Midsegment Theorem Quiz: Question 1 asks for the perimeter of parallelogram CDOG given side lengths of triangle ACT.

Explanation:
Let’s solve each question step by step.

Question 1:
We are given triangle $ \triangle ACT $, with:
- $ D $ = midpoint of $ \overline{AC} $
- $ O $ = midpoint of $ \overline{AT} $
- $ G $ = midpoint of $ \overline{CT} $

So points $ D, O, G $ are midpoints of the three sides of triangle $ ACT $.
Then quadrilateral $ CDOG $ is formed by connecting:
- $ C $ to $ D $ (half of $ AC $),
- $ D $ to $ O $,
- $ O $ to $ G $,
- $ G $ back to $ C $.

But more importantly, since $ D $ and $ O $ are midpoints of $ AC $ and $ AT $, segment $ DO $ is a midsegment of triangle $ ACT $, so:
- $ DO \parallel CT $ and $ DO = \frac{1}{2} CT $

Similarly:
- $ OG $ connects midpoints of $ AT $ and $ CT $ → $ OG \parallel AC $, and $ OG = \frac{1}{2} AC $
- $ DG $ connects midpoints of $ AC $ and $ CT $ → $ DG \parallel AT $, and $ DG = \frac{1}{2} AT $

Now look at quadrilateral $ CDOG $:
Vertices in order: $ C \to D \to O \to G \to C $

Let’s find the side lengths:
- $ CD $: Since $ D $ is midpoint of $ AC $, and $ AC = 10 $, then
$ CD = \frac{1}{2} \cdot AC = \frac{1}{2} \cdot 10 = 5 $

- $ DO $: As above, midsegment parallel to $ CT $, so
$ DO = \frac{1}{2} \cdot CT = \frac{1}{2} \cdot 22 = 11 $

- $ OG $: Midsegment parallel to $ AC $, so
$ OG = \frac{1}{2} \cdot AC = 5 $

- $ GC $: Since $ G $ is midpoint of $ CT $, and $ CT = 22 $, then
$ GC = \frac{1}{2} \cdot CT = 11 $

So sides of parallelogram $ CDOG $ are:
$ CD = 5 $, $ DO = 11 $, $ OG = 5 $, $ GC = 11 $

Perimeter = $ 5 + 11 + 5 + 11 = 32 $

✔ Check: Opposite sides equal → it's a parallelogram. Perimeter = 32.

Question 2:
Triangle $ ABC $ is equilateral, perimeter = 36 cm.
So each side = $ \frac{36}{3} = 12 $ cm.

Points $ D, E, F $ are midpoints:
- $ D $ = midpoint of $ AB $
- $ E $ = midpoint of $ BC $
- $ F $ = midpoint of $ AC $

We’re asked for length of $ \overline{EF} $

Look at triangle $ ABC $:
Segment $ EF $ connects midpoints of $ BC $ and $ AC $. So in triangle $ ABC $, $ EF $ is a midsegment, parallel to $ AB $, and half its length:

$$
EF = \frac{1}{2} \cdot AB = \frac{1}{2} \cdot 12 = 6 \text{ cm}
$$

Alternatively, triangle $ DEF $ is the medial triangle — all sides are half the original triangle’s sides, so each side of $ \triangle DEF $ is 6 cm.

So $ EF = 6 $ cm.

Final Answer:
1. c. 32
2. a. 6