Question

kerry plans to build a picnic table. she needs to create a drawing of the picnic table surface. complete the following on a piece of paper and/or in the answer box below: • draw a regular hexagon. label the vertices of your hexagon a, b, c, d, e, and f. label the center of your hexagon o. explain the steps you used in your drawing. • classify △aof according to its sides and/or angles. use mathematics to justify your answers. • classify △abd according to its sides and/or angles. use mathematics to justify your answer. • classify △abc according to its sides and/or angles. use mathematics to justify your answer.

Explanation:

Step1: Draw a regular hexagon

1. Use a compass. Set the compass width to a desired length for the side - length of the hexagon.
2. Draw a circle with the compass. Mark a point on the circle as vertex \(A\).
3. Without changing the compass width, place the compass on point \(A\) and mark an arc that intersects the circle at point \(B\).
4. Repeat this process from point \(B\) to get point \(C\), from \(C\) to get \(D\), from \(D\) to get \(E\), and from \(E\) to get \(F\). Connect the six points \(A,B,C,D,E,F\) to form a regular hexagon. Mark the center of the circle (which is also the center of the hexagon) as \(O\).

Step2: Classify \(\triangle AOF\)

1. In a regular hexagon, the radius of the circum - circle (distance from the center to a vertex) is equal to the side - length of the hexagon. So, \(OA = OF=AF\) (since the distance from the center to a vertex is the same for all vertices and is equal to the side - length of the regular hexagon).
2. Since \(OA = OF = AF\), \(\triangle AOF\) is an equilateral triangle. All angles in an equilateral triangle are \(60^{\circ}\) (by the angle - sum property of a triangle, \(\angle AOF=\angle OAF=\angle OFA = 60^{\circ}\)).

Step3: Classify \(\triangle ABD\)

1. In a regular hexagon, the interior angle \(\angle BAF=120^{\circ}\). The central angle \(\angle AOB=\angle BOC=\angle COD=\angle DOE=\angle EOF=\angle FOA = 60^{\circ}\).
2. \(AB = AD\) (sides of the hexagon and symmetry). Let the side - length of the hexagon be \(a\). \(BD\) can be found using the law of cosines in \(\triangle ABD\). In \(\triangle ABD\), \(\angle BAD = 120^{\circ}\), \(AB = AD=a\). By the law of cosines \(BD^{2}=AB^{2}+AD^{2}-2\cdot AB\cdot AD\cdot\cos\angle BAD=a^{2}+a^{2}-2a\cdot a\cdot(-\frac{1}{2})=3a^{2}\), so \(BD=\sqrt{3}a\). Since \(AB = AD

eq BD\) and \(\angle BAD = 120^{\circ}\), \(\triangle ABD\) is an isosceles obtuse - angled triangle.

Step4: Classify \(\triangle ABC\)

1. In a regular hexagon, \(AB = BC\) (sides of the hexagon). \(\angle ABC = 120^{\circ}\) (interior angle of a regular hexagon).
2. Since \(AB = BC\) and \(\angle ABC=120^{\circ}\), \(\triangle ABC\) is an isosceles obtuse - angled triangle.

Answer:

• Drawing steps for regular hexagon are as described above.
• \(\triangle AOF\) is an equilateral triangle.
• \(\triangle ABD\) is an isosceles obtuse - angled triangle.
• \(\triangle ABC\) is an isosceles obtuse - angled triangle.