. + We can now tell that \text{m}\angle BAE + is equal to \text{m}\angle AEB. Meaning \triangle ABE is isosceles, and we know the length - of \color{blue}{BE}. + of BE.

- line( TR_A.points[0], TR_A.points[1], {stroke: "blue"} ); - line( TR_A.points[1], POINT_E, {stroke: "blue"} ); + line(TR_A.points[0], TR_A.points[1], { stroke: BLUE, strokeWidth: 3 }); + line(TR_A.points[1], POINT_E, { stroke: BLUE, strokeWidth: 3 }); - TR_D.labels = { "sides": [ "", AB, "" ] }; + TR_D.labels = { "sides": ["", AB, ""] }; TR_D.drawLabels();

- \dfrac{AB}{BD} = \dfrac{\color{red}{AC}}{CD} + \dfrac{AB}{BD} = \dfrac{\pink{AC}}{CD}

- \color{red}{AC} = \dfrac{AB \cdot CD}{BD} + \pink{AC} = \dfrac{AB \cdot CD}{BD}

- \color{red}{AC} = AC + \pink{AC} = AC

@@ -218,79 +208,71 @@ }()
-
- What is the length of the side AB? (Round to 1 decimal place). -
+

+ What is the length of side AB? (Round to 1 decimal place) +

AB

There aren't any similar triangles in the problem figure, but we can make - a new traingle with a few lines.
+ a new triangle with a few lines.
1. Draw a line parallel to the line - \color{red}{AB}, through the point C.
+ \pink{AB}, through the point C.
2. Extend the line AD out to meet it at a new point E.

- style({ - stroke: "purple", - }); + style({ stroke: "purple" }); TR_D.draw(); TR_D.drawLabels();

- Now we have a very useful triangle - \triangle CDE which is similar to \triangle ABD. + Now we have a very useful triangle \triangle CDE + which is similar to \triangle ABD.

- Since \color{red}{AB} and CE are parallel, we - know that \angle BAE - is equal to \angle AEB. + Since \pink{AB} and CE are parallel, we + know that \angle BAE + is congruent to \angle AEB.

- style({ - stroke: "green", - }); - // draw the arc demonstrating congruent angles farther out if the angle is small + style({ stroke: GREEN }); var angle_mod_d = ((180 - TR_D.angles[2]) / 180) + 0.2; arc(TR_D.points[2], angle_mod_d, 180, 180 + TR_D.angles[2]); + line([(angle_mod_d - 0.1) * cos((180 + TR_D.angles[2] / 2) * PI/180) + TR_D.points[2][0], + (angle_mod_d - 0.1) * sin((180 + TR_D.angles[2] / 2) * PI/180) + TR_D.points[2][1]], + [(angle_mod_d + 0.1) * cos((180 + TR_D.angles[2] / 2) * PI/180) + TR_D.points[2][0], + (angle_mod_d + 0.1) * sin((180 + TR_D.angles[2] / 2) * PI/180) + TR_D.points[2][1]]);

- It's now clear that \angle CAE - is equal to \angle AEC. + We can now tell that \text{m}\angle CAE + is equal to \text{m}\angle AEC. Meaning \triangle ACE is isosceles, and we know the length - of \color{blue}{CE}. + of \blue{CE}.

- line( TR_A.points[0], TR_A.points[2], {stroke: "blue"} ); - line( TR_A.points[2], POINT_E, {stroke: "blue"} ); - TR_D.labels = { "sides": [ "", "", AC ] }; + line(TR_A.points[0], TR_A.points[2], { stroke: BLUE, strokeWidth: 3 }); + line(TR_A.points[2], POINT_E, { stroke: BLUE, strokeWidth: 3 }); + TR_D.labels = { "sides": ["", "", AC] }; TR_D.drawLabels();
- -

- \dfrac{\color{red}{AB}}{BD} = \dfrac{AC}{CD} -

-

- \color{red}{AB} = \dfrac{ AC \cdot BD }{ CD } -

-

- \color{red}{AB} = AB -

+

\dfrac{\pink{AB}}{BD} = \dfrac{AC}{CD}

+

\pink{AB} = \dfrac{ AC \cdot BD }{ CD }

+

\pink{AB} = AB

- roundTo( 1, TEMP_AC * TEMP_BD / TEMP_AB ) - [ [ 1, 0 ], [] ] - [ [ 2 ], [1 ] ] + roundTo(1, TEMP_AC * TEMP_BD / TEMP_AB) + [[1, 0], []] + [[2], [1]] TR_A.points[2] POINT_D @@ -309,67 +291,66 @@ }()
-
- What is the length of the side CD? (Round to 1 decimal place). -
+

+ What is the length of side CD? (Round to 1 decimal place) +

CD

There aren't any similar triangles in the problem figure, but we can make - a new traingle with a few lines.
+ a new triangle with a few lines.
1. Draw a line parallel to the line AB, through the point C.
2. Extend the line AD out to meet it at a new point E.

- style({ - stroke: "purple", - }); + style({ stroke: "purple" }); TR_D.draw(); TR_D.drawLabels(); - // redraw the red line for the problem - line( QUESTION_POINT_Q, QUESTION_POINT_R, { stroke: "red" } ); + // redraw the pink line for the problem + line(QUESTION_POINT_Q, QUESTION_POINT_R, { stroke: PINK, strokeWidth: 3 });

- Now we have a very useful triangle - \triangle CDE which is similar to \triangle ABD. + Now we have a very useful triangle \triangle CDE + which is similar to \triangle ABD.

Since AB and CE are parallel, we know that - \angle BAE - is equal to \angle AEC. + \angle BAE + is congruent to \angle AEC.

- style({ - stroke: "green", - }); - // draw the arc demonstrating congruent angles farther out if the angle is small + style({ stroke: GREEN }); var angle_mod_d = ((180 - TR_D.angles[2]) / 180) + 0.2; arc( TR_D.points[2], angle_mod_d, 180, 180 + TR_D.angles[2] ); + line([(angle_mod_d - 0.1) * cos((180 + TR_D.angles[2] / 2) * PI/180) + TR_D.points[2][0], + (angle_mod_d - 0.1) * sin((180 + TR_D.angles[2] / 2) * PI/180) + TR_D.points[2][1]], + [(angle_mod_d + 0.1) * cos((180 + TR_D.angles[2] / 2) * PI/180) + TR_D.points[2][0], + (angle_mod_d + 0.1) * sin((180 + TR_D.angles[2] / 2) * PI/180) + TR_D.points[2][1]]);

- It's now clear that \angle CAE - is equal to \angle AEC. + We can now see that \text{m}\angle CAE + is equal to \text{m}\angle AEC. Meaning \triangle ACE is isosceles, and we know the length - of \color{blue}{CE}. + of \blue{CE}.

- line( TR_A.points[0], TR_A.points[2], {stroke: "blue"} ); - line( TR_A.points[2], POINT_E, {stroke: "blue"} ); - TR_D.labels = { "sides": [ "", AC, "" ] }; + line(TR_A.points[0], TR_A.points[2], { stroke: BLUE, strokeWidth: 3 }); + line(TR_A.points[2], POINT_E, { stroke: BLUE, strokeWidth: 3 }); + TR_D.labels = { "sides": ["", AC, ""] }; TR_D.drawLabels();
-

\dfrac{ AB}{BD} = \dfrac{AC}{\color{red}{CD}}

-

\color{red}{CD} = \dfrac{AC \cdot BD}{AB}

-

\color{red}{CD} = CD

+

\dfrac{ AB}{BD} = \dfrac{AC}{\pink{CD}}

+

\pink{CD} = \dfrac{AC \cdot BD}{AB}

+

\pink{CD} = CD

@@ -390,73 +371,71 @@ function(){ var trD = new Triangle([0,0], [], 3, {}, [POINT_D, TR_A.points[1], POINT_E]); - trD.labels = {"points": ["", "", "E"] }; + trD.labels = { "points": ["", "", "E"] }; return trD; }()
-
+

What is the length of the side BD? (Round to 1 decimal place). -

+

BD

- There aren't any similar triangles in the problem figure, but we can - make a new traingle with a few lines.
+ There aren't any similar triangles in the problem figure, but we can make + a new triangle with a few lines.
1. Draw a line parallel to the line AC, through the point B.
2. Extend the line AD out to meet it at a new point E.

- style({ - stroke: "purple", - }); + style({ stroke: "purple" }); TR_D.draw(); TR_D.drawLabels(); - // redraw the red line for the problem - line( QUESTION_POINT_Q, QUESTION_POINT_R, { stroke: "red" } ); + line(QUESTION_POINT_Q, QUESTION_POINT_R, { stroke: PINK, strokeWidth: 3 });

- Now we have a very useful triangle - \triangle BDE which is similar to \triangle ACD. + Now we have a very useful triangle \triangle BDE + which is similar to \triangle ACD.

Since AC and BE are parallel, we know that - \angle CAE - is equal to \angle AEB. + \angle CAE + is congruent to \angle AEB.

- style({ - stroke: "green", - }); - // draw the arc demonstrating congruent angles farther out if the angle is small + style({ stroke: GREEN }); var angle_mod_d = ((180 - TR_D.angles[2]) / 180) + 0.2; arc(TR_D.points[2], angle_mod_d, 180 + TR_D.angles[2], 180 + TR_D.angles[2] * 2); + line([(angle_mod_d - 0.1) * cos((180 + TR_D.angles[2] * 1.5) * PI/180) + TR_D.points[2][0], + (angle_mod_d - 0.1) * sin((180 + TR_D.angles[2] * 1.5) * PI/180) + TR_D.points[2][1]], + [(angle_mod_d + 0.1) * cos((180 + TR_D.angles[2] * 1.5) * PI/180) + TR_D.points[2][0], + (angle_mod_d + 0.1) * sin((180 + TR_D.angles[2] * 1.5) * PI/180) + TR_D.points[2][1]]);

- It's now clear that \angle BAE - is equal to \angle AEB. + We can now tell that \text{m}\angle BAE + is equal to \text{m}\angle AEB. Meaning \triangle ABE is isosceles, and we know the length - of \color{blue}{BE}. + of \blue{BE}.

- line(TR_A.points[0], TR_A.points[1], {stroke: "blue"}); - line(TR_A.points[1], POINT_E, {stroke: "blue"}); + line(TR_A.points[0], TR_A.points[1], { stroke: BLUE, strokeWidth: 3 }); + line(TR_A.points[1], POINT_E, { stroke: BLUE, strokeWidth: 3 }); TR_D.labels = { "sides": ["", AB, ""] }; TR_D.drawLabels();
-

\dfrac{AB}{\color{red}{BD}} = \dfrac{AC}{CD}

-

\color{red}{BD} = \dfrac{AB \cdot CD}{AC}

-

\color{red}{BD} = BD

+

\dfrac{AB}{\pink{BD}} = \dfrac{AC}{CD}

+

\pink{BD} = \dfrac{AB \cdot CD}{AC}

+

\pink{BD} = BD