# 1

Construct a right triangle ABC by first constructing segment BC, and at B construct a perpendicular line to segment BC. Construct a point at any point on the perpendicular line to BC and name it A. Join point A to C and form the required right triangle as shown below.

After constructing the triangle ABC, use the length of the sides of the triangle to construct equilateral triangles having lengths equal to one of the sides of the triangle as shown in figure 1 below. To make this more clear, let us take side BC of triangle ABC and use point B as the center of the circle and segment BC as the radius of circle. Using this points construct circle 1 then using point C as the center of the circle and CB as the radius, construct circle2. Finally construct where the two circles meet and name it point D. Carry out this procedure for all sides of the triangle ABC to come up with figure 1.

click here to see the gsp file that created figure 1 and click here for the GSP script. From the diagram above it can be noted that the sum of areas triangle (ABF and BDC) created by the two short sides of triangle ABC is equal to the area of triangle created by the longest side (AEC). To justify what I'm saying, click animation button on the gsp file or click here for the proof.

#2

Let us construct any triangle ABC and then construct any two points on any of the two sides of the triangle. For my case I constructed point F and D on segment AB and BC respectively. join point F to vertex C and also D to vertex A to construct the two segment FC and AD. Construct the point of intersection of the two segment. Now using vertex B and the point of intersection of these segment, construct a line from B through the intersection point to meet segment AC at E. Then finally construct segment BE and hide the line. To see the gsp file that constructed the triangle click here and here for the gsp script.

It appears that there is a relationship between the product of segment AE*DC*FB and segment EC*BD*AF. The product seems to be equal, try to click the animation button of the gsp file link to see what happens. It is noted that as the triangle changes its size, the above products gives the same value. Therefore we may say that this relationship

AE*DC*FB =EC*BD*AF holds. It is also noted that AE/EC, DC/BD and FB/AF gives values which do not change as the triangle changes.

#3

Use GSP to construct regular polygons with n sides for each of the following values: n = 3, 4, 6, 8, and 12.

For n = 3, this is equilateral triangle. To construct this, first construct a line and in it construct two points A and B. From point A construct a line segment to point B and then hide the line initially constructed. Using point A as the center of the circle and segment AB as the radius of the circle, construct circle c1. Then using B as the center and segment BA as the radius of the circle, construct circle c2. Finally construct the point where the two circles meet and name it C. Join point C to A and B then finally hide the circles and any points in existence. The figure constructed is at a triangle with all sides equal (equilateral triangle) as shown below.

To play with the gsp file that constructed this image click here and here for the gsp script.

For n = 4, this is a four sided figure with all its four sides equal. To construct this, first draw a line and on it construct two points A and B. Construct a line through Point A which is not parallel to AB. Use point A as the center and segment AB as the radius to construct a circle to meets the line at D. Then use point B to construct a line parallel to AD. Also using B as the center and BA as the radius, construct a circle which will meet the parallel line at C. Finally construct segment BC and join point D to C as shown below. To play with the gsp file which constructed this image, click here and here for the gsp script. By using the animation button ,it will be noted the angles inside the polygon keeps on changing until it becomes a square at a given point.

For n = 6, This is a six sided regular figure which can be constructed by following the following steps. First note that the angle at the vertex of a six sided regular figure is 120 degrees. With this in mind, start at point A and construct a line through point F. With point A and F in place, construct a segment AF as shown below. Now at point A as the center and segment AF as the radius, construct a circle which meets the line at a given point (A1 hidden). Using this point and the center of the circle, construct equilateral triangle of side length AF by intersecting circle c1 center A and c2 center A1 at B.

Using one of the sides of this equilateral triangle that meets segment AF at A, construct a point at the vertex of this triangle and name it B. Then from here extend segment AB i.e construct a line through AB. Using point B as the center of the circle and AF as the radius, construct another point A2 (hidden) by intersecting point the line and the circle. With these two points B and A2 construct equilateral triangle by intersecting circle C3 (A2 as center and AF as radius) and C4 (B as the center and AF as the radius) and name this point S. Now at S carry out the similar steps until finally end at point F. To play with the gsp file that created the image, click here and here for the gsp script.

For n = 8 this is a eight sided figure with all its sides equal, note that the angle at the vertex is 135 degrees. Which means that if I construct 45 degrees from 180 degrees, I will get this vertex angle. With this in mind, start at A and construct a line l1 (hidden) through B to create segment AB as shown below. Now at A construct a perpendicular line to l1 to form a right angle at A. Bisect this angle and using A as the center and AB as the radius construct a circle C5 which will meet the angle bisector at point point C. Now from C carry out the same process until we end at point B.

It can be noted from the diagram above sides are the same as the figure changes and the angle at the vertex remains constant (135 degrees) at all times. To verify what I'm saying click here for the animation gsp file and here for the gsp script.

For n = 12 this is a figure with twelve equal sides with vertex angle of 150 degrees. The construction of this figure is very easy but it will take some time! First start with point A

and B, then construct a line l3 (hidden) form A passing through B, taking point B as the center and AB as the radius of the circle, construct a circle circle C6 to intersect line l3 at point B1 (hidden) Now using B and B1 as base of the triangle, construct equilateral triangle. This can be done by constructing two circles, one with center at B and radius AB where as the other to have center at B1 and AB as the radius. This two circles will meet at a given point let us call this point AA. Now using point AA, B and B1 construct angle bisector. Then using point B as the radius construct a gain circle of radius AB to intersect the angle bisector at point C. From point C carry out the same procedure to finally construct point L and A for that matter. To play with the gsp file that created the above image, click here and here for the gsp script.

# 4

Pythagoras theorem states that for any right triangle ABC, right at point A and sides AB (c), AC (b) are the short sides and CB (a) the longest side, then then relationship

exists where a = CB, b = CA and

c = AB

It can be noted from the above image that the sum of the squares of the two shorter sides of a right angled triangle is equal to the square of the longest side (hypotenuse). To see what I'm illustrating, click here to play with the gsp file that created the image or click here for the prove and here for the gsp script.

# 5

(a) In this investigation that the altitude on the hypotenuse of a right triangle is the geometric mean of the segment of the hypotenuse can be investigated with the aid of the diagram below.

In the diagram above, the perpendicular CD has split right triangle ABC into two pieces, triangle ADC and triangle DC, that are smaller, similar copies of the original. We now invoke the proportionality of similar figures. Equating the ratios of hypotenuse to longer leg in triangle ADC and triangle ABC, we conclude

AC/AD = AB/AC or simply b/x = c/b by cross multiplication b^2 = cx now by letting b = m and c = y, then m^2 = xy. From the above image, the calculated values agree with this equation. To play with the file that created this image click here

(b)

In this investigation of constructing a square with the same area as a given triangle. Construct triangle BCD, and then construct a perpendicular from D meeting BC at point E, let DE be the height of the triangle and BC be the base of that triangle. Therefore we can state that the area of the triangle is 0.5(BC)*(DE). Construct segment

DE and find the mid-point F as shown above. Construct a rectangle with length GH = BC and HJ = EF as shown above. Using the rectangle, construct a square having area equal to that of BCDE. Extend line BE to the right and mark of EF with length equal to that of ED. Note that ED is equal to EF as shown below. Construct segment IF and find its midpoint. Using the midpoint as the center, construct the circle as shown below. Extend line HJ to R and construct the point of intersection of the line and the circle. Using line JR construct the square JRUS as shown below.

It can be noted that the area of the rectangle and square remains the same as we vary the area of the rectangle. It will be noted that that the area of the triangle, rectangle and square are equal. Click here to see the animation gsp file that created the above image and click here for the proof.

References: Dunham, William. Journey Through Genius: The great Theorems of Mathematics.

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