Chapter Test · nothing is marked until you submit

Similarity and Proportion: Chapter Test

20 multiple-choice questions and 10 core practice problems, drawn from across the chapter and mixed together.

Core practice

10 problems from across the chapter. 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)

0 of 10 done

Progress saved in this browser.

Problem 1 of 10

Problem 1 A point and two horizontal levels

A dilation has center $O(4,-3)$ and positive scale factor $\frac23$ .

(a) Find the image of $P(-5,6)$ .

(b) Find the image of the entire line $y=9$ .

(c) State which points of the plane are fixed by this dilation.
1. Hint 1
  Measure horizontal and vertical displacements from the stated center.
2. Hint 2
  For the line, every point has the same vertical displacement from the center. Its horizontal coordinates still range over the whole line.
3. Hint 3
  Distinguish a fixed point from a line that maps onto itself.
Answer

(a)
$\begin{gathered}P'=(-2,3)\end{gathered}$
(b) The image is the line
$\begin{gathered}y=5\end{gathered}$
(c) Only $O(4,-3)$ is fixed.

Full solution

(a) The displacement from O to P is nine units left and nine units up.
Scale each displacement, then add the center back.
$\begin{gathered}x'=4+\frac23(-5-4)\\x'=-2\\y'=-3+\frac23(6+3)\\y'=3\end{gathered}$

(b) Every point on the original horizontal line has vertical coordinate nine.
$\begin{gathered}y'=-3+\frac23(9+3)\\y'=5\end{gathered}$

The horizontal rule has a nonzero coefficient, so all real horizontal coordinates occur as images.
The image is therefore the entire horizontal line with vertical coordinate five.

(c) The factor is positive and different from one.
The center is fixed, while the positive distance of any other point from the center changes.
Thus only O is fixed.

Answer

(a)
$\begin{gathered}P'=(-2,3)\end{gathered}$
(b) The image is the line
$\begin{gathered}y=5\end{gathered}$
(c) Only $O(4,-3)$ is fixed.

Key idea

A dilation scales displacements from its center, and an image of a whole line includes every transformed point on that line.
Problem 2 Two angle records

Nondegenerate triangle $KLM$ has angles
$\begin{gathered}m\angle K=34^\circ\\m\angle L=83^\circ\end{gathered}$
Nondegenerate triangle $RST$ has angles
$\begin{gathered}m\angle R=63^\circ\\m\angle S=34^\circ\end{gathered}$
The lengths are
$\begin{gathered}KL=7.5\text{ cm}\\ST=13.5\text{ cm}\end{gathered}$
(a) Find the angles at M and T. Establish similarity and write a correct ordered statement starting with KLM.

(b) Find the factor mapping triangle RST to triangle KLM.
1. Hint 1
  Complete each triangle's angle set before matching vertices.
2. Hint 2
  Pair equal angle measures, then identify which side corresponds to KL.
3. Hint 3
  The requested factor has lengths from KLM above their matches from RST.
Answer

(a)
$\begin{gathered}m\angle M=63^\circ\\m\angle T=83^\circ\\\triangle KLM\sim\triangle STR\end{gathered}$
The similarity follows by AA. (b) The RST-to-KLM factor is $\frac59$ .

Full solution

(a) Use the triangle-angle sum separately.
$\begin{gathered}m\angle M=180^\circ-34^\circ-83^\circ\\m\angle M=63^\circ\\m\angle T=180^\circ-63^\circ-34^\circ\\m\angle T=83^\circ\end{gathered}$

K matches S, L matches T, and M matches R.
Two distinct equal angle pairs establish AA, giving
$\begin{gathered}\triangle KLM\sim\triangle STR\end{gathered}$

(b) Side KL matches ST.
The requested direction is from the larger RST triangle to KLM.
$\begin{gathered}k=\frac{7.5}{13.5}\\k=\frac59\end{gathered}$

Answer

(a)
$\begin{gathered}m\angle M=63^\circ\\m\angle T=83^\circ\\\triangle KLM\sim\triangle STR\end{gathered}$
The similarity follows by AA. (b) The RST-to-KLM factor is $\frac59$ .

Key idea

Match equal angles before choosing the direction of a corresponding-side ratio.
Problem 3 An interior division and two labels

In nondegenerate triangle $ABC$ , $AD$ internally bisects $\angle BAC$ and $D$ lies strictly between $B,C$ .

The lengths are
$\begin{gathered}AB=4t-1\\AC=5t+2\\BD=12\\DC=16\end{gathered}$
Find t and both adjacent side lengths. Check that the resulting outer side lengths form a nondegenerate triangle.

Text description of this figure

AD is the given internal angle bisector, shown by two equal arcs. D lies strictly inside BC. The schematic drawing identifies which opposite-side part is beside each adjacent side. The algebraic and numeric givens are in the statement, and no solved value is shown.
1. Hint 1
  Pair each base part with the adjacent side sharing its outer endpoint.
2. Hint 2
  Set the adjacent-side ratio equal to the part-to-part ratio in the same order.
3. Hint 3
  After solving, substitute into both labels and check the longest outer side against the other two.
Answer

$\begin{gathered}t=10\\AB=39\\AC=52\end{gathered}$
The outer sides 39, 52 and 28 form a nondegenerate triangle.

Full solution

The internal bisector pairs AB with BD and AC with DC.
$\begin{gathered}\frac{4t-1}{5t+2}=\frac{12}{16}\\4(4t-1)=3(5t+2)\\16t-4=15t+6\\t=10\end{gathered}$

Substitute into the original adjacent-side expressions.
$\begin{gathered}AB=4(10)-1\\AB=39\\AC=5(10)+2\\AC=52\end{gathered}$

The two base parts add to the whole side.
$\begin{gathered}BC=12+16\\BC=28\end{gathered}$

All outer lengths are positive.
Their largest side is shorter than the sum of the other two.
$\begin{gathered}52<39+28\end{gathered}$

The adjacent-side ratio reduces to three fourths, matching twelve sixteenths, so the ratio also checks.

Answer

$\begin{gathered}t=10\\AB=39\\AC=52\end{gathered}$
The outer sides 39, 52 and 28 form a nondegenerate triangle.

Key idea

An angle-bisector ratio must agree in endpoint order and still describe compatible positive side lengths.
Problem 4 Two triangles inside a boundary

Triangle $ABC$ has side lengths
$\begin{gathered}AB=26\text{ cm}\\AC=35\text{ cm}\\BC=41\text{ cm}\end{gathered}$
Let D, E and F be the midpoints of AB, AC and BC. Let G, H and I be the midpoints of DE, EF and FD, respectively.

(a) Find GH, HI and IG.

(b) Write the natural ordered similarity from GHI to the outer triangle, and give the factor in that direction. Explain the side matching.
1. Hint 1
  Each new midpoint triangle has sides half as long as the previous triangle's corresponding sides.
2. Hint 2
  Track which original outer side is parallel to each side of GHI.
3. Hint 3
  A vertex match is determined by the two matching sides that meet there.
Answer

(a)
$\begin{gathered}GH=8.75\text{ cm}\\HI=10.25\text{ cm}\\IG=6.5\text{ cm}\end{gathered}$
(b)
$\begin{gathered}\triangle GHI\sim\triangle ACB\end{gathered}$
The inner-to-outer factor is $4$ .

Full solution

(a) The first midpoint triangle gives
$\begin{gathered}DE=\frac12BC\\DE=20.5\text{ cm}\\EF=\frac12AB\\EF=13\text{ cm}\\FD=\frac12AC\\FD=17.5\text{ cm}\end{gathered}$

The second midpoint triangle pairs GH with FD, HI with DE, and IG with EF.
$\begin{gathered}GH=\frac12(17.5)\\GH=8.75\text{ cm}\\HI=\frac12(20.5)\\HI=10.25\text{ cm}\\IG=\frac12(13)\\IG=6.5\text{ cm}\end{gathered}$

(b) Thus GH matches AC, HI matches CB, and IG matches BA.
The common endpoints give G to A, H to C, and I to B.
Hence
$\begin{gathered}\triangle GHI\sim\triangle ACB\end{gathered}$

Each inner length is one quarter of its outer match.
Reversing to the requested inner-to-outer mapping gives factor four.

Answer

(a)
$\begin{gathered}GH=8.75\text{ cm}\\HI=10.25\text{ cm}\\IG=6.5\text{ cm}\end{gathered}$
(b)
$\begin{gathered}\triangle GHI\sim\triangle ACB\end{gathered}$
The inner-to-outer factor is $4$ .

Key idea

Repeated half-length factors multiply, while the ordered vertex matching must still follow side endpoints.
Problem 5 Across the bank

In a top-view river survey, A, B and C lie on the same accessible straight bank in that order. Target T is across the river. The sight line contains T, B and D in that order, with D on dry land.

The perpendiculars and measured lengths are
$\begin{gathered}AT\perp AC\\CD\perp AC\\AB=46.0\text{ m}\\BC=20.0\text{ m}\\CD=13.5\text{ m}\end{gathered}$
Establish an ordered similarity and estimate AT to the nearest tenth of a meter. Explain why the result is the perpendicular crossing rather than the sight-line length TB.

Text description of this figure

A, B and C are in order on the accessible straight bank. The sight line passes through T, B and D in that order. Right-angle marks give AT and CD perpendicular to the bank. Only the supplied measurements in the statement may be used; AT is not labeled with an answer and the drawing is schematic.
1. Hint 1
  The bank perpendiculars give one equal angle pair; identify the pair at the crossing point.
2. Hint 2
  Match the inaccessible perpendicular with the accessible perpendicular, and keep bank intervals in that same order.
3. Hint 3
  Keep the unrounded calculation until reporting the final tenth.
Answer

$\begin{gathered}\triangle TAB\sim\triangle DCB\\AT\approx31.1\text{ m}\end{gathered}$
This is the perpendicular crossing AT, not the slanted sight line TB.

Full solution

The right angles at A and C match.
The angles TAB and DCB are therefore equal, and the angles TBA and DBC are vertical angles.
These two pairs give AA with
$\begin{gathered}\triangle TAB\sim\triangle DCB\end{gathered}$

The matching is T to D, A to C, and B to B.
Thus AT matches DC and AB matches CB.
$\begin{gathered}\frac{AT}{13.5}=\frac{46.0}{20.0}\\AT=13.5\left(\frac{46.0}{20.0}\right)\\AT=31.05\text{ m}\end{gathered}$

Round the final model calculation to the requested tenth.
$\begin{gathered}AT\approx31.1\text{ m}\end{gathered}$

The calculated side is AT because it corresponds to the measured perpendicular CD.
AT is perpendicular to the bank, while TB follows the slanted sight line to B.

Answer

$\begin{gathered}\triangle TAB\sim\triangle DCB\\AT\approx31.1\text{ m}\end{gathered}$
This is the perpendicular crossing AT, not the slanted sight line TB.

Key idea

Established angles determine the correspondence, and the correspondence identifies which physical distance a ratio measures.
Problem 6 Two interior segments

In nondegenerate triangle ABC, AD internally bisects angle BAC and D lies strictly between B and C. Point E lies strictly between A and B, point F strictly between A and C, and EF is parallel to BC.

Let G be the intersection of AD with EF. The known lengths are
$\begin{gathered}AB=18\text{ cm}\\AC=24\text{ cm}\\AE=12\text{ cm}\\EF=21\text{ cm}\end{gathered}$
(a) Find AF and BC.

(b) Find EG and GF. Explain why the angle relationship needed for this calculation holds at A in the smaller triangle.

Text description of this figure

In triangle ABC, E is inside AB and F inside AC. Matching arrows give EF parallel to BC. The internal bisector AD passes from A through G on EF to D on BC. The small triangle AEF uses the same rays at A as the whole triangle. No calculated part length is shown; the diagram is schematic.
1. Hint 1
  Use the same small-to-whole fraction for corresponding sides of the parallel cut.
2. Hint 2
  The rays AE and AB coincide, as do AF and AC. The ray AG is the given bisector ray.
3. Hint 3
  In the smaller triangle, compare EG with GF using its adjacent sides and then the whole EF.
Answer

(a)
$\begin{gathered}AF=16\text{ cm}\\BC=31.5\text{ cm}\end{gathered}$
(b)
$\begin{gathered}EG=9\text{ cm}\\GF=12\text{ cm}\end{gathered}$
AG bisects angle EAF because its rays at A are the original bisected angle rays.

Full solution

(a) The parallel cut gives the same small-to-whole ratio on both sides and the base.
$\begin{gathered}\frac{AE}{AB}=\frac{12}{18}\\\frac{AE}{AB}=\frac23\\AF=\frac23(24)\\AF=16\text{ cm}\end{gathered}$

Use that same fraction for EF and BC.
$\begin{gathered}\frac{21}{BC}=\frac23\\BC=31.5\text{ cm}\end{gathered}$

(b) Rays AE and AF are AB and AC, and ray AG is AD.
Therefore the equal angles BAD and DAC are also the equal angles EAG and GAF.
G is the interior bisector intersection in triangle AEF.

The internal angle-bisector ratio in AEF gives
$\begin{gathered}\frac{EG}{GF}=\frac{12}{16}\\EG:GF=3:4\end{gathered}$

The whole EF contains seven ratio parts.
$\begin{gathered}EG=21\left(\frac37\right)\\EG=9\text{ cm}\\GF=21-9\\GF=12\text{ cm}\end{gathered}$

Answer

(a)
$\begin{gathered}AF=16\text{ cm}\\BC=31.5\text{ cm}\end{gathered}$
(b)
$\begin{gathered}EG=9\text{ cm}\\GF=12\text{ cm}\end{gathered}$
AG bisects angle EAF because its rays at A are the original bisected angle rays.

Key idea

A parallel cut controls the size of the smaller triangle, while the unchanged vertex rays retain the angle-bisector relationship.
Problem 7 A division to retain

You are supplied with actual segments a and b of lengths eleven and sixteen millimeters, respectively, and an unmarked segment AB.

On paper, construct an interior point X on AB so that
$\begin{gathered}AX:XB=11:16\end{gathered}$
Use a compass and an unmarked straightedge, and retain the construction marks. Explain why your construction gives the requested ratio without reading a quotient from a ruler.

Text description of this figure

The first supplied segment a has length eleven millimeters and a single equality tick. Segment b has length sixteen millimeters and a double equality tick. A separate segment AB is shown without a numerical length or division point. The requested point must be constructed on the student's paper.
1. Hint 1
  Transfer the two supplied lengths in their given order rather than numerically measuring part of AB.
2. Hint 2
  A parallel in an auxiliary triangle can transfer the consecutive side-part ratio to AB.
3. Hint 3
  Keep the compass marks that identify each copied length and the angle-copy marks establishing the parallel.
Answer

Copy a then b as consecutive AU and UV on an auxiliary ray from A. Join VB; construct the parallel through U meeting AB at X. The result has
$\begin{gathered}AX:XB=11:16\end{gathered}$

Full solution

Draw an auxiliary ray from A that is not on line AB.
Copy the supplied first length a as AU using the compass.

From U, copy the supplied second length b farther along the same ray as UV.
Thus A, U and V occur in that order, with
$\begin{gathered}AU=11\text{ mm}\\UV=16\text{ mm}\end{gathered}$

Join V to B.
Use the earlier angle-copy construction to construct the parallel to VB through U.
Call its intersection with AB point X, retaining both the length-copy and angle-copy marks.

U is strictly inside AV because both copied lengths are positive.
The parallel cut places X strictly inside AB.
The side-splitter theorem in triangle AVB gives
$\begin{gathered}\frac{AX}{XB}=\frac{AU}{UV}\\\frac{AX}{XB}=\frac{11}{16}\end{gathered}$

The construction therefore has the requested ordered ratio.
It copies actual supplied lengths and uses a constructed parallel, without measuring a decimal quotient on AB.

Answer

Copy a then b as consecutive AU and UV on an auxiliary ray from A. Join VB; construct the parallel through U meeting AB at X. The result has
$\begin{gathered}AX:XB=11:16\end{gathered}$

Key idea

Consecutive copied lengths and a constructed parallel transfer an ordered part-to-part ratio exactly.
Problem 8 A boundary on a print

A triangular plan has one uniform drawing-to-real scale of $1:6400$ . Its boundary side lengths are $2.8$ , $4.1$ and $5.6$ cm.

A printer offers these two coordinate rules about the origin.

Rule A:
$\begin{gathered}(x,y)\longmapsto(1.25x,1.25y)\end{gathered}$
Rule B:
$\begin{gathered}(x,y)\longmapsto(1.25x,1.5y)\end{gathered}$
Which rule is a dilation of the whole plane? For that rule, find the new print-to-real scale and the real perimeter of the boundary in meters. Explain why the other rule has no single similarity scale.
1. Hint 1
  A dilation applies one positive length factor in every direction.
2. Hint 2
  Real lengths stay fixed while a uniform print changes every drawing length by its printer factor.
3. Hint 3
  The perimeter includes all three original boundary sides. Keep drawing and real units consistent.
Answer

Rule A is a dilation. Its new print-to-real scale is $1:5120$ , and the real perimeter is $800$ m. Rule B has no single similarity scale because its horizontal and vertical factors differ.

Full solution

Rule A multiplies both coordinate displacements from the origin by the same positive factor.
Rule B has horizontal factor 1.25 and vertical factor 1.5, so it is a directional stretch rather than a dilation.

For Rule A, each new drawing length is 1.25 times its original length while the real length is unchanged.
$\begin{gathered}\frac{L_{\rm print}}{L_{\rm real}}=\frac{1.25}{6400}\\\frac{L_{\rm print}}{L_{\rm real}}=\frac1{5120}\end{gathered}$

The new print-to-real scale is therefore one to 5120.
Add every boundary side on the original plan.
$\begin{gathered}P_{\rm original}=2.8+4.1+5.6\\P_{\rm original}=12.5\text{ cm}\end{gathered}$

The original uniform length factor also scales the entire boundary sum.
$\begin{gathered}P_{\rm real}=6400(12.5)\\P_{\rm real}=80000\text{ cm}\\P_{\rm real}=800\text{ m}\end{gathered}$

As a check, Rule A prints a perimeter of 15.625 centimeters.
Multiplying that by 5120 gives the same real perimeter of 80000 centimeters.

Answer

Rule A is a dilation. Its new print-to-real scale is $1:5120$ , and the real perimeter is $800$ m. Rule B has no single similarity scale because its horizontal and vertical factors differ.

Key idea

A uniform reprint changes the drawing ratio while leaving real boundary lengths unchanged.
Problem 9 Two sets of sufficient data

For each case, justify that the triangles are similar using results established before the triangle similarity criteria. Give an ordered statement and identify any scale factor your argument needs.

Case (a): Nondegenerate triangles KLM and RST have
$\begin{gathered}m\angle K=m\angle R\\m\angle K=41^\circ\\m\angle L=m\angle S\\m\angle L=66^\circ\\KM=11\\RT=17.6\end{gathered}$
Case (b): Nondegenerate triangles UVW and XYZ have
$\begin{gathered}UV=13\\UW=20\\XY=16.9\\XZ=26\\m\angle VUW=m\angle YXZ\\m\angle VUW=49^\circ\end{gathered}$
In either case, do not assume that an unmeasured third-side ratio already holds.
1. Hint 1
  A positive dilation can make a chosen corresponding side pair equal while preserving the angle data.
2. Hint 2
  In case (a), the selected side is not between the two selected angle vertices.
3. Hint 3
  In case (b), the two measured sides meet at the given equal angle. Name the earlier congruence result that becomes available after resizing.
Answer

(a) Factor $\frac85$ and AAS congruence give
$\begin{gathered}\triangle KLM\sim\triangle RST\end{gathered}$
(b) Factor $1.3$ and SAS congruence give
$\begin{gathered}\triangle UVW\sim\triangle XYZ\end{gathered}$

Full solution

Case (a): Match KM with RT by choosing
$\begin{gathered}k=\frac{17.6}{11}\\k=\frac85\end{gathered}$

Dilate KLM by this factor.
Its image of KM now has length 17.6, matching RT.
The two given angle pairs remain equal because dilation preserves angles.

KM is not the side between K and L, so the preserved two angle pairs and matched nonincluded side give earlier AAS congruence with RST.
Its rigid-motion match completes the similarity
$\begin{gathered}\triangle KLM\sim\triangle RST\end{gathered}$

An alternative is to first derive equal third angles and then use ASA with the selected side; that route must include its additional angle-sum step.

Case (b): Both measured pairs require the same positive factor.
$\begin{gathered}\frac{16.9}{13}=1.3\\\frac{26}{20}=1.3\end{gathered}$

Dilate UVW by 1.3.
The selected image sides become 16.9 and 26, and their included angle stays 49 degrees.
Earlier SAS congruence gives a rigid-motion match with XYZ, so
$\begin{gathered}\triangle UVW\sim\triangle XYZ\end{gathered}$

Neither argument assumes its unmeasured third-side ratio.
The ratio follows from the established dilation and rigid-motion match.

Answer

(a) Factor $\frac85$ and AAS congruence give
$\begin{gathered}\triangle KLM\sim\triangle RST\end{gathered}$
(b) Factor $1.3$ and SAS congruence give
$\begin{gathered}\triangle UVW\sim\triangle XYZ\end{gathered}$

Key idea

A dilation followed by earlier congruence establishes similarity without assuming the side ratios that are still to be proved.
Problem 10 Two conclusions from one configuration

In nondegenerate triangle ABC, D lies strictly between B and C and E lies strictly between B and A.

The outer side lengths are
$\begin{gathered}AB=30\\AC=42\\BC=36\end{gathered}$
The side parts are
$\begin{gathered}BD=15\\DC=21\\BE=12.5\\EA=17.5\end{gathered}$
(a) Prove that AD internally bisects angle BAC.

(b) Prove that DE is parallel to AC.

Explain which two kinds of segment ratio support the two conclusions, even though both comparisons reduce to the same numerical value.

Text description of this figure

The outer triangle is ABC. D is an interior point of BC and E an interior point of BA. Segment AD and segment DE have been drawn, but no angle halves are marked equal and no parallel arrows appear. The given side-part and outer-side lengths are in the statement. The schematic drawing leaves both requested conclusions to be established.
1. Hint 1
  For part (a), compare the parts of BC with the adjacent sides AB and AC in the same endpoint order.
2. Hint 2
  For part (b), regard B as the common vertex of the whole triangle and compare the interior divisions on BC and BA.
3. Hint 3
  Distinguish an opposite-side-to-adjacent-side comparison from a comparison of the two divided sides.
Answer

(a) AD internally bisects angle BAC. (b)
$\begin{gathered}DE\parallel AC\end{gathered}$
The first conclusion compares opposite-side parts with adjacent sides; the second compares first and leftover parts on BC and BA.

Full solution

(a) Compare the opposite-side parts with the adjacent sides.
$\begin{gathered}\frac{BD}{DC}=\frac{15}{21}\\\frac{BD}{DC}=\frac57\\\frac{AB}{AC}=\frac{30}{42}\\\frac{AB}{AC}=\frac57\end{gathered}$

The ratios agree, D is interior to BC, and the outer triangle is nondegenerate.
The angle-bisector ratio converse therefore proves that AD internally bisects angle BAC.

(b) View the whole triangle in the order BCA.
D is inside BC and E is inside BA.
Their first-part-to-leftover ratios are
$\begin{gathered}\frac{BD}{DC}=\frac57\\\frac{BE}{EA}=\frac{12.5}{17.5}\\\frac{BE}{EA}=\frac57\end{gathered}$

The two divided-side ratios agree, so the side-splitter converse gives
$\begin{gathered}DE\parallel CA\end{gathered}$

The first comparison uses the adjacent whole sides AB and AC to identify an internal angle bisector.
The second uses the interior divisions of BC and BA to identify a parallel cut.
Equal numerical ratios do not make those two theorem hypotheses interchangeable.

Answer

(a) AD internally bisects angle BAC. (b)
$\begin{gathered}DE\parallel AC\end{gathered}$
The first conclusion compares opposite-side parts with adjacent sides; the second compares first and leftover parts on BC and BA.

Key idea

Choose a converse by the geometric roles of its ratios, retaining every interior-point and nondegeneracy condition.

Multiple choice

Core practice