Super-Hard SAT Math Question 71 | Free SAT Prep | Aniko

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Question 71·200 Super-Hard SAT Math Questions·Geometry and Trigonometry

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In triangle $PQR$ and triangle $STU$ , sides $QR$ and $TU$ each have a length of 15, and angles $Q$ and $T$ each measure $40^\circ$ .

Consider each statement separately (each statement is added to the information above).

Which statement(s) are sufficient to prove that triangle $PQR$ is congruent to triangle $STU$ ?

I. $PR = SU = 10$

II. $PQ = ST$

III. $PR = SU = 20$

First see whether the added fact completes SAS or ASA/AAS, which always proves congruence. If the added fact creates SSA, don’t automatically reject it—check whether the given numbers force a unique triangle (for an acute angle, compare the opposite side to $b\sin(A)$ and to $b$ where $b$ is the adjacent known side).

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Try to finish a standard congruence test

With $QR=TU$ and $\angle Q=\angle T$ , ask what extra fact would create SAS or ASA/AAS.

Be careful with SSA

If the added information gives two sides and an angle that is not between them, that’s SSA. For an acute given angle, SSA can be ambiguous.

Use a comparison to decide whether SSA is unique

For acute angle $40^\circ$ with adjacent side 15, compare the other given side to $15\sin(40^\circ)$ and to 15 to see if there are 0, 1, or 2 possible triangles.

Desmos Guide

Fix the base and the given angle

Place $Q=(0,0)$ and $R=(15,0)$ .

Graph the ray from $Q$ that makes a $40^\circ$ angle with $\overline{QR}$ :

$y=(\tan(40^\circ))x\{x\ge 0\}$

Any valid point $P$ must lie on this ray.

Model statement I ( $PR=10$ ) and inspect intersections

Graph the circle centered at $R$ with radius 10:

$(x-15)^2+y^2=10^2$

Count how many times this circle intersects the $40^\circ$ ray. If there is more than one intersection, then the triangle is not uniquely determined by the SSA information.

Model statement III ( $PR=20$ ) and inspect intersections

Graph the circle centered at $R$ with radius 20:

$(x-15)^2+y^2=20^2$

Count how many times this circle intersects the $40^\circ$ ray. If there is exactly one intersection, then the triangle is uniquely determined by the SSA information.

Connect the visuals to congruence and conclude

Statement II corresponds to an SAS setup (two sides with the included $40^\circ$ angle), which guarantees congruence.

From the intersection counts for statements I and III, decide whether each SSA setup is ambiguous or unique. Therefore, the sufficient statements are II and III only.

Step-by-step Explanation

List the given information

You know in the two triangles:

$QR = TU = 15$
$\angle Q = \angle T = 40^\circ$

Now test each statement separately to see whether it forces a unique triangle (and therefore congruence).

Test statement II by checking for SAS

If II is added, then $PQ = ST$ .

Now each triangle would have:

Two sides: $PQ$ and $QR$ (matching $ST$ and $TU$ )
The included angle: $\angle Q$ (matching $\angle T$ )

That is an SAS configuration, which is a valid congruence test.

Recognize that statements I and III form SSA

Statements I and III both add $PR = SU$ .

But the known angle is $\angle Q$ (or $\angle T$ ), and side $PR$ (or $SU$ ) is opposite that angle, not adjacent to it. So the data would be SSA.

For acute $\angle Q=40^\circ$ with adjacent side $QR=15$ , use the benchmark

h = 15\sin(40^\circ).

If $h < PR < 15$ , there can be two different triangles.
If $PR \ge 15$ , there is exactly one triangle.

Apply the SSA test and decide which statements work

Compute

h = 15\sin(40^\circ) \approx 15(0.643) \approx 9.6.

With I ( $PR=10$ ): $9.6 < 10 < 15$ , so SSA is ambiguous (two possible triangles) and it does not guarantee congruence.
With III ( $PR=20$ ): $20 \ge 15$ , so SSA determines a unique triangle and it does guarantee congruence.

Combining this with the SAS result from statement II, the sufficient statements are II and III only.

Strategy

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Step-by-step Explanation

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Desmos Guide

Step-by-step Explanation