Triangle Midsegment Theorem Worksheet: Solve for unknown side lengths using the midsegment theorem in various triangle diagrams.

Here are the step-by-step solutions for each problem on the worksheet. The key rule to remember for all of these is the Triangle Midsegment Theorem:

* A midsegment connects the midpoints of two sides of a triangle.
* The midsegment is always half the length of the third side (the base).
* Conversely, the base is always twice the length of the midsegment.

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Problem 1

Given: Triangle $EGI$. $FH$ is the midsegment.
* $FH = 2x - 9$
* $EI = 2x + 2$

Step 1: Set up the equation. Since $EI$ is the base and $FH$ is the midsegment, $EI = 2 \cdot FH$.
$$2x + 2 = 2(2x - 9)$$

Step 2: Solve for $x$.
$$2x + 2 = 4x - 18$$
Subtract $2x$ from both sides:
$$2 = 2x - 18$$
Add 18 to both sides:
$$20 = 2x$$
Divide by 2:
$$x = 10$$

Step 3: Find the lengths.
* $FH = 2(10) - 9 = 20 - 9 = 11$
* $EI = 2(10) + 2 = 20 + 2 = 22$
*(Check: Is $22$ twice $11$? Yes.)*

Answers:
$FH = 11$
$EI = 22$

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Problem 2

Given: Triangle $ABE$. $CD$ is the midsegment parallel to side $AB$.
* $CD = 8$
* $AB = x$

Step 1: Use the theorem. The base ($AB$) is twice the midsegment ($CD$).
$$AB = 2 \cdot CD$$
$$x = 2 \cdot 8$$
$$x = 16$$

Answer:
$AB = 16$

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Problem 3

Given: Triangle $EHI$. $FG$ is the midsegment.
* Base $HI = 38$
* Midsegment $FG = x$

Step 1: Use the theorem. The midsegment is half the base.
$$FG = \frac{HI}{2}$$
$$x = \frac{38}{2}$$
$$x = 19$$

Answer:
$FG = 19$

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Problem 4

Given: Triangle $PST$. $QR$ is the midsegment.
* $QR = x - 3$
* Base $ST = x + 2$

Step 1: Set up the equation. Base $ST = 2 \cdot QR$.
$$x + 2 = 2(x - 3)$$

Step 2: Solve for $x$.
$$x + 2 = 2x - 6$$
Subtract $x$ from both sides:
$$2 = x - 6$$
Add 6 to both sides:
$$8 = x$$

Step 3: Find the lengths.
* $QR = 8 - 3 = 5$
* $ST = 8 + 2 = 10$
*(Check: Is $10$ twice $5$? Yes.)*

Answers:
$QR = 5$
$ST = 10$

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Problem 5

Given: Triangle $GHM$. $KL$ is the midsegment.
* Base $GH = 2x + 8$
* Midsegment $KL = -2 + 2x$

Step 1: Set up the equation. Base $GH = 2 \cdot KL$.
$$2x + 8 = 2(-2 + 2x)$$

Step 2: Solve for $x$.
Distribute the 2 on the right side:
$$2x + 8 = -4 + 4x$$
Subtract $2x$ from both sides:
$$8 = -4 + 2x$$
Add 4 to both sides:
$$12 = 2x$$
Divide by 2:
$$6 = x$$

Step 3: Find the lengths.
* $GH = 2(6) + 8 = 12 + 8 = 20$
* $KL = -2 + 2(6) = -2 + 12 = 10$
*(Check: Is $20$ twice $10$? Yes.)*

Answers:
$GH = 20$
$KL = 10$

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Problem 6

Given: Triangle $LOP$. $MN$ is the midsegment.
* Midsegment $MN = 8$
* Base $OP = x$

Step 1: Use the theorem. The base is twice the midsegment.
$$OP = 2 \cdot MN$$
$$x = 2 \cdot 8$$
$$x = 16$$

Answer:
$OP = 16$

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Problem 7

Given: Triangle $JSL$. $MN$ is the midsegment.
* Midsegment $MN = 5$
* Base $SL = x$

Step 1: Use the theorem. The base is twice the midsegment.
$$SL = 2 \cdot MN$$
$$x = 2 \cdot 5$$
$$x = 10$$

Answer:
$SL = 10$

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Problem 8

Given: Triangle $RWX$. $SU$ is the midsegment.
* Midsegment $SU = 12$
* Base $WX = x$

Step 1: Use the theorem. The base is twice the midsegment.
$$WX = 2 \cdot SU$$
$$x = 2 \cdot 12$$
$$x = 24$$

Answer:
$WX = 24$

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Final Answer:
1. FH = 11, EI = 22
2. AB = 16
3. FG = 19
4. QR = 5, ST = 10
5. GH = 20, KL = 10
6. OP = 16
7. SL = 10
8. WX = 24