in each figure below, $ \overline{de} $ is a midsegment of triangle $ abc $. find the value of $ x $

2.
\( \begin{tikzpicture}scale=0.8
\coordinate (a) at (0,0);
\coordinate (b) at (1,2);
\coordinate (c) at (3,0);
\coordinate (e) at (1,0);
\coordinate (d) at (2,1);
\draw (a)--(b)--(c)--(a);
\draw (e)--(d);
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ode at (0.5,1.5) {$x$};
\
ode at (1.5,0.5) {$7$};
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ode at (a) {$a$};
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ode at (b) {$b$};
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ode at (d) {$d$};
\end{tikzpicture}

3.
$ \begin{tikzpicture}scale=0.8
\coordinate (a) at (0,0);
\coordinate (b) at (2,2);
\coordinate (c) at (3,0);
\coordinate (e) at (1,0);
\coordinate (d) at (1,1);
\draw (a)--(b)--(c)--(a);
\draw (d)--(e);
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ode at (0.5,0) {$x$};
\
ode at (2,0) {$8$};
\
ode at (a) {$a$};
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ode at (b) {$b$};
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ode at (c) {$c$};
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ode at (e) {$e$};
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ode at (d) {$d$};
\end{tikzpicture}

4.
\( \begin{tikzpicture}scale=0.8
\coordinate (a) at (0,0);
\coordinate (b) at (0,2);
\coordinate (c) at (3,0);
\coordinate (d) at (0,1);
\coordinate (e) at (1,1);
\draw (a)--(b)--(c)--(a);
\draw (d)--(e);
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ode at (0.5,1) {$x$};
\
ode at (1.5,0) {$34$};
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ode at (a) {$a$};
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ode at (b) {$b$};
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ode at (c) {$c$};
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ode at (d) {$d$};
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ode at (e) {$e$};
\end{tikzpicture}

5. \( \overline{df} $ is a median of $ \triangle cde $. if $ ec = (8x) $ feet and $ cf = (x + 9) $ feet, find $ x $ and $ ef $.
$ \begin{tikzpicture}scale=0.8
\coordinate (e) at (0,0);
\coordinate (f) at (1,0);
\coordinate (c) at (2,0);
\coordinate (d) at (1,2);
\draw (e)--(f)--(c)--(d)--(e);
\draw (d)--(f);
\
ode at (e) {$e$};
\
ode at (f) {$f$};
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ode at (c) {$c$};
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ode at (d) {$d$};
\end{tikzpicture}

6. in \( \triangle rst $, $ \overline{ru} $ and $ \overline{sv} $ are medians. $ su = 3x + 16 $, $ tu = 8 + 4x $, and $ rv = 7x - 20 $, find $ vt $.
$ \begin{tikzpicture}scale=0.8
\coordinate (r) at (0,2);
\coordinate (s) at (0,0);
\coordinate (t) at (3,2);
\coordinate (v) at (1.5,2);
\coordinate (u) at (1.5,1);
\draw (r)--(s)--(t)--(r);
\draw (r)--(u);
\draw (s)--(v);
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ode at (r) {$r$};
\
ode at (s) {$s$};
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ode at (t) {$t$};
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ode at (v) {$v$};
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ode at (u) {$u$};
\end{tikzpicture}

find the value of \( n $ in each triangle.

7.
\( \begin{tikzpicture}scale=0.8
\coordinate (a) at (0,0);
\coordinate (b) at (2,0);
\coordinate (c) at (2,2);
\coordinate (d) at (1,1);
\draw (a)--(b)--(c)--(a);
\draw (d)--(b);
\
ode at (1,0.5) {$9.5$};
\
ode at (0.5,1) {$2n - 23$};
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ode at (b) {};
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ode at (c) {};
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ode at (d) {};
\end{tikzpicture}

8.
\( \begin{tikzpicture}scale=0.8
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\draw (a)--(b)--(c)--(a);
\draw (d)--(b);
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ode at (0.5,1) {$8n + 10$};
\
ode at (1.5,0.5) {$5n$};
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ode at (a) {};
\
ode at (b) {};
\
ode at (c) {};
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ode at (d) {};
\end{tikzpicture}

Question

in each figure below, $ \overline{de} $ is a midsegment of triangle $ abc $. find the value of $ x $

2.
\( \begin{tikzpicture}scale=0.8
    \coordinate (a) at (0,0);
    \coordinate (b) at (1,2);
    \coordinate (c) at (3,0);
    \coordinate (e) at (1,0);
    \coordinate (d) at (2,1);
    \draw (a)--(b)--(c)--(a);
    \draw (e)--(d);
    \
ode at (0.5,1.5) {$x$};
    \
ode at (1.5,0.5) {$7$};
    \
ode at (a) {$a$};
    \
ode at (b) {$b$};
    \
ode at (c) {$c$};
    \
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    \
ode at (d) {$d$};
\end{tikzpicture}

3.
$ \begin{tikzpicture}scale=0.8
    \coordinate (a) at (0,0);
    \coordinate (b) at (2,2);
    \coordinate (c) at (3,0);
    \coordinate (e) at (1,0);
    \coordinate (d) at (1,1);
    \draw (a)--(b)--(c)--(a);
    \draw (d)--(e);
    \
ode at (0.5,0) {$x$};
    \
ode at (2,0) {$8$};
    \
ode at (a) {$a$};
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ode at (b) {$b$};
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ode at (c) {$c$};
    \
ode at (e) {$e$};
    \
ode at (d) {$d$};
\end{tikzpicture}

4.
\( \begin{tikzpicture}scale=0.8
    \coordinate (a) at (0,0);
    \coordinate (b) at (0,2);
    \coordinate (c) at (3,0);
    \coordinate (d) at (0,1);
    \coordinate (e) at (1,1);
    \draw (a)--(b)--(c)--(a);
    \draw (d)--(e);
    \
ode at (0.5,1) {$x$};
    \
ode at (1.5,0) {$34$};
    \
ode at (a) {$a$};
    \
ode at (b) {$b$};
    \
ode at (c) {$c$};
    \
ode at (d) {$d$};
    \
ode at (e) {$e$};
\end{tikzpicture}

5. \( \overline{df} $ is a median of $ \triangle cde $. if $ ec = (8x) $ feet and $ cf = (x + 9) $ feet, find $ x $ and $ ef $.
$ \begin{tikzpicture}scale=0.8
    \coordinate (e) at (0,0);
    \coordinate (f) at (1,0);
    \coordinate (c) at (2,0);
    \coordinate (d) at (1,2);
    \draw (e)--(f)--(c)--(d)--(e);
    \draw (d)--(f);
    \
ode at (e) {$e$};
    \
ode at (f) {$f$};
    \
ode at (c) {$c$};
    \
ode at (d) {$d$};
\end{tikzpicture}

6. in \( \triangle rst $, $ \overline{ru} $ and $ \overline{sv} $ are medians. $ su = 3x + 16 $, $ tu = 8 + 4x $, and $ rv = 7x - 20 $, find $ vt $.
$ \begin{tikzpicture}scale=0.8
    \coordinate (r) at (0,2);
    \coordinate (s) at (0,0);
    \coordinate (t) at (3,2);
    \coordinate (v) at (1.5,2);
    \coordinate (u) at (1.5,1);
    \draw (r)--(s)--(t)--(r);
    \draw (r)--(u);
    \draw (s)--(v);
    \
ode at (r) {$r$};
    \
ode at (s) {$s$};
    \
ode at (t) {$t$};
    \
ode at (v) {$v$};
    \
ode at (u) {$u$};
\end{tikzpicture}

find the value of \( n $ in each triangle.

7.
\( \begin{tikzpicture}scale=0.8
    \coordinate (a) at (0,0);
    \coordinate (b) at (2,0);
    \coordinate (c) at (2,2);
    \coordinate (d) at (1,1);
    \draw (a)--(b)--(c)--(a);
    \draw (d)--(b);
    \
ode at (1,0.5) {$9.5$};
    \
ode at (0.5,1) {$2n - 23$};
    \
ode at (a) {};
    \
ode at (b) {};
    \
ode at (c) {};
    \
ode at (d) {};
\end{tikzpicture}

8.
\( \begin{tikzpicture}scale=0.8
    \coordinate (a) at (0,0);
    \coordinate (b) at (2,2);
    \coordinate (c) at (3,0);
    \coordinate (d) at (1,1);
    \draw (a)--(b)--(c)--(a);
    \draw (d)--(b);
    \
ode at (0.5,1) {$8n + 10$};
    \
ode at (1.5,0.5) {$5n$};
    \
ode at (a) {};
    \
ode at (b) {};
    \
ode at (c) {};
    \
ode at (d) {};
\end{tikzpicture}

Accepted Answer

### Problem 2 (Triangle Midsegment)
# Explanation:
## Step1: Recall Midsegment Theorem
The midsegment of a triangle is parallel to the third side and half its length. Here, $ DE $ is a midsegment, so $ AB = 2 \times DE $. Given $ DE = 7 $, so $ x = 2 \times 7 $.
## Step2: Calculate $ x $
$ x = 14 $
# Answer: $ x = 14 $

### Problem 3 (Triangle Midsegment)
# Explanation:
## Step1: Apply Midsegment Theorem
Midsegment $ DE $ implies $ DE \parallel BC $ and $ AD = DB $, $ AE = EC $? Wait, no, midsegment connects midpoints, so $ AE = EC $? Wait, no, midsegment: if $ D $ is midpoint of $ AB $, $ E $ midpoint of $ AC $, then $ DE \parallel BC $ and $ DE = \frac{1}{2}BC $. But here, $ AE = x $, $ EC = 8 $, and $ D $ is midpoint? Wait, maybe $ DE $ is midsegment, so $ AE = EC $? No, wait, midsegment: the segment connecting midpoints of two sides. So if $ D $ is midpoint of $ AB $, $ E $ midpoint of $ AC $, then $ AE = EC $. Wait, but here $ AE = x $, $ EC = 8 $, so $ x = 8 $? No, maybe I misread. Wait, the figure: $ A $---$ E $---$ C $, with $ AE = x $, $ EC = 8 $, and $ D $ on $ AB $, $ E $ on $ AC $, $ DE $ midsegment. So midsegment theorem: $ DE \parallel BC $, so $ AE = EC $? No, midpoint of $ AC $ would be $ AE = EC $, so $ x = 8 $? Wait, no, maybe $ DE $ is midsegment, so $ AD = DB $ and $ AE = EC $, so $ AE = EC $, so $ x = 8 $? Wait, no, maybe the problem is that $ DE $ is midsegment, so $ AB = 2DE $, but maybe the figure is different. Wait, maybe the correct approach: midsegment connects midpoints, so $ AE = EC $, so $ x = 8 $? Wait, no, maybe I made a mistake. Wait, let's re-express: midsegment theorem states that the midsegment is parallel to the third side and half its length. So if $ DE $ is midsegment, then $ E $ is midpoint of $ AC $, so $ AE = EC $, so $ x = 8 $. Wait, but the problem says "Find the value of $ x $". So $ x = 8 $? Wait, no, maybe the other way: if $ DE $ is midsegment, then $ AE = EC $, so $ x = 8 $.
## Step2: Confirm
Since $ DE $ is midsegment, $ E $ is midpoint of $ AC $, so $ AE = EC $, hence $ x = 8 $.
# Answer: $ x = 8 $

### Problem 4 (Triangle Midsegment)
# Explanation:
## Step1: Use Midsegment Theorem
Midsegment $ DE $ is parallel to $ AC $ and $ DE = \frac{1}{2}AC $. Given $ AC = 34 $, so $ x = \frac{34}{2} $.
## Step2: Calculate $ x $
$ x = 17 $
# Answer: $ x = 17 $

### Problem 5 (Median of Triangle)
# Explanation:
## Step1: Recall Median Definition
A median of a triangle connects a vertex to the midpoint of the opposite side. So $ DF $ is median, so $ F $ is midpoint of $ EC $, hence $ EF = FC $, and $ EC = EF + FC = 2FC $. Given $ EC = 8x $, $ CF = x + 9 $, so $ 8x = 2(x + 9) $.
## Step2: Solve for $ x $
$ 8x = 2x + 18 $
$ 8x - 2x = 18 $
$ 6x = 18 $
$ x = 3 $
## Step3: Find $ EF $
Since $ EF = CF $ (F is midpoint), $ CF = x + 9 = 3 + 9 = 12 $, so $ EF = 12 $.
# Answer: $ x = 3 $, $ EF = 12 $ feet

### Problem 6 (Medians of Triangle)
# Explanation:
## Step1: Recall Median Properties
Medians of a triangle intersect at the centroid, which divides each median into a ratio of $ 2:1 $ (longer segment:shorter segment). So for median $ RU $, $ SU = 2 \times TU $, and for median $ SV $, $ RV = VT $ (wait, no: centroid divides each median into $ 2:1 $, so from vertex to centroid is $ 2/3 $, centroid to midpoint is $ 1/3 $. So for median $ RU $: $ SU = 2 \times TU $ (since $ S $ is vertex, $ U $ is midpoint of $ RT $, so centroid is on $ RU $, so $ SU:TU = 2:1 $). So $ 3x + 16 = 2(8 + 4x) $.
## Step2: Solve for $ x $
$ 3x + 16 = 16 + 8x $
$ 16 - 16 = 8x - 3x $
$ 5x = 0 $? No, that can't be. Wait, maybe I mixed up. Wait, median $ RU $: $ R $ to $ U $ (midpoint of $ ST $), so centroid is on $ RU $, so $ SU = 2 \times TU $? Wait, $ S $ is vertex, $ U $ is midpoint of $ RT $, so the median is $ SV $? Wait, no, the problem says: In $ \triangle RST $, $ \overline{RU} $ and $ \overline{SV} $ are medians. So $ RU $ is median: $ U $ is midpoint of $ ST $, so $ RU $ connects $ R $ to midpoint $ U $ of $ ST $. $ SV $ is median: $ V $ is midpoint of $ RT $, so $ SV $ connects $ S $ to midpoint $ V $ of $ RT $. The centroid is the intersection of $ RU $ and $ SV $. So for median $ RU $: the centroid divides $ RU $ into $ 2:1 $, so $ SU = 2 \times TU $? Wait, no, $ RU $ is from $ R $ to $ U $ (midpoint of $ ST $), so the centroid is closer to $ U $, so $ R $ to centroid is $ 2/3 $ of $ RU $, centroid to $ U $ is $ 1/3 $. Similarly for $ SV $: $ S $ to centroid is $ 2/3 $ of $ SV $, centroid to $ V $ is $ 1/3 $. But the problem gives $ SU = 3x + 16 $, $ TU = 8 + 4x $, $ RV = 7x - 20 $. Wait, maybe $ U $ is midpoint of $ RT $, so $ RU $ is median, so $ RU $ connects $ R $ to midpoint $ U $ of $ RT $, so $ RU $ is median, so $ RU $ is split by centroid into $ 2:1 $. But $ SU $ and $ TU $: maybe $ S $ is on $ RT $? No, $ R, S, T $ are vertices. Wait, maybe the labels are wrong. Wait, let's re-express: $ \triangle RST $, with medians $ RU $ (U on ST) and $ SV $ (V on RT). So $ U $ is midpoint of $ ST $, $ V $ is midpoint of $ RT $. Then $ RU $ and $ SV $ intersect at centroid $ G $. Then $ SG:GV = 2:1 $ and $ RG:GU = 2:1 $. But the problem says $ SU = 3x + 16 $, $ TU = 8 + 4x $, $ RV = 7x - 20 $. Wait, maybe $ U $ is midpoint of $ RT $, so $ RU $ is median, so $ RU $ connects $ R $ to midpoint $ U $ of $ RT $, so $ RU $ is median, so $ RU $ is split into $ RG:GU = 2:1 $. But $ SU $ and $ TU $: maybe $ S $ is on $ RU $? No, $ S $ is a vertex. Wait, maybe the problem has a typo, or I misread. Wait, maybe $ SU = TU $? No, centroid divides median into $ 2:1 $. Wait, maybe $ RU $ is a median, so $ U $ is midpoint of $ ST $, so $ RU $ is from $ R $ to $ U $ (midpoint of $ ST $), so $ RU $ is median, so length $ RU = 3x + 16 + 8 + 4x $? No, $ SU = 3x + 16 $, $ TU = 8 + 4x $, so $ RU = SU + TU = 3x + 16 + 8 + 4x = 7x + 24 $. But since $ U $ is midpoint of $ ST $, no, $ U $ is midpoint of $ RT $? Wait, I think I made a mistake. Let's try again: If $ RU $ is a median, then $ U $ is midpoint of $ ST $, so $ RU $ connects $ R $ to midpoint $ U $ of $ ST $. Then $ SV $ is median, $ V $ is midpoint of $ RT $, so $ SV $ connects $ S $ to midpoint $ V $ of $ RT $. Then the centroid is the intersection of $ RU $ and $ SV $. Then $ RV = VT $ because $ V $ is midpoint of $ RT $, so $ RV = VT $. Wait, $ RV = 7x - 20 $, so $ VT = RV = 7x - 20 $. Now, for the median $ RU $, the centroid divides it into $ 2:1 $, so $ RG:GU = 2:1 $. But $ SU $ and $ TU $: maybe $ S $ is on $ RU $? No, $ S $ is a vertex. Wait, maybe the problem is that $ RU $ is a median, so $ U $ is midpoint of $ RT $, so $ RU $ connects $ R $ to midpoint $ U $ of $ RT $, so $ RU $ is median, so $ RU $ is split into $ RG:GU = 2:1 $. Then $ SU $ and $ TU $ are parts of $ RU $? So $ SU = 3x + 16 $, $ TU = 8 + 4x $, so $ RU = SU + TU = 3x + 16 + 8 + 4x = 7x + 24 $. Since $ U $ is midpoint of $ RT $, $ RU $ is median, so $ RU $ is split by centroid into $ 2:1 $, so $ RG = \frac{2}{3}RU $, $ GU = \frac{1}{3}RU $. But we also know that $ SV $ is median, so $ V $ is midpoint of $ RT $, so $ RV = VT $, so $ VT = RV = 7x - 20 $. But we need another equation. Wait, maybe $ SU = TU $? No, that would be if $ U $ is midpoint, but $ SU $ and $ TU $ are parts of $ RU $. Wait, maybe the centroid is the intersection, so $ SU = 2 \times TU $ (since centroid divides median into $ 2:1 $, so from vertex $ S $ to centroid is $ 2 $, centroid to midpoint $ U $ is $ 1 $). So $ 3x + 16 = 2(8 + 4x) $. Solving: $ 3x + 16 = 16 + 8x $ → $ 5x = 0 $ → $ x = 0 $, which is impossible. So maybe $ TU = 2 \times SU $? Then $ 8 + 4x = 2(3x + 16) $ → $ 8 + 4x = 6x + 32 $ → $ -2x = 24 $ → $ x = -12 $, also impossible. So maybe the median is $ SV $, so $ V $ is midpoint of $ RT $, so $ RV = VT $, so $ VT = RV = 7x - 20 $. And $ RU $ is median, so $ U $ is midpoint of $ ST $, so $ SU = TU $, so $ 3x + 16 = 8 + 4x $ → $ x = 8 $. Then $ RV = 7(8) - 20 = 56 - 20 = 36 $, so $ VT = 36 $. Let's check: if $ x = 8 $, then $ SU = 3(8) + 16 = 40 $, $ TU = 8 + 4(8) = 40 $, so $ SU = TU $, so $ U $ is midpoint of $ ST $, which makes sense for median $ RU $. Then $ RV = 7x - 20 = 36 $, so $ VT = RV = 36 $ (since $ V $ is midpoint of $ RT $). Yes, that works. So $ x = 8 $, $ VT = 36 $.
## Step2: Solve for $ x $
Assume $ SU = TU $ (since $ U $ is midpoint of $ ST $, so $ SU = TU $):
$ 3x + 16 = 8 + 4x $
$ x = 8 $
## Step3: Find $ VT $
Since $ V $ is midpoint of $ RT $, $ VT = RV = 7x - 20 $. Substitute $ x = 8 $:
$ VT = 7(8) - 20 = 56 - 20 = 36 $
# Answer: $ VT = 36 $

### Problem 7 (Triangle Midsegment)
# Explanation:
## Step1: Apply Midsegment Theorem
The midsegment of a triangle is parallel to the third side and half its length. Here, the midsegment length is $ 9.5 $, and the third side is $ 2n - 23 $. So $ 9.5 = \frac{1}{2}(2n - 23) $.
## Step2: Solve for $ n $
Multiply both sides by 2: $ 19 = 2n - 23 $
Add 23 to both sides: $ 42 = 2n $
Divide by 2: $ n = 21 $
# Answer: $ n = 21 $

### Problem 8 (Triangle Midsegment)
# Explanation:
## Step1: Use Midsegment Theorem
The midsegment (length $ 5n $) is parallel to the third side (length $ 8n + 10 $) and half its length. So $ 5n = \frac{1}{2}(8n + 10) $.
## Step2: Solve for $ n $
Multiply both sides by 2: $ 10n = 8n + 10 $
Subtract $ 8n $: $ 2n = 10 $
Divide by 2: $ n = 5 $
# Answer: $ n = 5 $

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Problem 2 (Triangle Midsegment)

Step1: Recall Midsegment Theorem

Step2: Calculate \( x \)

Step1: Apply Midsegment Theorem

Step2: Confirm

Step1: Use Midsegment Theorem

Step2: Calculate \( x \)

Answer:

Problem 3 (Triangle Midsegment)