if $ef = fg = 98$ and $gh = 81$, what is $eh$?
$eh = \square$

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Accepted Answer

# Explanation:
## Step1: Identify triangle properties
Since $ EF = FG = 98 $, triangle $ EFG $ is isosceles with $ EF = FG $. The line $ FH $ is perpendicular to $ EG $ (right angle at $ H $), so by the perpendicular bisector theorem, $ EH = GH $ if $ FH $ bisects $ EG $, but wait, actually, in an isosceles triangle, the altitude from the apex ( $ F $) to the base ( $ EG $) bisects the base. Wait, no, here $ EF = FG $, so the apex is $ F $, and the base is $ EG $. Wait, but the diagram shows $ FH $ perpendicular to $ EG $ at $ H $, so $ H $ is the midpoint? Wait, no, wait the problem: $ EF = FG = 98 $, $ GH = 81 $. Wait, maybe $ EH = GH $? Wait, no, maybe I misread. Wait, $ EF = FG $, so triangle $ EFG $ is isosceles with $ EF = FG $, so the sides $ EF $ and $ FG $ are equal. Then $ FH $ is perpendicular to $ EG $, so triangle $ FHE $ and $ FHG $ are right triangles. Since $ EF = FG $ and $ FH $ is common, by HL congruence, $ \triangle FHE \cong \triangle FHG $, so $ EH = GH $? Wait, but $ GH = 81 $, so $ EH = 81 $? Wait, no, that can't be. Wait, maybe I made a mistake. Wait, the problem says $ EF = FG = 98 $, $ GH = 81 $, find $ EH $. Wait, maybe $ EG $ is a line, and $ F $ is a point such that $ EF = FG = 98 $, and $ FH $ is perpendicular to $ EG $ at $ H $. Then in right triangle $ FHG $, $ FG = 98 $, $ GH = 81 $, so we can find $ FH $, but wait, no, we need $ EH $. Wait, but if $ \triangle FHE \cong \triangle FHG $, then $ EH = GH $. Wait, but that would mean $ EH = 81 $. Wait, maybe that's the case. Wait, let's check: $ EF = FG $, $ FH \perp EG $, so $ \angle FHE = \angle FHG = 90^\circ $, $ FH = FH $, $ EF = FG $, so by HL, $ \triangle FHE \cong \triangle FHG $, so $ EH = GH = 81 $. So $ EH = 81 $.

Wait, but maybe I misread the problem. Wait, the problem says $ EF = FG = 98 $, $ GH = 81 $, what is $ EH $. So according to the congruence, $ EH = GH = 81 $.

## Step1: Recognize congruent triangles
Since $ EF = FG $, $ \angle FHE = \angle FHG = 90^\circ $, and $ FH $ is common, $ \triangle FHE \cong \triangle FHG $ (HL).
## Step2: Conclude $ EH = GH $
From congruence, corresponding sides $ EH = GH $. Given $ GH = 81 $, so $ EH = 81 $.

# Answer:
$ 81 $

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QUESTION IMAGE

Question

Explanation:

Step1: Identify triangle properties

Step1: Recognize congruent triangles

Step2: Conclude \( EH = GH \)

Answer: