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aivan3 [116]
3 years ago
11

Prove that the segments joining the midpoint of consecutive sides of an isosceles trapezoid form a rhombus.

Mathematics
1 answer:
sergiy2304 [10]3 years ago
5 0

Answer:

See explanation

Step-by-step explanation:

a) To prove that DEFG is a rhombus, it is sufficient to prove that:

  1. All the sides of the rhombus are congruent:  |DG|\cong |GF| \cong |EF| \cong |DE|
  2. The diagonals are perpendicular

Using the distance formula; d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

|DG|=\sqrt{(0-(-a-b))^2+(0-c)^2}

\implies |DG|=\sqrt{a^2+b^2+c^2+2ab}

|GF|=\sqrt{((a+b)-0)^2+(c-0)^2}

\implies |GF|=\sqrt{a^2+b^2+c^2+2ab}

|EF|=\sqrt{((a+b)-0)^2+(c-2c)^2}

\implies |EF|=\sqrt{a^2+b^2+c^2+2ab}

|DE|=\sqrt{(0-(-a-b))^2+(2c-c)^2}

\implies |DE|=\sqrt{a^2+b^2+c^2+2ab}

Using the slope formula; m=\frac{y_2-y_1}{x_2-x_1}

The slope of EG is m_{EG}=\frac{2c-0}{0-0}

\implies m_{EG}=\frac{2c}{0}

The slope of EG is undefined hence it is a vertical line.

The slope of  DF is m_{DF}=\frac{c-c}{a+b-(-a-b)}

\implies m_{DF}=\frac{0}{2a+2b)}=0

The slope of DF is zero, hence it is a horizontal line.

A horizontal line meets a vertical line at 90 degrees.

Conclusion:

Since |DG|\cong |GF| \cong |EF| \cong |DE| and DF \perp FG , DEFG is a rhombus

b) Using the slope formula:

The slope of DE is m_{DE}=\frac{2c-c}{0-(-a-b)}

m_{DE}=\frac{c}{a+b)}

The slope of FG is m_{FG}=\frac{c-0}{a+b-0}

\implies m_{FG}=\frac{c}{a+b}

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-5u(t)=\begin{cases}0&\text{for }t

If we shift the argument by 1 and scale by -5, we have

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