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$8.8
$26,460
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Answer:
Step-by-step explanation:
17) HI ≅ UH ; GH ≅ TU ; GI ≅ TH
ΔHGI ≅ ΔUTH by Side Side Side congruent
∠G ≅ ∠T ; GI ≅ TH ; ∠GIH ≅ ∠THU
ΔHGI ≅ ΔUTH by Angel Side Angle congruent
19) IJ ≅ KD ; IK ≅ KC ; KJ ≅ CD
ΔIJK ≅ ΔKDC by Side Side Side congruent
∠J ≅ ∠D ; IJ ≅ KD ; ∠I ≅ ∠DKC
ΔIJK ≅ ΔKDC by Angle Side Angle congruent
to put them together and add them.
we know that
The measurement of the exterior angle is the semi-difference of the arcs which comprises
In this problem
∠FGH is the exterior angle
∠FGH=
∠FGH=
-----> equation A
--------> equation B
Substitute equation B in equation A
![100\°=(arc\ FEH-[360\°-arc\ FEH])](https://tex.z-dn.net/?f=100%5C%C2%B0%3D%28arc%5C%20FEH-%5B360%5C%C2%B0-arc%5C%20FEH%5D%29)
therefore
<u>The answer is</u>
The measure of arc FEH is equal to 
I think it is B -2 < x < 2
