Question

discussion
a. repeat problem 2 with each of the following. does the pythagorean theorem work in these cases? if it fails, how?
a. an acute triangle (crossed out) the
b. an obtuse triangle
b. in what parts of problem 5 did you not use the pythagorean theorem? how did you find those distances?
c. find as many geoboard isosceles triangles as possible whose legs share a vertex at the origin and whose bases are not horizontal, vertical, or at a 45° angle. (use euclidean distance for this puzzle. limit yourself to examples that can be found on an 11 × 11 geoboard. problem 6 provides a hint.) record your findings on graph or dot paper.

Explanation:

Part A

a. Acute Triangle

Step1: Recall Pythagorean Theorem

The Pythagorean theorem states that for a right - triangle with legs of length \(a\) and \(b\) and hypotenuse of length \(c\), \(a^{2}+b^{2}=c^{2}\). For an acute triangle with side lengths \(a\), \(b\), and \(c\) (where \(c\) is the longest side), we have the inequality \(a^{2}+b^{2}>c^{2}\). So the Pythagorean theorem (the equation \(a^{2}+b^{2}=c^{2}\)) does not hold in the same way as in a right - triangle. The Pythagorean theorem is specific to right - triangles. In an acute triangle, the sum of the squares of the two shorter sides is greater than the square of the longest side.

Step2: Conclusion for Acute Triangle

The Pythagorean theorem (the equality \(a^{2}+b^{2}=c^{2}\)) does not work for an acute triangle. Instead, the relationship is \(a^{2}+b^{2}>c^{2}\) (where \(c\) is the longest side of the acute triangle).

b. Obtuse Triangle

Step1: Recall the inequality for Obtuse Triangle

For an obtuse triangle with side lengths \(a\), \(b\), and \(c\) (where \(c\) is the longest side, opposite the obtuse angle), the relationship between the sides is given by the inequality \(a^{2}+b^{2}

Step2: Conclusion for Obtuse Triangle

The Pythagorean theorem does not work for an obtuse triangle. The correct relationship between the sides of an obtuse triangle (with \(c\) as the longest side) is \(a^{2}+b^{2}

Part B

Since we don't have the details of Problem 5, we can provide a general approach. If Problem 5 involved non - right - triangle figures (like acute or obtuse triangles, or other geometric shapes like quadrilaterals that are not right - angled), we would not use the Pythagorean theorem. For example, if we were finding the distance between two points in a non - right - angled context, we might use the distance formula (which is derived from the Pythagorean theorem, but if the figure is not a right - triangle, we can't directly apply \(a^{2}+b^{2}=c^{2}\) in the same way). Or if we were dealing with shapes like circles, we would use properties of circles (radius, diameter, circumference formulas) instead of the Pythagorean theorem. If we were finding distances in a one - dimensional context (like along a straight line on a number line), we would use the absolute difference of the coordinates (\(d=\vert x_2 - x_1\vert\)) instead of the Pythagorean theorem.

Part C

To find geoboard isosceles triangles with legs sharing a vertex at the origin and bases not horizontal, vertical, or at \(45^{\circ}\):

1. Let the two legs of the isosceles triangle have endpoints \((x_1,y_1)\) and \((x_2,y_2)\) with the origin \((0,0)\) as the common vertex. The length of the first leg \(L_1=\sqrt{(x_1 - 0)^{2}+(y_1 - 0)^{2}}=\sqrt{x_1^{2}+y_1^{2}}\), and the length of the second leg \(L_2=\sqrt{(x_2 - 0)^{2}+(y_2 - 0)^{2}}=\sqrt{x_2^{2}+y_2^{2}}\). For an isosceles triangle, \(L_1 = L_2\), so \(x_1^{2}+y_1^{2}=x_2^{2}+y_2^{2}\).

1. Let's find an example:

• Let the first leg go from \((0,0)\) to \((2,3)\). The length of this leg is \(\sqrt{2^{2}+3^{2}}=\sqrt{4 + 9}=\sqrt{13}\).
• Let the second leg go from \((0,0)\) to \((3,2)\). The length of this leg is \(\sqrt{3^{2}+2^{2}}=\sqrt{9 + 4}=\sqrt{13}\).
• Now, we need to find the base. The base is the distance between \((2,3)\) and \((3,2)\). Using the distance formula \(d=\sqrt{(x_2 - x_1)^{2}+(y_2 - y_1)^{2}}\), we have \(d=\sqrt{(3 - 2)^{2}+(2 - 3)^{2}}=\sqrt{1+( - 1)^{2}}=\sqrt{2}\). The base is not horizontal, vertical, or at \(45^{\circ}\) (the slope of the base is \(\frac{2 - 3}{3 - 2}=- 1\), but the base is a line segment between \((2,3)\) and \((3,2)\), and the triangle is isosceles with legs from the origin to \((2,3)\) and \((3,2)\)).

Another example:

• Leg 1: from \((0,0)\) to \((1,4)\), length \(=\sqrt{1^{2}+4^{2}}=\sqrt{1 + 16}=\sqrt{17}\).
• Leg 2: from \((0,0)\) to \((4,1)\), length \(=\sqrt{4^{2}+1^{2}}=\sqrt{16 + 1}=\sqrt{17}\).
• Base: distance between \((1,4)\) and \((4,1)\) is \(\sqrt{(4 - 1)^{2}+(1 - 4)^{2}}=\sqrt{9+( - 3)^{2}}=\sqrt{9 + 9}=\sqrt{18}=3\sqrt{2}\). The base is not horizontal, vertical, or at \(45^{\circ}\) (slope of base is \(\frac{1 - 4}{4 - 1}=-1\), but the base segment is between \((1,4)\) and \((4,1)\)).

We can continue to find more such triangles by choosing pairs of points \((x,y)\) and \((y,x)\) (where \(x
eq…

Answer:

Step1: Recall the inequality for Obtuse Triangle

For an obtuse triangle with side lengths \(a\), \(b\), and \(c\) (where \(c\) is the longest side, opposite the obtuse angle), the relationship between the sides is given by the inequality \(a^{2}+b^{2}

Step2: Conclusion for Obtuse Triangle

The Pythagorean theorem does not work for an obtuse triangle. The correct relationship between the sides of an obtuse triangle (with \(c\) as the longest side) is \(a^{2}+b^{2}

Part B

Since we don't have the details of Problem 5, we can provide a general approach. If Problem 5 involved non - right - triangle figures (like acute or obtuse triangles, or other geometric shapes like quadrilaterals that are not right - angled), we would not use the Pythagorean theorem. For example, if we were finding the distance between two points in a non - right - angled context, we might use the distance formula (which is derived from the Pythagorean theorem, but if the figure is not a right - triangle, we can't directly apply \(a^{2}+b^{2}=c^{2}\) in the same way). Or if we were dealing with shapes like circles, we would use properties of circles (radius, diameter, circumference formulas) instead of the Pythagorean theorem. If we were finding distances in a one - dimensional context (like along a straight line on a number line), we would use the absolute difference of the coordinates (\(d=\vert x_2 - x_1\vert\)) instead of the Pythagorean theorem.

Part C

To find geoboard isosceles triangles with legs sharing a vertex at the origin and bases not horizontal, vertical, or at \(45^{\circ}\):

1. Let the two legs of the isosceles triangle have endpoints \((x_1,y_1)\) and \((x_2,y_2)\) with the origin \((0,0)\) as the common vertex. The length of the first leg \(L_1=\sqrt{(x_1 - 0)^{2}+(y_1 - 0)^{2}}=\sqrt{x_1^{2}+y_1^{2}}\), and the length of the second leg \(L_2=\sqrt{(x_2 - 0)^{2}+(y_2 - 0)^{2}}=\sqrt{x_2^{2}+y_2^{2}}\). For an isosceles triangle, \(L_1 = L_2\), so \(x_1^{2}+y_1^{2}=x_2^{2}+y_2^{2}\).

1. Let's find an example:

• Let the first leg go from \((0,0)\) to \((2,3)\). The length of this leg is \(\sqrt{2^{2}+3^{2}}=\sqrt{4 + 9}=\sqrt{13}\).
• Let the second leg go from \((0,0)\) to \((3,2)\). The length of this leg is \(\sqrt{3^{2}+2^{2}}=\sqrt{9 + 4}=\sqrt{13}\).
• Now, we need to find the base. The base is the distance between \((2,3)\) and \((3,2)\). Using the distance formula \(d=\sqrt{(x_2 - x_1)^{2}+(y_2 - y_1)^{2}}\), we have \(d=\sqrt{(3 - 2)^{2}+(2 - 3)^{2}}=\sqrt{1+( - 1)^{2}}=\sqrt{2}\). The base is not horizontal, vertical, or at \(45^{\circ}\) (the slope of the base is \(\frac{2 - 3}{3 - 2}=- 1\), but the base is a line segment between \((2,3)\) and \((3,2)\), and the triangle is isosceles with legs from the origin to \((2,3)\) and \((3,2)\)).

Another example:

• Leg 1: from \((0,0)\) to \((1,4)\), length \(=\sqrt{1^{2}+4^{2}}=\sqrt{1 + 16}=\sqrt{17}\).
• Leg 2: from \((0,0)\) to \((4,1)\), length \(=\sqrt{4^{2}+1^{2}}=\sqrt{16 + 1}=\sqrt{17}\).
• Base: distance between \((1,4)\) and \((4,1)\) is \(\sqrt{(4 - 1)^{2}+(1 - 4)^{2}}=\sqrt{9+( - 3)^{2}}=\sqrt{9 + 9}=\sqrt{18}=3\sqrt{2}\). The base is not horizontal, vertical, or at \(45^{\circ}\) (slope of base is \(\frac{1 - 4}{4 - 1}=-1\), but the base segment is between \((1,4)\) and \((4,1)\)).

We can continue to find more such triangles by choosing pairs of points \((x,y)\) and \((y,x)\) (where \(x
eq y\) and \(x,y\in\{1,2,\cdots,10\}\) since the geoboard is \(11\times11\) with coordinates from \((0,0)\) to \((10,10)\)) to ensure that the legs from the origin are equal in length, and then checking that the base is not horizontal, vertical, or at \(45^{\circ}\).

Final Answers (Summarized)

Part A

a. The Pythagorean theorem (equality \(a^{2}+b^{2}=c^{2}\)) does not work. For an acute triangle with longest side \(c\), \(a^{2}+b^{2}>c^{2}\).
b. The Pythagorean theorem (equality \(a^{2}+b^{2}=c^{2}\)) does not work. For an obtuse triangle with longest side \(c\), \(a^{2}+b^{2}

Part B

(Depends on Problem 5 details, general: non - right - triangle related parts, e.g., 1D distances, non - right - angled shapes; distances found via absolute difference (1D) or other geometric properties).

Part C

Examples include triangles with vertices \((0,0)\), \((2,3)\), \((3,2)\) and \((0,0)\), \((1,4)\), \((4,1)\) (and more by choosing \((x,y)\) and \((y,x)\) with \(x
eq y\) on the \(11\times11\) geoboard).