Why three side lengths fix a triangle completely: the SSS rule.
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Three side lengths leave a triangle no freedom at all.
Two students must copy a triangular frame that is too big to trace.
Meera and Rabia have been asked to make a cardboard cutout identical to a triangular frame they have in school. They see that the frame is too big to simply be traced on a piece of paper and replicated.
Building the triangle from three sides seems to give two different answers.
Rabia thinks about how to draw the scaled-down triangle with sides , , and . First, she would draw a line segment having one of the lengths, say .
The 4-6-8 construction, with the circles cutting at E above and F below.
The circles from A and B meet in two places, creating two triangles that share the base AB.
Why the two triangles from the construction are congruent after all.
Problem. Examine whether and constructed from the intersections of the two circles are congruent.
The SSS test: what to check, what it lets you conclude.
What you need: All three pairs of corresponding sides must be equal.
Conclusion:
Conditions and scope: The three lengths must be matched to their correct partners, not just be the same set of numbers in any random pairing. No angle information is needed. Lengths must be in matching units.
Four quick decisions about whether three sides settle congruence.
Triangle ABC has sides AB = 5 cm, BC = 7 cm, and CA = 9 cm. Triangle PQR has sides PQ = 5 cm, QR = 7 cm, and RP = 9 cm. Are these two triangles congruent?