How to find if trapezoids are similar - Geometry

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Question

Given Trapezoid with bases $\overline{TR}$ and $\overline{AP}$ . The trapezoid has midsegment $\overline{MN}$ , where and are the midpoints of $\overline{TP}$ and $\overline{RA}$ , respectively.

True, false, or undetermined: Trapezoid $TRNM \sim$ Trapezoid .

Answer

The fact that the trapezoid is isosceles is actually irrelevant. Since and are the midpoints of legs $\overline{TP}$ and $\overline{RA}$ , it holds by definition that

$\overline{TM} \cong \overline{MP}$

and

$\overline{RN} \cong \overline{NA}$

It follows that $\frac{TM}{MP} = \frac{RN}{NA} = 1$

However, the bases of each trapezoid are noncongruent, by definition, so, in particular,

Assume that $\overline{AP}$ is the longer base - this argument works symmetrically if the opposite is true. Let - equivalently, .

Since the bases are of unequal length, . The length of the midsegment $\overline{MN}$ is the arithmetic mean of the lengths of these two bases, so

$MN = \frac{TR + AP}{2} = \frac{TR + TR + x }{2} = \frac{ 2 \cdot TR + x }{2} = TR + \frac{x}{2}$

Since , it follows that

$\frac{TR}{MN} e 1$ ,

and

$\frac{TR}{MN} e \frac{TM}{MP} = \frac{RN}{NA}$ .

This disproves similarity, since one condition is that corresponding sides must be in proportion.

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Question

Given Trapezoid with bases $\overline{TR}$ and $\overline{AP}$ . The trapezoid has midsegment $\overline{MN}$ , where and are the midpoints of $\overline{TP}$ and $\overline{RA}$ , respectively.

True, false, or undetermined: Trapezoid $TRNM \sim$ Trapezoid .

Answer

The fact that the trapezoid is isosceles is actually irrelevant. Since and are the midpoints of legs $\overline{TP}$ and $\overline{RA}$ , it holds by definition that

$\overline{TM} \cong \overline{MP}$

and

$\overline{RN} \cong \overline{NA}$

It follows that $\frac{TM}{MP} = \frac{RN}{NA} = 1$

However, the bases of each trapezoid are noncongruent, by definition, so, in particular,

Assume that $\overline{AP}$ is the longer base - this argument works symmetrically if the opposite is true. Let - equivalently, .

Since the bases are of unequal length, . The length of the midsegment $\overline{MN}$ is the arithmetic mean of the lengths of these two bases, so

$MN = \frac{TR + AP}{2} = \frac{TR + TR + x }{2} = \frac{ 2 \cdot TR + x }{2} = TR + \frac{x}{2}$

Since , it follows that

$\frac{TR}{MN} e 1$ ,

and

$\frac{TR}{MN} e \frac{TM}{MP} = \frac{RN}{NA}$ .

This disproves similarity, since one condition is that corresponding sides must be in proportion.

Compare your answer with the correct one above

Question

Given Trapezoid with bases $\overline{TR}$ and $\overline{AP}$ . The trapezoid has midsegment $\overline{MN}$ , where and are the midpoints of $\overline{TP}$ and $\overline{RA}$ , respectively.

True, false, or undetermined: Trapezoid $TRNM \sim$ Trapezoid .

Answer

The fact that the trapezoid is isosceles is actually irrelevant. Since and are the midpoints of legs $\overline{TP}$ and $\overline{RA}$ , it holds by definition that

$\overline{TM} \cong \overline{MP}$

and

$\overline{RN} \cong \overline{NA}$

It follows that $\frac{TM}{MP} = \frac{RN}{NA} = 1$

However, the bases of each trapezoid are noncongruent, by definition, so, in particular,

Assume that $\overline{AP}$ is the longer base - this argument works symmetrically if the opposite is true. Let - equivalently, .

Since the bases are of unequal length, . The length of the midsegment $\overline{MN}$ is the arithmetic mean of the lengths of these two bases, so

$MN = \frac{TR + AP}{2} = \frac{TR + TR + x }{2} = \frac{ 2 \cdot TR + x }{2} = TR + \frac{x}{2}$

Since , it follows that

$\frac{TR}{MN} e 1$ ,

and

$\frac{TR}{MN} e \frac{TM}{MP} = \frac{RN}{NA}$ .

This disproves similarity, since one condition is that corresponding sides must be in proportion.

Compare your answer with the correct one above

Question

Given Trapezoid with bases $\overline{TR}$ and $\overline{AP}$ . The trapezoid has midsegment $\overline{MN}$ , where and are the midpoints of $\overline{TP}$ and $\overline{RA}$ , respectively.

True, false, or undetermined: Trapezoid $TRNM \sim$ Trapezoid .

Answer

The fact that the trapezoid is isosceles is actually irrelevant. Since and are the midpoints of legs $\overline{TP}$ and $\overline{RA}$ , it holds by definition that

$\overline{TM} \cong \overline{MP}$

and

$\overline{RN} \cong \overline{NA}$

It follows that $\frac{TM}{MP} = \frac{RN}{NA} = 1$

However, the bases of each trapezoid are noncongruent, by definition, so, in particular,

Assume that $\overline{AP}$ is the longer base - this argument works symmetrically if the opposite is true. Let - equivalently, .

Since the bases are of unequal length, . The length of the midsegment $\overline{MN}$ is the arithmetic mean of the lengths of these two bases, so

$MN = \frac{TR + AP}{2} = \frac{TR + TR + x }{2} = \frac{ 2 \cdot TR + x }{2} = TR + \frac{x}{2}$

Since , it follows that

$\frac{TR}{MN} e 1$ ,

and

$\frac{TR}{MN} e \frac{TM}{MP} = \frac{RN}{NA}$ .

This disproves similarity, since one condition is that corresponding sides must be in proportion.

Compare your answer with the correct one above

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