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SSS and SAS Similarity: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 Three ratios in the stated order

Triangle $ABC$ has lengths
$\begin{aligned}AB&=7\\BC&=10\\CA&=12.\end{aligned}$
Triangle $DEF$ has lengths
$\begin{aligned}DE&=10.5\\EF&=15\\FD&=18.\end{aligned}$
Determine whether $\triangle ABC\sim\triangle DEF$ is justified. State the criterion and the factor from the first triangle to the second.
1. Hint 1
  Read the proposed matching from the triangle names.
2. Hint 2
  Compute every target-to-original side ratio before naming SSS.
Answer

$\triangle ABC\sim\triangle DEF$ by SSS, with factor $1.5$ .

Full solution

The side pairs in the stated order are $AB$ with $DE$ , $BC$ with $EF$ , and $CA$ with $FD$ .
$\begin{aligned}\frac{10.5}{7}&=1.5,\\\frac{15}{10}&=1.5,\\\frac{18}{12}&=1.5.\end{aligned}$

All three ratios agree.
Both positive triples satisfy the triangle inequality, so SSS similarity applies with factor $1.5$ .

Reconstructed from the data after the solution establishes the result.

Answer

$\triangle ABC\sim\triangle DEF$ by SSS, with factor $1.5$ .

Key idea

SSS requires all three ratios in one fixed matching.
Problem 2 Place the angle between the selected sides

In triangle $ABC$ ,
$\begin{aligned}AB&=5\\AC&=9\\\angle BAC&=57^\circ.\end{aligned}$
In triangle $PQR$ ,
$\begin{aligned}PQ&=8\\PR&=14.4\\\angle QPR&=57^\circ.\end{aligned}$
Prove an ordered similarity statement and state the scale factor from $ABC$ to the second triangle. Explain why the angle is in the required position.
1. Hint 1
  Locate the common endpoint of the two given sides in each triangle.
2. Hint 2
  Compare the two target-to-original ratios that surround those endpoints.
Answer

$\triangle ABC\sim\triangle PQR$ by SAS, with factor $1.6$ ; the equal angles at $A$ and $P$ are included.

Full solution

The given sides meet at $A$ in the first triangle and at $P$ in the second.
Thus the two $57^\circ$ angles are included.

Both corresponding side ratios equal $1.6$ .
$\begin{aligned}\frac{PQ}{AB}&=\frac 85,\\&=1.6,\\\frac{PR}{AC}&=\frac{14.4}{9},\\&=1.6.\end{aligned}$

SAS similarity gives $A$ to $P$ , $B$ to $Q$ , and $C$ to $R$ , so the ordered statement is $\triangle ABC\sim\triangle PQR$ .

Reconstructed from the data after the solution establishes the result.

Answer

$\triangle ABC\sim\triangle PQR$ by SAS, with factor $1.6$ ; the equal angles at $A$ and $P$ are included.

Key idea

For SAS, the two compared sides must meet at the equal angle.
Problem 3 Recover an unstated correspondence

The sides of triangle $ABC$ are
$\begin{aligned}AB&=9\\BC&=14\\CA&=16.\end{aligned}$
The sides of triangle $PQR$ are
$\begin{aligned}PQ&=21\\QR&=24\\RP&=13.5.\end{aligned}$
No correspondence is supplied. Decide whether the triangles are similar, write a valid ordered statement if they are, and state the factor from $ABC$ to the second triangle.
1. Hint 1
  A positive dilation preserves the order of side lengths.
2. Hint 2
  Match shortest, middle, and longest sides, then use the endpoints shared by two matched sides.
Answer

$\triangle ABC\sim\triangle RPQ$ by SSS, with factor $1.5$ .

Full solution

Sort the first lengths as $9, 14, 16$ and the second as $13.5, 21, 24$ .
The three target-to-original ratios are all $1.5$ .

Thus $AB$ matches $RP$ , $BC$ matches $PQ$ , and $CA$ matches $QR$ .
Vertex $B$ is shared by $AB$ and $BC$ ; its match $P$ is shared by $RP$ and $PQ$ .
The remaining matching is $A$ to $R$ and $C$ to $Q$ .

Both triples satisfy the triangle inequality.
SSS therefore gives $\triangle ABC\sim\triangle RPQ$ , with factor $1.5$ .

Reconstructed from the data after the solution establishes the result.

Answer

$\triangle ABC\sim\triangle RPQ$ by SSS, with factor $1.5$ .

Key idea

Side matching determines vertex matching through shared endpoints.
Problem 4 Use a third side after SAS is established

Triangle $ABC$ has
$\begin{aligned}AB&=5\text{ cm}\\AC&=8\text{ cm}\\BC&=7\text{ cm}\end{aligned}$
, and included angle $A=60^\circ$ . Triangle $DEF$ has
$\begin{aligned}DE&=10\text{ cm}\\DF&=16\text{ cm}\end{aligned}$
, and included angle $D=60^\circ$ .

Justify the ordered similarity and find $EF$ .
1. Hint 1
  The two known target side lengths surround the known angle.
2. Hint 2
  After proving similarity, identify the side that matches BC.
Answer

$\triangle ABC\sim\triangle DEF$ by SAS; $EF=14$ cm.

Full solution

The two side ratios are
$\begin{aligned}10/5&=2\\16/8&=2.\end{aligned}$
The equal $60^\circ$ angles are included between those side pairs, so SAS gives $\triangle ABC\sim\triangle DEF$ .

The third side $BC$ now corresponds to $EF$ .
$\begin{aligned}EF&=2\cdot 7,\\&=14\text{ cm}.\end{aligned}$

Reconstructed from the data after the solution establishes the result.

Answer

$\triangle ABC\sim\triangle DEF$ by SAS; $EF=14$ cm.

Key idea

An unknown third side follows from similarity; it is not a missing premise to assume.
Problem 5 Compare and complete the side data

Two triangles have side lengths $8, 13, 16$ and $12, 19.5, 25$ , with no correspondence specified.

(a) Decide whether the original pair is similar, considering all possible correspondences.

(b) Keep the lengths $12$ and $19.5$ in the second triangle. Replace only its length $25$ so that the triangles become similar.
1. Hint 1
  Compare the sorted side lengths; a positive factor cannot reverse their size order.
2. Hint 2
  Use the common factor from the first two size ranks to find the required third length.
Answer

Part (a). They are not similar. Part (b). Replace $25$ by $24$ ; the factor is $1.5$ .

Full solution

(a) The positive triples both form triangles.
Sorted side matching is necessary for any similarity.
The first two ratios are
$\begin{aligned}12/8&=1.5\\19.5/13&=1.5\end{aligned}$
, but $25/16$ differs from $1.5$ .
Thus no correspondence works.

(b) The retained side pairs require factor $1.5$ .
The third side must therefore be
$1.5\cdot 16=24.$

The repaired second triple $12, 19.5, 24$ satisfies the triangle inequality and all three ratios are $1.5$ , so SSS now establishes similarity.

Reconstructed from the data after the solution establishes the result.

Answer

Part (a). They are not similar. Part (b). Replace $25$ by $24$ ; the factor is $1.5$ .

Key idea

Repair a failed SSS test by enforcing the same factor on the remaining side.
Problem 6 Determine the algebraic side condition

Nondegenerate triangles $ABC$ and $DEF$ have equal included angles
$\begin{aligned}A&=D,\\&=48^\circ.\end{aligned}$
Their sides satisfy
$\begin{aligned}AB&=2x+3\\AC&=15\\DE&=27\\DF&=27.\end{aligned}$
All lengths are positive.

Find $x$ and the factor from $ABC$ to $DEF$ that make SAS similarity hold in the stated order.
1. Hint 1
  Both target sides have the same length. Think about what the same scale factor requires of their original sides.
2. Hint 2
  Use the numerical pair AC and DF to determine the factor.
Answer

$x=6$ and the factor is $\frac 95$ , or $1.8$ .

Full solution

The numerical side pair fixes the scale factor.
$\begin{aligned}k&=\frac{27}{15},\\&=\frac 95.\end{aligned}$

The other original side must also have length $15$ , because its image length is the same $27$ and its factor must be the same.
$\begin{aligned}2x+3&=15,\\x&=6.\end{aligned}$

The resulting original side lengths are positive and the equal angles are included.
Thus SAS similarity holds in the stated order with factor $1.8$ .

Answer

$x=6$ and the factor is $\frac 95$ , or $1.8$ .

Key idea

Under one positive scale factor, equal original lengths give equal image lengths, and equal image lengths require equal original lengths.
Problem 7 Assess a named SAS argument

Maya knows
$\begin{aligned}AB&=9\\BC&=6\\DE&=22.5\\EF&=15\\\angle A&=\angle D,\\&=35^\circ.\end{aligned}$
She says the two side ratios are $2.5$ , so the triangles are similar by SAS.

Identify the flaw, state which angle equality would allow SAS for the order $ABC$ to $DEF$ , and explain whether the flaw itself proves that the triangles are dissimilar.

The marked angle location is the issue in the argument.

Text description of this figure

The known side pairs are AB/DE and BC/EF. They meet at B and E. The equal marked 35-degree angles are at A and D, so they are nonincluded for those side pairs. The diagram is schematic. The arcs locate the given angles; the numeric 35-degree measure is in the statement, and no unmeasured third side or selected SSA branch is depicted to scale.
1. Hint 1
  Trace the two compared sides to the vertex where they meet.
2. Hint 2
  A failed argument and a false conclusion are different things.
Answer

Maya used nonincluded angles. SAS would need $\angle ABC\cong\angle DEF$ . Her flawed argument does not itself prove dissimilarity.

Full solution

The ratios $22.5/9$ and $15/6$ both equal $2.5$ , but the sides $AB,BC$ meet at $B$ , while $DE,EF$ meet at $E$ .
The given angles at $A,D$ are not the included ones.

SAS would be justified by the added equality $\angle ABC\cong\angle DEF$ .

Without it, the listed argument does not establish the claimed similarity.
This is insufficient evidence for her criterion, not a proof that the triangles cannot be similar.

Answer

Maya used nonincluded angles. SAS would need $\angle ABC\cong\angle DEF$ . Her flawed argument does not itself prove dissimilarity.

Key idea

An equal angle belongs in SAS only when it is between the sides used in the ratios.
Problem 8 Use SAS at an intersection

Points $A,O,C$ are collinear in that order, and $B,O,D$ are collinear in that order. No parallelism is assumed. The lengths are
$\begin{aligned}AO&=7\\CO&=10.5\\BO&=12\\DO&=18\\CD&=27.\end{aligned}$
Prove a similarity statement and find $AB$ .

The equal vertical angles are included between the two measured side pairs.

Text description of this figure

A,O,C and B,O,D are collinear in order. AO,CO,BO,DO are 7, 10.5, 12, 18 respectively. CD is 27 and AB is unknown. No parallel arrows are drawn because no parallelism is assumed. All drawn lengths are schematic.
1. Hint 1
  The side pairs AO/CO and BO/DO meet at O.
2. Hint 2
  The two angles at O are vertical angles, so they are equal.
3. Hint 3
  After proving SAS similarity, match the two outer sides.
Answer

$\triangle AOB\sim\triangle COD$ by SAS; $AB=18$ .

Full solution

The corresponding sides around $O$ have common target-to-original ratio $1.5$ .
$\begin{aligned}\frac{CO}{AO}&=\frac{10.5}{7},\\&=1.5,\\\frac{DO}{BO}&=\frac{18}{12},\\&=1.5.\end{aligned}$

The included angles $AOB$ and $COD$ are equal vertical angles.
Therefore $\triangle AOB\sim\triangle COD$ by SAS.

The outer side $AB$ corresponds to $CD$ .
$\begin{aligned}AB&=\frac{27}{1.5},\\&=18.\end{aligned}$

Answer

$\triangle AOB\sim\triangle COD$ by SAS; $AB=18$ .

Key idea

Two proportional sides around equal vertical angles can establish similarity without a parallelism premise.
Problem 9 Justify a relationship for unmeasured sides

Nondegenerate triangles $ABC$ and $DEF$ have
$\begin{aligned}AB&=4\\AC&=10\\DE&=14\\DF&=35\end{aligned}$
, and equal included angles
$\begin{aligned}A&=D,\\&=52^\circ.\end{aligned}$
Neither $BC$ nor $EF$ has been measured.

Explain why $EF/BC=7/2$ must hold, using a dilation and an earlier congruence criterion rather than assuming the missing side ratio.
1. Hint 1
  Scale the two measured sides of ABC to the measured lengths in DEF.
2. Hint 2
  Name the congruence criterion that applies after those two lengths match and the included angle is preserved.
Answer

Dilating by $\frac 72$ gives a SAS-congruent copy of $DEF$ , so $EF/BC=\frac 72$ .

Full solution

Both measured side pairs require the same factor.
$\begin{aligned}\frac{14}{4}&=\frac 72,\\\frac{35}{10}&=\frac 72.\end{aligned}$

Dilate $ABC$ by $7/2$ .
The two selected side lengths become $14$ and $35$ , while the included angle stays $52^\circ$ .
Earlier SAS congruence gives a rigid-motion match with $DEF$ .

Rigid motions preserve the scaled third length.
The image of $BC$ therefore has length $EF$ , giving
$EF=\frac 72\,BC.$

The side $BC$ is positive, so dividing gives the requested ratio.
The third-side relation is a consequence of SAS, not a premise used to prove it.

Reconstructed from the data after the solution establishes the result.

Answer

Dilating by $\frac 72$ gives a SAS-congruent copy of $DEF$ , so $EF/BC=\frac 72$ .

Key idea

Dilation and congruence can justify a ratio for a side that has never been measured.
Problem 10 Separate possible triangles from similar triangles

One triangle has side lengths $6, 10, 13$ . A second has side lengths $12, 26,q$ . Here, $q>0$ . No vertex correspondence is specified.

(a) Find the full range of $q$ for which the second triangle is nondegenerate.

(b) Find every value of $q$ that makes the triangles similar, considering possible correspondences.

(c) Give one permitted value of $q$ that refutes the claim “Two proportional side pairs guarantee similarity.”
1. Hint 1
  For part (a), use the triangle inequality with q as the varying side.
2. Hint 2
  For part (b), compare the ratio 26/12 with ratios of side pairs in the first triangle.
3. Hint 3
  For part (c), keep q inside its valid range but different from every similarity value.
Answer

Part (a).
$14<q<38.$
Part (b).
$q=20.$
Part (c). For example, $q=18$ gives two valid triangles with two proportional side pairs but no similarity.

Full solution

(a) A positive third side lies strictly between the difference and sum of the other two sides.
$\begin{aligned}q&>26-12=14,\\q&<26+12=38.\end{aligned}$

(b) The known ratio $26/12$ is $13/6$ .
Among the first triangle's three side lengths, only the pair $13$ and $6$ has this ratio.
Thus any similarity must match $6$ to $12$ and $13$ to $26$ , with factor $2$ .
The remaining length must be
$\begin{aligned}q&=2\cdot 10,\\&=20.\end{aligned}$

This value lies in the permitted interval, and all three side pairs have factor $2$ , so SSS proves similarity.
No other correspondence can produce another value.

(c) Choose $q=18$ , which lies strictly between $14$ and $38$ .
The sorted side triples are $6, 10, 13$ and $12, 18, 26$ .
Their corresponding ratios are $2$ , $1.8$ , and $2$ , so no positive common factor works.

Two matching side pairs are proportional, but the triangles are not similar.
This one valid pair of triangles refutes the universal claim.

Reconstructed from the data after the solution establishes the result.

Answer

Part (a).
$14<q<38.$
Part (b).
$q=20.$
Part (c). For example, $q=18$ gives two valid triangles with two proportional side pairs but no similarity.

Key idea

Triangle existence, a complete similarity test, and a two-side partial match answer different questions.

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