Question

elp

segment BD and EG are intersected by AH. Points of intersection are C and F. angle ACD is x+8+2x. angle EFH is 2(2x-5).

Find the students error and show work to fix it.

The students work:

2(2x-5)+x+8+2x=180
4x-10+x+8+2x=180
7x-10+8=180
7x-2=180
+2 +2
7x=182
divide
- -
7 7
x=26

mEFH=2(x-5)
=2[2(26)-5]
=2[52-5]
=2[47]
EFH=94
=

Answer 1

What's wrong: the student copied the expression for m∠EFH incorrectly in the final step. They wrote m∠EFH = 2(x−5) (missing the extra 2 in 2(2x−5)), but then evaluated using 2(2·26 − 5). That is inconsistent. The algebra to find x was done correctly.

Correct work:

1. Angles ACD and EFH are a linear pair, so they sum to 180°:
2(2x−5) + (x + 8 + 2x) = 180.

2. Simplify:
4x − 10 + 3x + 8 = 180
7x − 2 = 180
7x = 182
x = 26.

3. Find each angle:
m∠EFH = 2(2x − 5) = 2(2·26 − 5) = 2(52 − 5) = 2·47 = 94°.
m∠ACD = x + 8 + 2x = 3x + 8 = 3·26 + 8 = 78 + 8 = 86°.

4. Check: 94° + 86° = 180° (supplementary), as required.

So the student’s numerical answer for the angle (94°) is correct, but they made a copying/notation error by writing 2(x−5) instead of the given 2(2x−5).