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c. What kind of triangle is shown in the first practical activity at the top of the previous page?

Mathematics

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Answer

f

360∘\mathbf{360^\circ}

Step-by-step solution

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c) Without the image of the "previous page", it's common for the first practical activity demonstrating the sum of angles in a triangle to use a general acute triangle. This type of triangle helps illustrate the principle without the special properties of right or isosceles triangles.
The kind of triangle is likely an acute triangle\boxed{\text{acute triangle}}.

Part d

Yes, the paper-folding experiment will work equally well for an obtuse triangle. The sum of angles in any Euclidean triangle, regardless of its type (acute, right, or obtuse), is always 180∘180^\circ.

e) As an AI, I cannot perform physical experiments. However, based on geometric principles, the prediction from part (d) is correct. The paper-folding experiment would successfully demonstrate that the sum of the angles in an obtuse triangle is 180∘180^\circ.

Part f

360∘\mathbf{360^\circ}

To show that the sum of the angles of a quadrilateral is 360∘360^\circ, you can use the method of dividing the quadrilateral into two triangles.

  1. Draw any quadrilateral, for example, ABCDABCD.

  2. Draw a diagonal connecting two non-adjacent vertices, for instance, from AA to CC.

    This divides the quadrilateral into two triangles: △ABC\triangle ABC and △ADC\triangle ADC.

  3. Recall that the sum of the angles in any triangle is 180∘180^\circ.

    For △ABC\triangle ABC: ∠BAC+∠ABC+∠BCA=180∘\angle BAC + \angle ABC + \angle BCA = 180^\circ.
    For △ADC\triangle ADC: ∠DAC+∠ADC+∠DCA=180∘\angle DAC + \angle ADC + \angle DCA = 180^\circ.

  4. The sum of the angles of the quadrilateral ABCDABCD is the sum of the angles of the two triangles.

    ∠DAB+∠ABC+∠BCD+∠CDA\angle DAB + \angle ABC + \angle BCD + \angle CDA

    This can be rewritten as:

    (∠DAC+∠BAC)+∠ABC+(∠BCA+∠DCA)+∠ADC(\angle DAC + \angle BAC) + \angle ABC + (\angle BCA + \angle DCA) + \angle ADC

    Rearranging the terms to group the angles of each triangle:

    (∠BAC+∠ABC+∠BCA)+(∠DAC+∠ADC+∠DCA)(\angle BAC + \angle ABC + \angle BCA) + (\angle DAC + \angle ADC + \angle DCA)

    Since each bracket sums to 180∘180^\circ:

    180∘+180∘=360∘180^\circ + 180^\circ = 360^\circ

    Thus, the sum of the angles of a quadrilateral is 360∘\boxed{\mathbf{360^\circ}}.

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