7. in △abc shown below, l is the midpoint of (overline{bc}), m is the midpoint of (overline{ab}), and n is the midpoint of (overline{ac}). if (mn = 8), (ml = 5), and (nl = 6), the perimeter of trapezoid (bmnc) is
(1) 35
(2) 31
(3) 28
(4) 26

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Accepted Answer

# Explanation:
## Step1: Recall Midline Theorem
In a triangle, the segment connecting the midpoints of two sides is parallel to the third side and half its length. So, $ MN \parallel BC $ and $ MN=\frac{1}{2}BC $, $ ML $ and $ NL $ are midsegments? Wait, $ M $ is midpoint of $ AB $, $ N $ midpoint of $ AC $, so $ MN $ is midline, $ MN = 8$, so $ BC=16 $? Wait no, trapezoid $ BMNC $: sides are $ BM $, $ MN $, $ NC $, $ CB $? Wait no, trapezoid $ BMNC $: let's identify the sides. $ M $ is midpoint of $ AB $, $ N $ midpoint of $ AC $, $ L $ midpoint of $ BC $. Given $ ML = 5$, $ NL = 6$, $ MN = 8$.

Wait, trapezoid $ BMNC $: the sides are $ BM $, $ MN $, $ NC $, and $ BC $? No, wait $ M $ is midpoint of $ AB $, so $ BM = MA $. $ N $ is midpoint of $ AC $, so $ NC = AN $. Now, $ ML $ and $ NL $: $ L $ is midpoint of $ BC $, so $ BL = LC $.

Wait, maybe $ BMNC $ has sides: $ BM $, $ MN $, $ NC $, and $ BC $? No, $ MN $ is parallel to $ BC $ (midline theorem), so $ BMNC $ is a trapezoid with $ MN \parallel BC $. Now, $ M $ is midpoint of $ AB $, so $ BM = \frac{AB}{2} $. $ N $ is midpoint of $ AC $, so $ NC = \frac{AC}{2} $. Also, $ ML = 5 $, $ NL = 6 $. Wait, maybe $ BM = NL = 6 $? Wait no, $ M $ is midpoint of $ AB $, $ L $ is midpoint of $ BC $, so $ ML $ is midline of $ \triangle ABC $? Wait, $ ML $ connects midpoints of $ AB $ and $ BC $, so $ ML \parallel AC $ and $ ML = \frac{AC}{2} $. Similarly, $ NL $ connects midpoints of $ AC $ and $ BC $, so $ NL \parallel AB $ and $ NL = \frac{AB}{2} $.

Ah! So $ ML \parallel AC $ and $ ML = \frac{AC}{2} = 5 $? Wait no, $ ML = 5 $, so $ AC = 10 $? Wait no, $ ML $ is midline of $ \triangle ABC $ (connecting midpoints of $ AB $ and $ BC $), so $ ML \parallel AC $ and $ ML = \frac{AC}{2} $, so $ AC = 2 \times ML = 10 $? But $ NL = 6 $, $ NL $ is midline of $ \triangle ABC $ (connecting midpoints of $ AC $ and $ BC $), so $ NL \parallel AB $ and $ NL = \frac{AB}{2} $, so $ AB = 12 $. Then $ BM = \frac{AB}{2} = 6 $, $ NC = \frac{AC}{2} = 5 $. Now, $ MN = 8 $ (midline, $ MN \parallel BC $, $ MN = \frac{BC}{2} $, so $ BC = 16 $). Now, trapezoid $ BMNC $: sides are $ BM = 6 $, $ MN = 8 $, $ NC = 5 $, and $ BC = 16 $? Wait no, that can't be. Wait, no, trapezoid $ BMNC $: the two parallel sides are $ MN $ and $ BC $, and the non-parallel sides are $ BM $ and $ NC $. Wait, but $ ML = 5 $, $ NL = 6 $. Wait, maybe $ BM = NL = 6 $, $ NC = ML = 5 $, because $ ML \parallel AC $ (so $ ML = NC $ since $ N $ is midpoint, $ NC = \frac{AC}{2} = ML $), and $ NL \parallel AB $ (so $ NL = BM = \frac{AB}{2} = NL $). Then $ BM = 6 $, $ NC = 5 $, $ MN = 8 $, $ BC = 16 $? Wait no, $ BC $ is $ BL + LC = 2BL $, but $ ML $ is midline, $ ML = \frac{AC}{2} = 5 $, so $ AC = 10 $, $ NC = 5 $. $ NL = \frac{AB}{2} = 6 $, so $ AB = 12 $, $ BM = 6 $. Then $ BC $: from midline $ MN = \frac{BC}{2} = 8 $, so $ BC = 16 $. Now, perimeter of trapezoid $ BMNC $ is $ BM + MN + NC + BC = 6 + 8 + 5 + 16 $? Wait no, that's 35? Wait no, wait the options are 35, 31, 28, 26. Wait maybe I messed up.

Wait, trapezoid $ BMNC $: sides are $ BM $, $ MN $, $ NC $, and $ BC $? No, wait $ M $ is midpoint of $ AB $, $ N $ midpoint of $ AC $, $ L $ midpoint of $ BC $. So $ ML $ is midline of $ \triangle ABC $ (AB and BC midpoints), so $ ML \parallel AC $ and $ ML = \frac{AC}{2} $, so $ AC = 2 \times ML = 10 $, so $ NC = \frac{AC}{2} = 5 $. Similarly, $ NL $ is midline of $ \triangle ABC $ (AC and BC midpoints), so $ NL \parallel AB $ and $ NL = \frac{AB}{2} $, so $ AB = 2 \times NL = 12 $, so $ BM = \frac{AB}{2} = 6 $. $ MN $ is midline of $ \triangle ABC $ (AB and AC midpoints), so $ MN = \frac{BC}{2} = 8 $, so $ BC = 16 $. Now, trapezoid $ BMNC $: the sides are $ BM = 6 $, $ MN = 8 $, $ NC = 5 $, and $ BC = 16 $? Wait, but that would be 6+8+5+16=35, which is option (1). But wait, maybe the trapezoid is $ BMNL $? No, the problem says trapezoid $ BMNC $. Wait, maybe I made a mistake. Wait, the trapezoid has sides $ BM $, $ ML $, $ LN $, $ NC $? No, $ ML = 5 $, $ NL = 6 $, $ MN = 8 $. Wait, maybe the perimeter is $ BM + MN + NC + BC $, but $ BC = BM + MN + NC $? No, no. Wait, let's re-express:

$ M $ is midpoint of $ AB $, so $ BM = MA $.

$ N $ is midpoint of $ AC $, so $ NC = AN $.

$ L $ is midpoint of $ BC $, so $ BL = LC $.

$ ML $ connects $ M $ (mid AB) and $ L $ (mid BC), so $ ML \parallel AC $ (midline theorem) and $ ML = \frac{AC}{2} $, so $ AC = 2 \times ML = 10 $, so $ NC = \frac{AC}{2} = 5 $.

$ NL $ connects $ N $ (mid AC) and $ L $ (mid BC), so $ NL \parallel AB $ (midline theorem) and $ NL = \frac{AB}{2} $, so $ AB = 2 \times NL = 12 $, so $ BM = \frac{AB}{2} = 6 $.

$ MN $ connects $ M $ (mid AB) and $ N $ (mid AC), so $ MN \parallel BC $ (midline theorem) and $ MN = \frac{BC}{2} = 8 $, so $ BC = 16 $.

Now, trapezoid $ BMNC $: the four sides are $ BM $, $ MN $, $ NC $, and $ BC $. Wait, but $ BM = 6 $, $ MN = 8 $, $ NC = 5 $, $ BC = 16 $. Then perimeter is $ 6 + 8 + 5 + 16 = 35 $. But wait, the options include 35 as (1). But maybe I messed up the trapezoid sides. Wait, maybe the trapezoid is $ BMNL $? No, the problem says $ BMNC $. Wait, another approach: $ M $ and $ N $ are midpoints, so $ MN \parallel BC $, so $ BMNC $ is trapezoid with $ MN \parallel BC $. The legs are $ BM $ and $ NC $, and the bases are $ MN $ and $ BC $. Now, $ ML = 5 $, $ NL = 6 $. Since $ L $ is midpoint of $ BC $, $ ML $ is midline of $ \triangle ABC $ (AB and BC midpoints), so $ ML = \frac{AC}{2} $, so $ AC = 10 $, $ NC = 5 $. $ NL = \frac{AB}{2} = 6 $, so $ AB = 12 $, $ BM = 6 $. $ MN = 8 $, $ BC = 16 $. Perimeter: $ BM + MN + NC + BC = 6 + 8 + 5 + 16 = 35 $. So the answer should be 35.

## Step1: Identify Midline Theorems
Using midline theorem: $ MN \parallel BC $, $ ML \parallel AC $, $ NL \parallel AB $.

## Step2: Calculate Side Lengths
- $ ML = \frac{AC}{2} = 5 \Rightarrow AC = 10 \Rightarrow NC = \frac{AC}{2} = 5 $
- $ NL = \frac{AB}{2} = 6 \Rightarrow AB = 12 \Rightarrow BM = \frac{AB}{2} = 6 $
- $ MN = \frac{BC}{2} = 8 \Rightarrow BC = 16 $

## Step3: Perimeter of Trapezoid $ BMNC $
Perimeter $ = BM + MN + NC + BC = 6 + 8 + 5 + 16 = 35 $

# Answer:
35

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Explanation:

Step1: Recall Midline Theorem

Answer: