SAT · Geometry and Trigonometry · TOPIC LESSON
Choose the right measure, build a geometric model, and track what changes when a figure is resized.
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BEFORE YOU START
Original educational lesson. Work through the examples before revealing the practice answers. This lesson supports SAT Math and Reading and Writing skill study; it does not reproduce official questions or guarantee a score.
LEARN THROUGH ORIGINAL PRACTICE
The teaching passages, worked examples and exercises in this lesson are original. They are not official College Board questions or past-exam reproductions. Blueprint and observed-practice analytics describe separate factual metadata; those source questions are not displayed here.
01 / UNDERSTAND & APPLY
A length measures a one-dimensional distance. Perimeter measures the complete boundary of a plane region; circumference is the perimeter of a circle. Area measures the interior of a plane region. Surface area measures the combined areas of the exposed faces of a solid, whereas volume measures the three-dimensional space it occupies.
A story about fencing suggests boundary length; covering a floor suggests area; wrapping a package suggests surface area; filling a tank suggests volume. The wording controls the model. A container described as open has no top surface to include, although its capacity still uses the space inside.
Label all dimensions before substituting. Use one length unit throughout a formula. Converting meters to centimeters multiplies a length by 100, an area by 100², and a volume by 100³. A numerical result without its unit can conceal a wrong calculation.
ORIGINAL WORKED EXAMPLE
A rectangular mat is 2.4 m long and 75 cm wide. Find its area and perimeter.
Answer Area 1.8 m²; perimeter 6.3 m.
ORIGINAL PRACTICE · 1
320 cm²
Since 1 m = 100 cm, 1 m² = 10,000 cm². Multiply 0.032 by 10,000. Multiplying only by 100 would convert a length, not an area.
02 / UNDERSTAND & APPLY
A rectangle has area length × width. A parallelogram has area base × perpendicular height: a triangular end can be shifted to form a rectangle without changing area. A triangle with the same base and height has half the parallelogram’s area.
The height must be perpendicular to the chosen base or its extension. A sloping side is generally not a height. An obtuse triangle can have a height outside the triangle. You may choose any side as the base, but must use that side’s matching perpendicular height.
For a trapezoid, use the average of the two parallel base lengths times their perpendicular separation. Do not average the nonparallel sides. Equal perimeter does not imply equal area: a 1-by-9 rectangle and a 5-by-5 square both have perimeter 20, but areas 9 and 25.
Arectangle = lw; Aparallelogram = bh; Atriangle = bh/2
In each height formula, h is perpendicular to b.
Atrapezoid = (b₁ + b₂)h/2
b₁ and b₂ are the parallel bases.
Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.
ORIGINAL WORKED EXAMPLE
A triangle has base 14 cm. The altitude meets that base 5 cm from the left endpoint. The left sloping side is 13 cm. Find the triangle’s area.
Answer 84 cm²
ORIGINAL PRACTICE · 1
66 square units
The average base is (7 + 15)/2 = 11. Multiplying by the height gives 11 × 6 = 66. Neither slanted side is needed.
03 / UNDERSTAND & APPLY
The radius runs from the center to the circle; the diameter is twice the radius. Circumference is linear in the radius, while area is quadratic. Always convert a given diameter to a radius before using πr².
An annulus is the region between concentric circles. Subtract the inner disk’s area from the outer disk’s area. Its boundary includes two circles, so the total boundary length adds their circumferences. Subtracting circumferences would answer a different question.
Keep π symbolic until the end when possible. If an answer asks for a multiple of π, the coefficient may be the requested number rather than the decimal value of the whole measurement.
C = 2πr = πd; A = πr²
d = 2r; a circular region has area, while its boundary has circumference.
Aannulus = π(R² − r²)
R is the outer radius, r the inner radius, and R > r.
ORIGINAL WORKED EXAMPLE
A flat circular border has outer radius 5 cm and inner radius 3 cm. Find its area and total boundary length.
Answer Area 16π cm²; boundary length 16π cm.
ORIGINAL PRACTICE · 1
81π square units
The radius is 18/2 = 9, so area is π(9²) = 81π. Using 18 as the radius would multiply the correct area by 4.
04 / UNDERSTAND & APPLY
A composite region can be split into pieces with nonoverlapping interiors or found by subtracting a missing region from a larger easy shape. Choose the decomposition that uses the fewest unknown lengths. Draw light auxiliary lines and derive any missing lengths from the given totals.
For area, internal dividing lines do not matter. For perimeter, trace the exposed boundary once. An internal shared edge must not be counted, but the sides of a cutout exposed to the outside must be counted. The perimeter of a shape with a hole includes the hole’s boundary if total boundary length is requested.
The diagram below shows a corner notch. Removing it reduces area but replaces two outside edge segments with two equally long inner edge segments. Consequently the perimeter happens to remain unchanged. A notch cut into the middle of one side can increase perimeter, so do not turn this observation into a universal rule.
Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.
ORIGINAL WORKED EXAMPLE
A 14 m by 10 m rectangular courtyard loses a 6 m by 4 m rectangle from its upper-right corner. Find the remaining area and perimeter.
Answer Area 116 m²; perimeter 48 m.
ORIGINAL PRACTICE · 1
Area 74; perimeter 40
Area is 80 − 6 = 74. Original perimeter is 36. The removed 3-unit segment is replaced by a 3-unit segment and two new 2-unit sides, so perimeter increases by 4 to 40.
05 / UNDERSTAND & APPLY
A net unfolds a solid’s surfaces into the plane. For a rectangular prism, there are two faces of each of three types. Add only the faces described as exposed or covered. An open-top box has one fewer face; a solid resting on a table may still have a bottom, even if a painting problem excludes it.
A cylinder’s curved side unfolds to a rectangle whose length is its circumference and whose width is its height. Thus lateral area is 2πrh. A closed cylinder also has two circular bases. Surface area is not the same as cross-sectional area.
When solids are attached along a face, that shared interface is internal. Add the separate surface areas and remove both copies of the shared face. For curved solids, use a supplied or justified surface-area formula; do not substitute a volume formula simply because the shape is familiar.
SAclosed rectangular prism = 2(lw + lh + wh)
Three pairs of rectangular faces.
SAclosed cylinder = 2πr² + 2πrh
Two circular ends plus the curved side.
SAsphere = 4πr²
All of the sphere’s curved exterior; compare with the disk area πr².
Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.
ORIGINAL WORKED EXAMPLE
A rectangular box has base dimensions 3 cm by 5 cm and height 8 cm. It has no lid. Find its surface area, including the bottom.
Answer 143 cm²
ORIGINAL PRACTICE · 1
60π square units
The two bases contribute 2π(3²) = 18π. The curved side contributes 2π(3)(7) = 42π. Total is 60π.
ORIGINAL PRACTICE · 2
40 square units
Each cube has 6 × 2² = 24 square units of surface area. The two shared face copies total 8 and become internal, so 48 − 8 = 40. The resulting 4-by-2-by-2 prism confirms the result.
06 / UNDERSTAND & APPLY
A prism has congruent parallel bases and a constant cross section along its height; a cylinder is the circular-base counterpart. The base can be a triangle, rectangle, or another region. First find the base area B, then multiply by the perpendicular distance between the bases.
Choose the base strategically. In a rectangular prism any face can be the base, giving the same product lwh. In a triangular prism, the triangle’s altitude helps find B, while the prism’s length is the distance through which that triangle extends. These are two different measurements.
Capacity questions may use inside dimensions rather than outside dimensions. If wall thickness is mentioned, determine the interior lengths before multiplying. A tank partly filled to depth h uses that fluid depth as the height when the cross section is constant.
Vprism or cylinder = Bh
B is base area, and h is perpendicular height.
Vrectangular prism = lwh; Vcylinder = πr²h
Volume has cubic units.
ORIGINAL WORKED EXAMPLE
A cylindrical tank has inside diameter 8 m and water depth 9 m. What volume of water does it hold?
Answer 144π m³
ORIGINAL PRACTICE · 1
198 cm³
The triangular base area is (1/2)(9)(4) = 18 cm². The stated 11 cm distance is perpendicular to the bases, so V = 18 × 11 = 198 cm³.
07 / UNDERSTAND & APPLY
A cone or pyramid tapers from a base to a point. Its volume is one third of the volume of a cylinder or prism with the same base area and perpendicular height. The slant height follows the exterior and is not the height used in the volume formula.
In the net-and-cone diagram above, the cone’s interior dashed segment is its perpendicular height; the sloping exterior segment is longer. Use the interior height in volume calculations.
A sphere’s volume depends on the cube of its radius. A hemisphere has half that volume. For a composite solid, add disjoint volumes; for a cavity, subtract the removed volume. Do not add surface areas to volumes, and do not assume attached solids have equal heights or radii unless stated.
Formulas can be read backward. Isolate the required power of the unknown first, then take the positive root because a length is positive. Keeping constants such as π uncanceled until the final numerical step can obscure simple cancellations.
Vcone = πr²h/3; Vpyramid = Bh/3
h is the perpendicular height to the base plane.
Vsphere = 4πr³/3
The diameter is twice r, not r itself.
ORIGINAL WORKED EXAMPLE
A cone has volume 96π cm³ and perpendicular height 8 cm. Find its radius.
Answer 6 cm
ORIGINAL PRACTICE · 1
18π cm³
The whole sphere has volume (4/3)π(3³) = 36π. Half is 18π. The curved surface area would have square units and answer a different question.
ORIGINAL PRACTICE · 2
120 cubic units
The base area is 6² = 36. Volume is (1/3)(36)(10) = 120. Using the side length 6 in place of the base area would be dimensionally wrong.
08 / UNDERSTAND & APPLY
If every length of a figure is multiplied by the same positive scale factor k, all perimeters multiply by k, all areas and surface areas by k², and all volumes by k³. This follows directly from the number of length factors in the formulas.
The word similar matters. If only a cylinder’s radius doubles while its height stays fixed, its volume multiplies by 4, not 8. A uniform scale factor would have doubled the height too. Compare each changed dimension before deciding which shortcut applies.
You can reverse the reasoning: an area ratio determines a length ratio by a square root; a volume ratio determines it by a cube root. Similar figures preserve angles, not numerical lengths. A 20% increase in every length is k = 1.20, not k = 0.20.
Length ratio k → area ratio k² → volume ratio k³
These relations require uniform scaling of corresponding lengths.
ORIGINAL WORKED EXAMPLE
Every dimension of a solid model increases by 20%. By what percent do its surface area and volume increase?
Answer Surface area increases 44%; volume increases 72.8%.
ORIGINAL PRACTICE · 1
4:9
The corresponding length ratio is the cube root of 8:27, or 2:3. Squaring gives the surface-area ratio 4:9.
ORIGINAL PRACTICE · 2
9:2
Volume contains radius squared and height once. Its factor is 3² × (1/2) = 9/2. The solids are not similar because the dimensions did not scale uniformly.
09 / UNDERSTAND & APPLY
When a dimension is unknown, name it, express the related lengths in the same variable, and substitute into the relevant formula. Solve the equation only after the geometric relationship is established. Dimensional analysis can expose a missing factor even when the algebra looks tidy.
Check each algebraic solution against the figure. Lengths must be positive, a cutout cannot be wider than its enclosing region, and a water depth cannot exceed a stated tank height. A negative solution can be valid algebraically but invalid geometrically.
A useful calculator workflow is to enter the complete expression with parentheses, keep intermediate values unrounded, and verify the final dimension by substitution. A graph of the equation can help identify possible roots, but the geometric constraints decide which roots to keep.
ORIGINAL WORKED EXAMPLE
A rectangular box has length x cm, width (x − 2) cm, and height 3 cm. Its volume is 144 cm³. Find x.
Answer x = 8 cm
ORIGINAL PRACTICE · 1
66 square units
Use the perimeter first: 2w + 2(w + 5) = 34 gives 4w = 24 and w = 6. Length is 11, so area is 6 × 11 = 66. The given 34 is not the area.
10 / UNDERSTAND & APPLY
Reread the target. A question can provide a volume and ask for radius, provide a surface area and ask for a ratio, or ask only for the coefficient of π. Write a final answer that names the requested quantity and uses its unit.
Estimate order of magnitude. A triangle inside a rectangle with the same base and height has half its area. A cone has one third of a matching cylinder’s volume. Removing a cavity must reduce volume. A solid’s linear dimensions getting larger cannot decrease volume when all dimensions scale together.
When rounding is requested, round once at the end. When an exact form is accepted, fractions, roots, and π avoid approximation error. Geometric sketches help organize information, but lengths and right angles must come from statements, labels, or valid deductions.
ORIGINAL PRACTICE · 1
Height 8; the radius was probably used without squaring
Use 200π = π(5²)h = 25πh, giving h = 8. Dividing by 5π instead would give 40 and forget that the base area contains r².
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