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Alenkasestr [34]
1 year ago
6

Consider AABC with vertices A(-2, -3) and B(-6, -6) and C(-8, 5).a) Draw AABC on graph paper. Is this a right triangle? JUSTIFY

your answer.b) Reflect AABC across AC to create AA'B'C'. Find the location of B'. What kind oftriangle is ABB'C:? JUSTIFY your answer.
Mathematics
1 answer:
Fiesta28 [93]1 year ago
6 0

Given data:

The graph of the given points is shown below.

The slope of the line passing through A(-2, -3) and B(-6, -6).

\begin{gathered} m=\frac{-6-(-3)}{-6-(-2)} \\ =\frac{-3}{-4} \\ =\frac{3}{4} \end{gathered}

The slope of the line passing through B(-6, -6) and C(-8, 5).

\begin{gathered} m^{\prime}=\frac{5-(-6)}{-8-(-6)} \\ =-\frac{11}{2} \end{gathered}

The slope of the line passing through A(-2, -3) and C(-8, 5).

\begin{gathered} m^{\doubleprime}=\frac{5-(-3)}{-8-(-2)} \\ =\frac{8}{-6} \\ =-\frac{4}{3} \end{gathered}

The product of the slope of line passing through AB and line passing through AC is,

\begin{gathered} m\times m^{\doubleprime}=\frac{3}{4}\times(-\frac{4}{3}_{}) \\ =-1 \end{gathered}

Thus, the line passing through AB and line passing through AC are perpendicular.

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7 0
2 years ago
The perimeter of equilateral triangle ABC is 81/3 centimeters, find the length of the radius and apothem.
MAXImum [283]

There is a typo error, the perimeter of equilateral triangle ABC is 81/√3 centimeters.

Answer:

Radius = OB= 27 cm

Apothem = 13.5 cm

A diagram is attached for reference.

Step-by-step explanation:

Given,

The perimeter of equilateral triangle ABC is 81/√3 centimeters.

Substituting this in the formula of perimeter of equilateral triangle =3\times\ side

3\times\ side =[tex]81\sqrt{3}

Side = \frac{81\sqrt{3} }{3} =27\sqrt{3} \ cm

Thus from the diagram , Side AB=BC=AC= 27\sqrt{3} \ cm

We know each angle of an equilateral triangle is 60°.

From the diagram, OB is an angle bisector.

Thus \angle OBC = 30°

Apothem is the line segment from the mid point of any side to the center the equilateral triangle.

Therefore considering ΔOBE, and applying tan function.

tan\theta =\frac{perpendicular}{base} \\tan\theta=\frac{OE}{BE} \\tan\theta=\frac{OE}{\frac{27\sqrt{3}}{2}  } \\tan30\times {\frac{27\sqrt{3} }{2} }= OE\\\frac{1}{\sqrt{3} } \times\frac{27\sqrt{3} }{2} =OE\\

Thus ,apothem  OE= 13.5 cm

Now for radius,

We consider ΔOBE

cos\theta=\frac{base}{hypotenuse} \\cos30= \frac{BE}{OB} \\Cos30 = \frac{\frac{27\sqrt{3} }{2}}{OB}  \\OB= \frac{\frac{27\sqrt{3} }{2}}{cos30} \\OB= \frac{\frac{27\sqrt{3} }{2}}{\frac{\sqrt{3} }{2} } \\OB =27 \ cm

Thus for

Perimeter of equilateral triangle ABC is 81/√3 centimeters,

The radius of equilateral triangle ABC is 27 cm

The apothem of equilateral triangle ABC is 13.5 cm

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