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PLEASE HELP!!! Find the length of the midsegment to the nearest tenth.

PLEASE HELP!!! Find the length of the midsegment to the nearest tenth.-example-1
User Tkazik
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1 Answer

16 votes
16 votes

Answer:

Midsegment is 35.

Explanation:

TL; DR:

The smaller triangle (with a base of 6x + 5) has side lengths that are twice as small as the entire triangle (the one with a base of 3x+55). So, 2(6x+5) = 3x + 55. x = 5, so the midsegment is 35.

Look at the photo attached below.

Here is the statement/reason:

1. ED ║ AB (Midsegments are parallel to the base)

Note: Then, CA and CB are transversals

2. Then, m∠CED = m∠EAB (corresponding angles)

3. m∠CDE = m∠DBA (corresponding angles)

4. ΔCED is similar to ΔCAB (angle angle)

5. CA = CE + EA

6. CA = 39 + 39 = 78 (Substitution, algebra)

7. ED/AB = CE/CA (similar triangles)

Note: You can also use CD and CB

8. (6x+5) / (3x+55) = 39 / 78 (substitution)

9. (6x + 5) / (3x + 55) = 1 / 2 (algebra)

10. 12x + 10 = 3x + 55, x = 5 (algebra)

11. ED = 6x + 5 (given)

12. ED = 6*5 + 5 = 35 (substitution, algebra)

An explanation of the algebra:

(6x+5) is twice as small as (3x + 55). So, we can write this equation:

2(6x+5) = 3x + 55

12x + 10 = 3x + 55

Subtract 3x from both sides.

9x + 10 = 55

Subtract 10 from both sides.

9x = 45

x = 5.

Now, put that into ED.

6x + 5 = 6 * 5 + 5 = 35

I hope this helps! Feel free to ask any questions!

PLEASE HELP!!! Find the length of the midsegment to the nearest tenth.-example-1
User Lemar
by
3.0k points