GRE · Geometry · TOPIC LESSON
Build reliable side, angle and area models before reaching for a formula.
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
A nondegenerate triangle has three positive side lengths and three positive interior angles totaling 180°. The largest side is opposite the largest angle, and equal sides face equal angles. Do not pair a side with an angle merely because they appear next to each other: “opposite” means that the side does not touch the angle’s vertex.
An exterior angle formed by extending one side is supplementary to the interior angle at that vertex. It therefore equals the sum of the other two interior angles. This is a shortcut derived from the 180° total, not an additional arbitrary formula.
ORIGINAL WORKED EXAMPLE
A triangle has an exterior angle of 124°. Its two remote interior angles are x° and (2x + 10)°. Find all three interior angles.
Answer 38°, 86°, and 56°.
03 / UNDERSTAND & APPLY
Any two sides must sum to more than the third. With two known sides a and b, the third side c must satisfy |a − b| < c < a + b. Both inequalities are strict: equality flattens the figure into a line rather than a triangle. Use this interval before counting integer possibilities or accepting an algebraic solution.
Knowing two sides alone does not normally determine the third side, the included angle, or the area. A triangle with sides 5 and 7 can open widely or narrowly while respecting its side constraints. Shape information such as a right angle, an included angle, or a third side can remove that freedom.
ORIGINAL WORKED EXAMPLE
Two sides of a triangle are 6 cm and 11 cm. How many integer values, in centimeters, are possible for the third side?
Answer 11 possible integer values.
04 / UNDERSTAND & APPLY
An isosceles triangle has at least two equal sides, and the angles opposite those sides are equal. If the equal sides meet at a vertex angle θ, each base angle is (180° − θ)/2. The reverse is useful too: two equal angles force their opposite sides to be equal.
An equilateral triangle has three equal sides and three 60° angles. Dropping a perpendicular from one vertex to the opposite side bisects that side and the vertex angle, creating two 30°–60°–90° triangles. In a general triangle, an altitude need not bisect either the base or the vertex angle; symmetry is the reason it does here.
ORIGINAL WORKED EXAMPLE
An isosceles triangle has equal sides of 13 cm and a base of 10 cm. Find its area.
Answer 60 cm².
05 / UNDERSTAND & APPLY
In a right triangle, the legs meet at the 90° angle and the hypotenuse lies opposite it. If a and b are legs and c is the hypotenuse, a² + b² = c². The theorem is conditional on a right angle. A slanted side in a drawing is not automatically a hypotenuse.
When finding a leg, subtract its partner’s square from the hypotenuse square; when finding the hypotenuse, add the two leg squares. Lengths use the positive square root. The triples 3–4–5, 5–12–13, and 8–15–17 and their positive multiples can save arithmetic. They are patterns to verify, not a reason to round nearby lengths to a familiar triple.
The converse also helps: if the longest side c satisfies c² = a² + b², the triangle is right. For example, sides 9, 12, and 15 meet this condition. Without that equality or another right-angle condition, do not apply the equation.
ORIGINAL WORKED EXAMPLE
A 10 m ladder rests against a vertical wall on level ground, with its foot 6 m from the wall. The wall and ground are perpendicular. How high does the ladder reach?
Answer 8 m.
06 / UNDERSTAND & APPLY
A 45°–45°–90° triangle has equal legs s and hypotenuse s√2. This follows from s² + s² = c². Given the hypotenuse c, each leg is c/√2 = c√2/2. Multiplying c by √2 instead would produce a leg longer than the hypotenuse.
A 30°–60°–90° triangle has sides s, s√3, and 2s opposite 30°, 60°, and 90°, respectively. The shortest leg is half the hypotenuse, and the long leg is √3 times the short leg. Label the angles first; rotated figures preserve the ratios.
These special-triangle facts and the Pythagorean theorem solve the required geometry without trigonometry. Keep radicals exact unless approximation is explicitly requested.
Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.
ORIGINAL WORKED EXAMPLE
In a 30°–60°–90° triangle, the side opposite 60° is 9√3 cm. Find the perimeter.
Answer 27 + 9√3 cm.
ORIGINAL WORKED EXAMPLE
A square has diagonal 14 cm. Find its area.
Answer 98 cm².
07 / UNDERSTAND & APPLY
Triangle area is one half of a selected base times its corresponding perpendicular height: A = bh/2. Any side can be the base, but its paired altitude must meet the line containing that side at 90°. In an obtuse triangle the altitude may fall outside the triangle. The formula is unchanged.
Do not substitute a sloping side for the height. If a base is fixed and the opposite vertex moves along a line parallel to that base, the perpendicular height and area stay fixed while the other sides can change. Equal bases and equal perpendicular heights imply equal areas even when the triangles look different.
An equilateral triangle with side s has height s√3/2 and area s²√3/4. Derive these from the bisected base and a special triangle whenever the formula is forgotten. All lengths must use consistent units before multiplication.
A = bh/2; P = a + b + c
Area is measured in square units; perimeter is a sum of lengths and uses linear units.
08 / UNDERSTAND & APPLY
A diagonal, altitude or coordinate difference can expose a right triangle that was not named in the prompt. For points (x₁, y₁) and (x₂, y₂), horizontal and vertical differences form perpendicular legs, so their distance is √[(x₂ − x₁)² + (y₂ − y₁)²]. The calculation is just the Pythagorean theorem in coordinates.
For a composite area, partition into nonoverlapping triangles or subtract a missing triangle from a rectangle. Add areas, but count perimeter only along the outside boundary. Internal diagonals used to split the region are not part of its perimeter.
ORIGINAL WORKED EXAMPLE
A triangle has vertices (1, 2), (9, 2), and (4, 7), with coordinates in centimeters. Find its area.
Answer 20 cm².
09 / UNDERSTAND & APPLY
QC does not always require finding every side. In a right triangle, the hypotenuse is greater than each individual leg but less than the sum of the legs. By contrast, perimeter alone leaves many areas possible. A side-length interval is useful evidence, whereas a drawing that “looks almost equilateral” is not.
To show an area comparison is undetermined, hold the given base fixed and vary the positive height. If the problem instead gives all three sides, that shape variation is no longer allowed. Counterexamples must retain every stated condition.
ORIGINAL WORKED EXAMPLE
QC. A triangle has a base of length 8. Quantity A: its area. Quantity B: 20.
Answer D: the relationship cannot be determined.
10 / UNDERSTAND & APPLY
Use the QC key stated at the start. Numeric entries ask for numbers only; the question supplies the unit.
ORIGINAL PRACTICE · 1
68
The two equal base angles total 180 − 44 = 136°, so each is 68°.
ORIGINAL PRACTICE · 2
B, C and E
The strict interval is 5 < c < 13. Thus 6, 10, and 12.5 work; the two endpoints flatten the triangle.
ORIGINAL PRACTICE · 3
15
The missing leg is √(17² − 8²) = √225 = 15. Subtract because the hypotenuse is already known.
ORIGINAL PRACTICE · 4
C. 6√2
Equal legs of 6 give c² = 36 + 36 = 72, so c = 6√2. The hypotenuse must exceed 6.
ORIGINAL PRACTICE · 5
C. 11√3
The short leg is half the hypotenuse, 11. The long leg opposite 60° is 11√3.
11 / UNDERSTAND & APPLY
Identify whether the target is linear or square units before choosing a formula.
ORIGINAL PRACTICE · 1
0.3
Convert 50 cm to 0.5 m first. Then A = (1/2)(1.2)(0.5) = 0.3 m².
ORIGINAL PRACTICE · 2
C
Each area is 7×9/2 = 31.5, even if their other sides differ.
ORIGINAL PRACTICE · 3
D
Bases 4 and 8 both satisfy 0 < base < 10 and produce valid isosceles triangles. One is below 6 and one is above it.
ORIGINAL PRACTICE · 4
36
The hypotenuse is √(81 + 144) = 15. Perimeter is 9 + 12 + 15 = 36, not area 54.
ORIGINAL PRACTICE · 5
A and C
The largest angle faces the longest side, and the right angle makes the Pythagorean theorem applicable. Unequal angles rule out equal sides; angle measures alone do not set an area.
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.