Question

1. find the value of x. 6. find the value of x. 3. if l, m, and n are the midpoints of the sides of δpor, pr = 46, pq = 40, and ln = 17, find each measure. a) lm = b) mn = c) qr = d) mr = 4. if f, g, and h are the midpoints of the sides of δcde, fg = 9, gh = 7, and cd = 24, find each measure. a) ce = b) de = c) fh = d) perimeter of δcde.

Explanation:

Problem 3:

Step1: Apply midsegment theorem

The midsegment of a triangle is half the length of the third side. For $LM$, it is half of $PR$.
$LM = \frac{1}{2}PR = \frac{1}{2} \times 46 = 23$

Step2: Find $MN$ via midsegment theorem

$MN$ is half of $PQ$.
$MN = \frac{1}{2}PQ = \frac{1}{2} \times 40 = 20$

Step3: Find $QR$ via midsegment theorem

$LN$ is half of $QR$, so rearrange to solve for $QR$.
$QR = 2 \times LN = 2 \times 17 = 34$

Step4: Find $MR$ via midpoint definition

$M$ is the midpoint of $QR$, so $MR = \frac{1}{2}QR$.
$MR = \frac{1}{2} \times 34 = 17$

Step1: Find $CE$ via midsegment theorem

$FG$ is the midsegment corresponding to $CE$, so $CE = 2 \times FG$.
$CE = 2 \times 9 = 18$

Step2: Find $DE$ via midsegment theorem

$GH$ is the midsegment corresponding to $DE$, so $DE = 2 \times GH$.
$DE = 2 \times 7 = 14$

Step3: Find $FH$ via midsegment theorem

$FH$ is the midsegment corresponding to $CD$, so $FH = \frac{1}{2}CD$.
$FH = \frac{1}{2} \times 24 = 12$

Step4: Calculate perimeter of $\triangle CDE$

Sum the lengths of all sides of $\triangle CDE$.
Perimeter $= CE + DE + CD = 18 + 14 + 24 = 56$

Answer:

a) $LM = 23$
b) $MN = 20$
c) $QR = 34$
d) $MR = 17$

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Problem 4: