12 multiple-choice questions, progressively harder.
For an SAS match, corresponding sides have lengths 4x−34x-34x−3 cm and 2x+92x+92x+9 cm. The other side pair and included angles already match. What value of xxx completes the match?
Solution
Correct answer: C
Set the corresponding lengths equal:
Subtract 2x and add 3, then divide by 2:
Both lengths are 21 cm, which is positive.
Two triangles already have matching included sides and one endpoint-angle pair. Their remaining endpoint angles measure (3x+5)∘(3x+5)^\circ(3x+5)∘ and 74∘74^\circ74∘. What xxx completes ASA?
Correct answer: A
The remaining endpoint angles must have equal measures.
Solve
Subtract 5 and divide by 3:
The given matching parts are:
Which ordered statement follows by SAS?
Correct answer: D
The included vertices D and V match.
DE matches VT, so E matches T.
DF matches VU, so F matches U.
Therefore
An ASA construction has a 141414 cm base and endpoint angles 52∘52^\circ52∘ and 79∘79^\circ79∘. What is the third angle?
Correct answer: B
The angle sum does not depend on the base length.
Subtract both endpoint angles:
A report lists these matching parts:
What would be needed to use SAS directly with those two side pairs?
The given side pairs meet at A and D, not B and E.
Direct SAS needs
The report has not supplied that pair.
After aligning two equal bases, the third vertices lie on opposite sides of the base line. Which rigid motion keeps both base endpoints fixed and changes the side of the third vertex?
Reflection in a line fixes every point on that line.
Thus both endpoints stay in place, while C goes to its mirror image:
A stretch is not a rigid motion.
In the ASA justification, why can the aligned angle rays not give two different third vertices in the same half-plane?
Two different proposed intersections would put two distinct lines through the same two distinct points.
Only one line can pass through such a pair.
The third-vertex images must therefore agree:
For triangles HJKHJKHJK and RSTRSTRST, you know
Which additional pair directly supplies the other endpoint-angle pair for ASA using these sides, without calculating another angle?
HJ joins H to J, and RS joins R to S.
To supply the other endpoint angles directly, match J with S:
The pair at K and T would require calculating the remaining angles before using ASA; it is not the direct endpoint pair requested.
Triangles KLMKLMKLM and PQRPQRPQR have these given matching parts:
Which correspondence is established by SAS?
K↔PK\leftrightarrow PK↔P
L↔QL\leftrightarrow QL↔Q
M↔RM\leftrightarrow RM↔R
K↔RK\leftrightarrow RK↔R
M↔PM\leftrightarrow PM↔P
K↔QK\leftrightarrow QK↔Q
L↔PL\leftrightarrow PL↔P
L↔RL\leftrightarrow RL↔R
The given angles identify L with P.
The side of length 11 then pairs K with Q, and the side of length 14 pairs M with R.
Hence
Triangles ABCABCABC and DEFDEFDEF have these given matching parts:
Which fact explains why ASA applies?
A and B are the endpoint angles of AB; D and E are the endpoint angles of DE. The known sides are included between the known angles.
ASA gives:
A base of positive length has intended endpoint angles 68∘68^\circ68∘ and 112∘112^\circ112∘, both drawn into the same half-plane. Why does this not construct a triangle?
The two endpoint angles total
Their new rays are parallel; equivalently, a supposed third angle would have measure zero.
Two triangles have corresponding side lengths 999 cm and 131313 cm and corresponding included angles 104∘104^\circ104∘. One drawing is rotated and the other is reflected. What follows?
SAS allows acute, right, or obtuse included angles strictly between zero and 180 degrees.
Rotation and reflection preserve the matching 9 cm sides, 13 cm sides, and included 104∘104^\circ104∘ angles.
Thus the triangles are congruent by SAS, with those sides and included vertices corresponding.
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