Khan/khan-exercises

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 Angles of a polygon
randRange( 5, 7 )
180 * ( SIDES - 2 ) {}

What is the sum of this polygon's interior angles?

init({ range: [ [ -5, 5 ], [ -1, 5 ] ], scale: [ 40, 40 ] }); graph.polygon = new Polygon( SIDES ); graph.polygon.draw(); CLONE = graph.polygon.clone();

There are a couple of ways to approach this problem.

Since this polygon has SIDES sides, we can draw SIDES triangles that all meet in the center.

We can combine all the triangles' angles, and then we must subtract 360^{\circ} because the circle in the middle is extra.

There are 180^{\circ} in a triangle.

\begin{align*}&SIDES \times 180^{\circ} - 360^{\circ} \\ &= SIDES * 180^{\circ} - 360^{\circ} \\ &= ANSWER^{\circ}\end{align*}

An alternative approach is shown below.

We can use four of the cardinalThrough20( SIDES ) sides to make two triangles, as shown in orange.

init({ range: [ [ -5, 5 ], [ -1, 5 ] ] }); graph.polygon = CLONE; graph.polygon.draw(); graph.polygon.drawDiagonals( randRange( 0, SIDES - 1 ) );

There is one side between the orange triangles, to make one additional triangle.

There are SIDES - 4 sides between the orange triangles, to make SIDES - 4 additional triangles.

We chopped this polygon into SIDES - 2 triangles, and each triangle's angles sum to 180^{\circ}.

SIDES - 2 \times 180^{\circ} = ANSWER^{\circ}

The sum of the polygon's interior angles is ANSWER^{\circ}.

What is the sum of this polygon's exterior angles?

init({ range: [ [ -6, 6 ], [ -2, 7 ] ] }); graph.polygon = new Polygon( SIDES ); graph.polygon.draw();
360 \Large{^\circ}

The exterior angles are shown above.

graph.polygon.drawExteriorAngles();
graph.polygon.animateExteriorAngles( randRange( 0, SIDES - 1 ) );

The exterior angles fit together to form a circle.

Therefore, the sum of the exterior angles is 360^{\circ}.