This 6-second lead II strip is irregular, so the 300 and 1500 methods do not work. Use the 6-second method to estimate the ventricular heart rate in beats per minute.
ECG rhythm strip interpretation worksheet with answers
1.Easy
lead II, 25 mm/s, 10 mm/mVstrip length 6 s (marks at 0, 3 and 6 s)
heart rate = bpm
2.Easy
Measure the PR interval on this lead II strip, in seconds.
lead II, 25 mm/s, 10 mm/mV1 small box = 0.04 s, 1 large box = 0.20 sstrip length 6 s
PR = s
3.Medium
Measure the QT interval on this lead II strip, in seconds.
lead II, 25 mm/s, 10 mm/mV1 small box = 0.04 s, 1 large box = 0.20 sstrip length 6 s
QT = s
4.Medium
Interpret this lead II rhythm strip. Which rhythm is it? Click your answer.
lead II, 25 mm/s, 10 mm/mV1 small box = 0.04 s, 1 large box = 0.20 sstrip length 6 s
- 1junctional rhythm
- 2normal sinus rhythm
- 3sinus bradycardia
- 4atrial fibrillation
5.Hard
This lead II strip shows a conduction problem between the atria and the ventricles. Which rhythm is it? Click your answer.
lead II, 25 mm/s, 10 mm/mV1 small box = 0.04 s, 1 large box = 0.20 sstrip length 6 s
- 1third-degree (complete) AV block
- 2atrial fibrillation
- 3second-degree AV block type I (Wenckebach)
- 4second-degree AV block type II
6.Hard
A monitored patient is found unresponsive and not breathing normally, and no carotid pulse can be felt. The monitor shows this organised rhythm. What is this situation called? Click your answer.
lead II, 25 mm/s, 10 mm/mV1 small box = 0.04 s, 1 large box = 0.20 sstrip length 6 s
- 1normal sinus rhythm
- 2pulseless electrical activity (PEA)
- 3asystole
- 4ventricular fibrillation
Answer key
- 1. heart rate = 90 bpm
- 2. PR = 0.28 s
- 3. QT = 0.44 s
- 4. junctional rhythm
- 5. second-degree AV block type I (Wenckebach)
- 6. pulseless electrical activity (PEA)
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