Constructions and Conjectures

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Use the original graphs, tables, and diagrams from released STAAR questions to practice reading labels, values, and relationships before answering the items below.

Archived released STAAR Geometry EOC · 2013 · Item 39 · 1 point

Worked example

Equal-radius arcs centered at L and N intersect at R and S. Line RS crosses LN at T. If LN = 14 cm and ∠RLN = 38°, find LT, ∠RTN, and ∠LNR.

1. RS is the perpendicular bisector of LN, so T is the midpoint.
2. LT = 14 ÷ 2 = 7 cm.
3. RS is perpendicular to LN, so ∠RTN = 90°.
4. RL = RN because both were drawn with the same radius. Triangle LRN is isosceles, so its base angles at L and N are equal.

LT = 7 cm; ∠RTN = 90°; ∠LNR = 38°

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TEKSForge practice · unscored

1. 1. Why must the compass opening be greater than half the distance between the endpoints?

   Show answer and reasoning

   So the arcs intersect at two distinct points.

   A smaller radius produces no intersection. A radius equal to half the segment length produces only one intersection, which is not enough to determine the bisector line.

2. 2. P lies on the constructed perpendicular bisector of AB. If PA = 11 cm, what is PB? Must triangle APB be equilateral?

   Show answer and reasoning

   PB = 11 cm; the triangle need not be equilateral.

   Equidistance guarantees PA = PB. It does not guarantee that AB is also 11 cm.

Released STAAR practice

Texas Education Agency · archived 2013 released EOC items · scored with TEA's answer key; step-by-step explanations written by TEKSForge because this release has no rationales. Original graphs, tables, and diagrams are shown where the released item includes one.

Archived released STAAR Geometry EOC · 2013 · Item 39 · 1 point

The diagram shows the arcs and segments used to construct $\triangle ABC$, given line $k$. Based on this construction, which statement is not true?

1. $\angle AWC$ is complementary to $\angle CWB$.
2. $\triangle CWB$ is a right triangle.
3. $\triangle ACB$ is isosceles.
4. $m\angle CAB = m\angle CBA$

Show STAAR answer and explanation

$\angle AWC$ is complementary to $\angle CWB$.

The equal-radius arcs construct the perpendicular bisector CW of AB. Both AWC and CWB measure 90°, so their sum is 180°; they are supplementary, not complementary.

$\triangle CWB$ is a right triangle.: CW is perpendicular to WB, so CWB is indeed a right triangle.

$\triangle ACB$ is isosceles.: C lies on the perpendicular bisector of AB, giving CA = CB; the triangle is isosceles.

$m\angle CAB = m\angle CBA$: Equal sides CA and CB imply equal base angles CAB and CBA.

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