GRE · Geometry · TOPIC LESSON
Track corresponding parts and scale each measurement by the right power.
BY THE END, YOU CAN
BEFORE YOU START
Original educational lesson. Work through the examples before revealing the practice answers. This original lesson supports current GRE General Test study; it does not reproduce official questions, certify a writing score or predict test content.
LEARN THROUGH ORIGINAL PRACTICE
All teaching passages, writing prompts, examples and exercises in this lesson are original. No official ETS questions, prompts or explanations are reproduced. Current structure and observed public sample metadata are separate from this original teaching; no official score or question probability is implied.
01 / UNDERSTAND & APPLY
All examples and practice in this lesson are original. For quantitative comparison (QC), use A if Quantity A is greater, B if Quantity B is greater, C if they are equal, and D if the relationship cannot be determined from the information given. A, B or C must hold for every permitted case. To establish D, give two allowed cases producing different relationships; an inability to calculate is not itself evidence for D.
A single-answer question requires one choice. For multiple-answer questions, obey the stated selection count or select every correct option if instructed. Test each option separately; do not stop after finding one. Multiple-answer questions receive credit only when the complete correct set is selected, with no incorrect choices. Numeric entry requires the requested value, with units handled in the calculation; round only when the question directs you to. A single numeric-entry box accepts an integer or decimal; a pair of fraction boxes accepts an integer numerator and a nonzero integer denominator. Fractions need not be reduced, but reducing makes checks easier. Do not enter a percent sign when the requested number is a percentage. Preserve exact fractions until the last step.
A calculator can check arithmetic but cannot decide the domain, interpretation or denominator for you. Estimate sign and size first, enter parentheses carefully, and compare the result with the original conditions. Cancel factors or exploit structure when doing so is faster and more reliable than decimal entry.
02 / UNDERSTAND & APPLY
Congruent figures have the same shape and size. A translation, rotation or reflection can match one to the other without stretching. Corresponding side lengths, angles, areas, and, for solids, volumes are equal. A reflected figure can still be congruent even though its orientation looks reversed.
Similar figures have the same shape but may differ in size. Corresponding angles are equal and all corresponding lengths share one positive scale factor k. Congruence is similarity with k = 1. If a figure is enlarged only horizontally and not vertically, the change generally destroys similarity.
Equal area does not establish either relationship. A 2 by 8 rectangle and a 4 by 4 square each have area 16, but their side proportions differ. Equal perimeter likewise does not fix shape. Use actual shape conditions rather than one aggregate measurement.
03 / UNDERSTAND & APPLY
When triangle ABC is stated to be similar to triangle DEF in that order, A corresponds to D, B to E, and C to F. Thus AB matches DE, BC matches EF, and AC matches DF. Matching vertices by page position is unsafe because either triangle may be rotated or reflected.
Choose one direction for every ratio, such as image/original. If DE/AB = k, then EF/BC and DF/AC must equal that same k. Crossing the direction in one fraction produces a false equation even though all the relevant side lengths appear somewhere in it.
Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.
ORIGINAL WORKED EXAMPLE
Triangle ABC is similar to triangle DEF in that order. AB = 6, BC = 8, DE = 15, and DF = 25. Find AC and EF.
Answer AC = 10 and EF = 20.
04 / UNDERSTAND & APPLY
For triangles, two matching angle measures establish similarity because the third pair must also match. Three proportional corresponding sides also establish similarity. Two proportional corresponding sides together with the equal included angle provide another sufficient condition. These are ways to recognize a valid model; constructing formal proofs is not required here.
One equal angle alone is insufficient. So are two pairs of proportional sides without the needed angle condition. For quadrilaterals, even all matching angles can be insufficient: all rectangles have four right angles, but a 2 by 3 rectangle is not similar to a 2 by 5 rectangle. Both angle correspondence and a uniform side scale must fit.
For triangle congruence, three matching sides or two matching sides with their included angle fix shape and size. Two corresponding angles and one corresponding side also suffice. Three equal angles alone establish similarity, not necessarily congruence. This difference prevents an angle-only diagram from silently fixing a length.
05 / UNDERSTAND & APPLY
If D lies on AB, E lies on AC, and DE is parallel to BC, then triangles ADE and ABC share angle A and have matching corresponding angles from the parallel lines. They are similar, with AD/AB = AE/AC = DE/BC. Use whole sides in the larger triangle: AB is AD + DB, not DB alone.
A frequent trap is comparing a small side to the leftover segment instead of to its corresponding full side. A ratio such as AD:DB = 2:3 means AD:AB = 2:5. The trapezoid left below the parallel segment is not itself the large triangle; subtract areas when the leftover region is requested.
ORIGINAL WORKED EXAMPLE
In triangle ABC, D lies on AB and E on AC, with DE parallel to BC. AD = 4, DB = 6, and BC = 15. The area of ABC is 75. Find DE and the area of quadrilateral DBCE.
Answer DE = 6; remaining area = 63 square units.
06 / UNDERSTAND & APPLY
Shadow problems use similar right triangles only under compatible conditions: vertical objects on the same level ground, light rays making the same angle with that ground, and measurements taken at the same relevant time and location. Heights correspond to heights, and horizontal shadow lengths correspond to horizontal shadow lengths.
Maps and scale drawings likewise convert lengths with a stated ratio. A scale “1 cm represents 8 m” is a unit conversion, not the dimensionless ratio 1:8. Express both lengths in the same unit before writing a dimensionless scale, and do not apply a map’s linear scale directly to its area.
ORIGINAL WORKED EXAMPLE
On level ground at the same time, a vertical 1.8 m post casts a 1.2 m shadow and a vertical pole casts a 7 m shadow. Assume the sunlight rays are parallel. Find the pole’s height.
Answer 10.5 m.
07 / UNDERSTAND & APPLY
If every length is multiplied by k, each perimeter is multiplied by k because it is a sum of lengths. Area is multiplied by k². For a triangle this follows from (kb)(kh)/2 = k²bh/2; for a rectangle it follows from (kl)(kw) = k²lw. Heights and diagonals are corresponding lengths too, so they use k, not k².
If corresponding areas have ratio a:b, the length ratio is √a:√b, taking positive roots. An area factor of 9 does not mean a length factor of 9; it means a length factor of 3. A percentage change must also be converted to a factor before squaring: a 20% increase in length means factor 1.2, hence area factor 1.44 and an area increase of 44%.
ORIGINAL WORKED EXAMPLE
Two similar polygons have areas 98 cm² and 200 cm². The smaller has perimeter 35 cm. Find the larger perimeter.
Answer 50 cm.
08 / UNDERSTAND & APPLY
For geometrically similar solids with length factor k, surface area scales by k² and volume by k³. A rectangular prism’s volume illustrates the third power: (kl)(kw)(kh) = k³lwh. Every dimension must be scaled uniformly for a similarity claim. Doubling only the height is a different change.
Given a volume ratio, take its positive cube root to find the length ratio; then square that length ratio if surface area is requested. Do not take a square root of a volume ratio just because the target is an area. First recover k, then use the power appropriate to the target.
ORIGINAL WORKED EXAMPLE
Two similar solids have volumes 64 cm³ and 216 cm³. The smaller has surface area 80 cm². Find the larger surface area.
Answer 180 cm².
09 / UNDERSTAND & APPLY
A rectangle whose length is doubled and width halved retains its area, since the product changes by 2×1/2 = 1. It is not generally similar to the original. Its perimeter may change because the sum of lengths responds differently from their product. Check for one common factor before applying similarity shortcuts.
Similarly, two cylinders with equal volumes need not be similar: one can be broad and shallow while the other is narrow and tall. Matching one radius ratio does not determine a height ratio unless similarity is supplied. A dimensionless scale factor cannot create missing shape information.
ORIGINAL WORKED EXAMPLE
QC. Two rectangles have areas 24 and 54. Quantity A: the perimeter of the 24-area rectangle. Quantity B: the perimeter of the 54-area rectangle.
Answer D: the relationship cannot be determined.
10 / UNDERSTAND & APPLY
For QC use A = first quantity greater, B = second greater, C = equal, D = relationship not determined. Use each item’s exact selection directions.
ORIGINAL PRACTICE · 1
20
PQ corresponds to XY, giving factor 15/9 = 5/3. QR corresponds to YZ, so YZ = 12×5/3 = 20.
ORIGINAL PRACTICE · 2
B. 4:9
The length ratio simplifies to 2:3. Squaring both terms gives area ratio 4:9.
ORIGINAL PRACTICE · 3
12
AD is 3 of the full 5 shares in AB, so the small-to-large scale is 3/5. DE = (3/5)20 = 12.
ORIGINAL PRACTICE · 4
A and B
Two matching angles fix the third pair, and three uniform side ratios fix the same shape. One angle and one aggregate perimeter condition separately leave many shapes possible.
ORIGINAL PRACTICE · 5
52.5
Each centimeter represents 15/2 = 7.5 m. The wall is 7×7.5 = 52.5 m long.
11 / UNDERSTAND & APPLY
If similarity is given, use corresponding-length factor k, area factor k² and volume factor k³. If it is not given, check whether other facts establish it.
ORIGINAL PRACTICE · 1
51
The length factor is 0.7. The area factor is 0.49, so the decrease is 1 − 0.49 = 0.51, or 51%.
ORIGINAL PRACTICE · 2
15
The linear ratio is 2:5 after taking cube roots. Larger height is 6×5/2 = 15 cm.
ORIGINAL PRACTICE · 3
C
Congruence preserves all corresponding lengths, regardless of rotation or reflection, so their sums are equal.
ORIGINAL PRACTICE · 4
D
Two equilateral triangles with sides 1 and 2 satisfy the angle condition and make the first area smaller. Reversing the side assignments makes it larger. One angle does not set size.
ORIGINAL PRACTICE · 5
B and D
Surface area uses k² = 16 and volume uses k³ = 64. Count how many dimensions are multiplied.
ORIGINAL PRACTICE · 6
C. They have equal areas but different perimeters
Both areas are 24. Their perimeters are 22 and 20, respectively, and no pairing of the side lengths has one common scale factor.
KEEP THE BIG IDEAS
READ WITH CONTEXT
The teaching notes, passages, worked explanations and practice are original. Official references identify the assessment framework and skill scope; they do not endorse this lesson or license official question reproduction.
GRE analytics distinguish current test structure from a small independently classified public sample selection. Writing instruction has no observed prompt-frequency denominator. These are not historical administered-exam frequencies.