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Use the graph of △ABC with midsegments DE, EF and DF. Show that EF is parallel to AC and that EF=1/2 AC

Use The Graph Of ABC With Midsegments DE EF And DF Show That EF Is Parallel To AC And That EF12 AC class=

Sagot :

According to the midsegment theorem, the midsegments are parallel to

and half the length of the opposite side.

The completed statement are as follows;

  • Because the slopes of [tex]\overline{EF}[/tex] and [tex]\overline{AC}[/tex] are both -4, [tex]\overline{EF}[/tex] ║ [tex]\overline{AC}[/tex]
  • EF = [tex]\underline{\sqrt{17}}[/tex] and AC = [tex]\underline{2 \cdot \sqrt{17} }[/tex]
  • Because [tex]\underline{\sqrt{17} }[/tex] = [tex]\underline{\frac{1}{2} \cdot 2 \cdot \sqrt{17} }[/tex], [tex]EF = \frac{1}{2} \cdot AC[/tex]

Reasons:

From the given graph of ΔABC, we have;

Coordinates of the points A, B, and C are; A(-5, 2), B(1, -2), and C(-3, -6)

The coordinates of the point D and E on [tex]\mathbf{\overline{DE}}[/tex] are; D(-4, -2), and E(-2, 0)

The coordinates of the point F is; F(-1, -4)

  • [tex]Slope, \, m =\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}[/tex]

[tex]\displaystyle Slope \ of \ line \ \overline{AC} = \mathbf{\frac{(-6) - 2}{-3 - (-5)} = \frac{-8}{2}} = -4[/tex]

[tex]\displaystyle Slope \ of \ line \ \overline{EF} = \frac{0 - (-4)}{-2 - (-1)} = \frac{4}{-1} = -4[/tex]

  • [tex]Length \ of \ segment,\ l = \sqrt{\left (x_{2}-x_{1} \right )^{2}+\left (y_{2}-y_{1} \right )^{2}}[/tex]

Length of EF  = √((-1 - (-2))² + (-4 - 0)²) = √(17)

Length of AC = √((-3 - (-5))² + (-6 - 2)²) = √(4 × 17) = 2·√(17)

Therefore, we have;

Because the slope of [tex]\mathbf{\overline {EF}}[/tex] and [tex]\mathbf{\overline {AC}}[/tex] are both , -4, [tex]\overline {EF}[/tex] ║ [tex]\overline {AC}[/tex]. EF = [tex]\underline{\sqrt{17}}[/tex], and AC

= [tex]\underline{\frac{1}{2} \cdot 2 \cdot \sqrt{17}}[/tex],. Because [tex]\underline{\sqrt{17}}[/tex] = [tex]\underline{\frac{1}{2} \cdot 2 \cdot \sqrt{17}}[/tex], [tex]EF =\mathbf{ \frac{1}{2} AC}[/tex]

Learn more about midsegment theorem of a triangle here:

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