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PSAT/NMSQT Math / LEVEL 4 · DIFFICULTY 3/5

Geometry & Trigonometry: Applications

Connect measurements, similarity and right-triangle ratios.

4 stages · 32 practice problems · two 8-question assessment forms

Choose a node to explore its lesson.

MINI QUEST PSAT 4.1Surface Area, Volume & Missing DimensionsRead the lesson
MINI QUEST PSAT 4.2Similar Solids & Scale FactorsRead the lesson
MINI QUEST PSAT 4.3Triangles, Similarity & Sufficient InformationRead the lesson
MINI QUEST PSAT 4.4Right-Triangle Sine, Cosine & TangentRead the lesson
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STAGE PSAT 4.1 · 8 PRACTICE PROBLEMS · 4 NEW-CONTEXT APPLICATIONS

Surface Area, Volume & Missing Dimensions

Goal: Choose area versus volume, include all exposed faces, and solve for a missing dimension of a solid.

Before you begin: Read the problem carefully and use the linked foundation lessons when a term or operation is unfamiliar.

TURN IT. LOOK INSIDE. EXPLAIN IT.

Rectangular prism · 3D explorer

Drag to rotate

A worked example you can change. Model values are separate from practice answers. Drag with a mouse or one finger. Use the turn buttons or arrow keys as an alternative.

Explore the solid in motion

$\displaystyle a = 4\, b = 3\, h = 6$

Dashed line: perpendicular height. Your marks use the colors you choose below.

Base length a: 4

Base width b: 3

Perpendicular height h: 6

Show a horizontal cross-sectionShow the two right triangles for the space diagonal

What should you notice?

Every horizontal layer has the same area. Doubling only the height doubles the number of layers and the volume.

Rotating the view changes what you see; it does not change the lengths or volume. Read the labelled dimensions, rather than measuring the screen.

Highlight parts and add your information

Choose a part here, or select it in the diagram. Mark what you know, what you need, or a reason. These are your scratch notes; the website does not grade or save them.

One small move: Select a part above. Find its given measurement in the question, then label that part below.

Part of the diagramWhat does your mark mean?Length, angle or reasoning

Use the problem’s stated values. A drawing can look shorter after rotation even when the actual length stays the same. Changing the model or reloading clears these notes.

From the model to the calculation

Find the base area $\displaystyle B = 4 \times 3 \approx 12$ square units.
V = Bh = $\displaystyle \left(4 \times 3\right) \times 6 \approx 72$ cubic units. Stacking equal-area layers explains the multiplication by height.
Check units: base area has square units; multiplying by a length gives cubic units. Rounded values are shown to three decimal places; keep $\displaystyle \pi$ or radicals for exact answers when requested.

Make sense of the idea

Volume measures space inside, in cubic units; surface area measures the outside, in square units. A closed rectangular box has area 2(lw + lh + wh), because each face has an opposite partner.

Choose a method

A cylinder has volume $\displaystyle \pi\, r^{2}\, h$ and closed surface area $\displaystyle 2\, \pi\, r^{2} + 2\, \pi$ rh. The two circles are the caps. A cone with the same base and height has one third of the cylinder volume.

Check and interpret

A sphere has volume $\displaystyle \left(\frac{4}{3}\right)\, \pi\, r^{3}$ . In a reverse problem, divide out known factors before finding a root. Read whether the task gives a radius or diameter and whether a container is open or closed.

WORKED EXAMPLE 1

A rectangular prism has dimensions 5, 4, 2 cm. Find its total surface area in cm².

lw = 20, lh = 10, wh = 8.
Total area = $\displaystyle 2\, \left(20 + 10 + 8\right) = 76$ cm².

WORKED EXAMPLE 2

A spherical ball has radius 5. Its volume is $\displaystyle c\, \pi$ . Find c as a fraction if needed.

$\displaystyle V = \left(\frac{4}{3}\right)\, \pi \times 5^{3}$ .
The coefficient is $\displaystyle c = \frac{500}{3}$ .

$\displaystyle \frac{500}{3}$

Common pitfalls

Possible mix-up: Surface area and volume use the same units.

Area uses squares; volume uses cubes.

Possible mix-up: A diameter can be placed directly into a radius formula.

First divide the diameter by 2.

Explain it to yourself

In surface area, volume & missing dimensions, what information determines your method, and how can you check whether your result fits the original conditions?

Preview the eight practice prompts

Problem 1
A rectangular prism has dimensions 3, 2, 2 cm. Find its total surface area in cm².
Practise this exact problem →
Problem 2
A cylinder has radius 2 cm and volume $\displaystyle 16\, \pi$ cm³. Find its height in cm.
Practise this exact problem →
Problem 3
A cone has radius 2 and height 9. Its volume is $\displaystyle c\, \pi$ . Find c.
Practise this exact problem →
Problem 4
A sphere has radius 2. Its volume is $\displaystyle c\, \pi$ . Find c as a fraction if needed.
Practise this exact problem →
Problem 5New context
A closed gift box has dimensions 4, 3, 2 cm. Find its total surface area in cm².
Practise this exact problem →
Problem 6New context
A cylindrical tank has radius 3 cm and volume $\displaystyle 45\, \pi$ cm³. Find its height in cm.
Practise this exact problem →
Problem 7New context
A cone-shaped hopper has radius 3 and height 12. Its volume is $\displaystyle c\, \pi$ . Find c.
Practise this exact problem →
Problem 8New context
A spherical ball has radius 3. Its volume is $\displaystyle c\, \pi$ . Find c as a fraction if needed.
Practise this exact problem →

Open stage PSAT 4.1 in the student workspace →

STAGE PSAT 4.2 · 8 PRACTICE PROBLEMS · 4 NEW-CONTEXT APPLICATIONS

Similar Solids & Scale Factors

Goal: Use k, $\displaystyle k^{2}$ and $\displaystyle k^{3}$ correctly and recover length factors from area or volume information.

Before you begin: Read the problem carefully and use the linked foundation lessons when a term or operation is unfamiliar.

TURN IT. LOOK INSIDE. EXPLAIN IT.

Rectangular prism · 3D explorer

Drag to rotate

A worked example you can change. Model values are separate from practice answers. Drag with a mouse or one finger. Use the turn buttons or arrow keys as an alternative.

Explore the solid in motion

$\displaystyle a = 4\, b = 3\, h = 6$

Dashed line: perpendicular height. Your marks use the colors you choose below.

Base length a: 4

Base width b: 3

Perpendicular height h: 6

Show a horizontal cross-sectionShow the two right triangles for the space diagonal

What should you notice?

Every horizontal layer has the same area. Doubling only the height doubles the number of layers and the volume.

Rotating the view changes what you see; it does not change the lengths or volume. Read the labelled dimensions, rather than measuring the screen.

Highlight parts and add your information

Choose a part here, or select it in the diagram. Mark what you know, what you need, or a reason. These are your scratch notes; the website does not grade or save them.

One small move: Select a part above. Find its given measurement in the question, then label that part below.

Part of the diagramWhat does your mark mean?Length, angle or reasoning

Use the problem’s stated values. A drawing can look shorter after rotation even when the actual length stays the same. Changing the model or reloading clears these notes.

From the model to the calculation

Find the base area $\displaystyle B = 4 \times 3 \approx 12$ square units.
V = Bh = $\displaystyle \left(4 \times 3\right) \times 6 \approx 72$ cubic units. Stacking equal-area layers explains the multiplication by height.
Check units: base area has square units; multiplying by a length gives cubic units. Rounded values are shown to three decimal places; keep $\displaystyle \pi$ or radicals for exact answers when requested.

Make sense of the idea

Similar figures have corresponding lengths in the same ratio k. Each area includes two length factors, so the area ratio is $\displaystyle k^{2}$ .

Choose a method

Volume includes three length factors, so similar solids have volume ratio $\displaystyle k^{3}$ . This rule needs similarity: changing only the height of a cylinder is not a uniform scale change.

Check and interpret

Reverse the process when given area or volume ratios. Take a square root of an area ratio, or a cube root of a volume ratio, to recover the length factor.

WORKED EXAMPLE 1

A figure is enlarged so every length is multiplied by 4. Its original area is 9. Find its new area.

Area factor = $\displaystyle 4^{2} = 16$ .
New area = $\displaystyle 9 \times 16 = 144$ .

144

WORKED EXAMPLE 2

Two similar storage tanks have volume ratio large:small = $\displaystyle 125 : 1$ . What is their surface-area ratio large:small? Enter the number before :1.

The length factor is $\displaystyle \sqrt[3]{125} = 5$ .
The surface-area ratio is $\displaystyle 25 : 1$ .

Common pitfalls

Possible mix-up: Tripling lengths triples area.

It multiplies area by 9 and volume by 27.

Possible mix-up: Any two solids use the $\displaystyle k^{3}$ rule.

They must be similar, with every corresponding length scaled equally.

Explain it to yourself

In similar solids & scale factors, what information determines your method, and how can you check whether your result fits the original conditions?

Preview the eight practice prompts

Problem 1
A figure is enlarged so every length is multiplied by 2. Its original area is 7. Find its new area.
Practise this exact problem →
Problem 2
A solid is scaled by 2 in EVERY length. Its original volume is 3. Find the new volume.
Practise this exact problem →
Problem 3
Two similar figures have area ratio large:small = $\displaystyle 4 : 1$ . A side of the smaller is 3. Find the corresponding side of the larger.
Practise this exact problem →
Problem 4
Two similar solids have volume ratio large:small = $\displaystyle 8 : 1$ . What is their surface-area ratio large:small? Enter the number before :1.
Practise this exact problem →
Problem 5New context
A printed design is enlarged so every length is multiplied by 3. Its original area is 8. Find its new area.
Practise this exact problem →
Problem 6New context
A similar model container is scaled by 3 in EVERY length. Its original volume is 4. Find the new volume.
Practise this exact problem →
Problem 7New context
Two similar display panels have area ratio large:small = $\displaystyle 9 : 1$ . A side of the smaller is 4. Find the corresponding side of the larger.
Practise this exact problem →
Problem 8New context
Two similar storage tanks have volume ratio large:small = $\displaystyle 27 : 1$ . What is their surface-area ratio large:small? Enter the number before :1.
Practise this exact problem →

Open stage PSAT 4.2 in the student workspace →

STAGE PSAT 4.3 · 8 PRACTICE PROBLEMS · 4 NEW-CONTEXT APPLICATIONS

Triangles, Similarity & Sufficient Information

Goal: Combine triangle and parallel-line angle facts, use corresponding sides, and decide whether information proves similarity or congruence.

Before you begin: Read the problem carefully and use the linked foundation lessons when a term or operation is unfamiliar.

NOTICE → MARK → REASON

Same shape, different size

Mark matching sides before comparing areas. This is a separate learning diagram; it does not supply the values for your practice question.

Highlight parts and add your information

Choose a part here, or select it in the diagram. Mark what you know, what you need, or a reason. These are your scratch notes; the website does not grade or save them.

One small move: Select a part above. Find its given measurement in the question, then label that part below.

Part of the diagramWhat does your mark mean?Length, angle or reasoning

Use the problem’s stated values. A drawing can look shorter after rotation even when the actual length stays the same. Changing the model or reloading clears these notes.

Make sense of the idea

A triangle has angle sum $\displaystyle 180^{\circ}$ . An exterior angle equals the sum of the two remote interior angles. With parallel lines, corresponding and alternate interior angles are equal.

Choose a method

AA proves triangles similar, but does not fix their size. Match corresponding vertices before setting up side proportions.

Check and interpret

SSS, SAS and ASA/AAS can establish congruence. Two angles alone establish shape, not size. A diagram that looks equal is not evidence unless the relationship is stated or proved.

WORKED EXAMPLE 1

A triangle has two interior angles $\displaystyle 32^{\circ}$ and $\displaystyle 52^{\circ}$ . Find the exterior angle at the third vertex.

The third interior angle is $\displaystyle 96^{\circ}$ .
Its exterior angle is $\displaystyle 180^{\circ} - 96^{\circ} = 84^{\circ}$ .

WORKED EXAMPLE 2

Parallel rails are cut by a transversal. One acute angle is $\displaystyle 43^{\circ}$ . Find the obtuse angle supplementary to a corresponding acute angle.

The corresponding acute angle is $\displaystyle 43^{\circ}$ .
Its supplementary obtuse angle is $\displaystyle 180^{\circ} - 43^{\circ} = 137^{\circ}$ .

137

Common pitfalls

Possible mix-up: AA proves congruence.

It proves similarity; triangles can have different sizes.

Possible mix-up: Sides in matching positions on the page correspond.

Use corresponding angles or named vertices, not drawing orientation.

Explain it to yourself

In triangles, similarity & sufficient information, what information determines your method, and how can you check whether your result fits the original conditions?

Preview the eight practice prompts

Problem 1
A triangle has two interior angles $\displaystyle 30^{\circ}$ and $\displaystyle 50^{\circ}$ . Find the exterior angle at the third vertex.
Practise this exact problem →
Problem 2
Triangles ABC and DEF are similar with A↔D, B↔E, C↔F. $\displaystyle AB = 3$ , $\displaystyle DE = 6$ , $\displaystyle BC = 5$ . Find EF.
Practise this exact problem →
Problem 3
Two triangles both have angles $\displaystyle 35^{\circ}$ and $\displaystyle 45^{\circ}$ . Must they be congruent? Enter 1 for yes or 0 for no.
Practise this exact problem →
Problem 4
Two parallel lines are cut by a transversal. One acute angle is $\displaystyle 40^{\circ}$ . Find the obtuse angle supplementary to a corresponding acute angle.
Practise this exact problem →
Problem 5New context
A triangular brace has two interior angles $\displaystyle 31^{\circ}$ and $\displaystyle 51^{\circ}$ . Find the exterior angle at the third vertex.
Practise this exact problem →
Problem 6New context
Two triangular signs are similar with A↔D, B↔E, C↔F. $\displaystyle AB = 4$ , $\displaystyle DE = 12$ , $\displaystyle BC = 6$ . Find EF.
Practise this exact problem →
Problem 7New context
Two triangular patterns both have angles $\displaystyle 36^{\circ}$ and $\displaystyle 46^{\circ}$ . Must they be congruent? Enter 1 for yes or 0 for no.
Practise this exact problem →
Problem 8New context
Parallel rails are cut by a transversal. One acute angle is $\displaystyle 41^{\circ}$ . Find the obtuse angle supplementary to a corresponding acute angle.
Practise this exact problem →

Open stage PSAT 4.3 in the student workspace →

STAGE PSAT 4.4 · 8 PRACTICE PROBLEMS · 4 NEW-CONTEXT APPLICATIONS

Right-Triangle Sine, Cosine & Tangent

Goal: Choose opposite, adjacent and hypotenuse relative to a marked angle; find missing lengths and use complementary angles.

Before you begin: Read the problem carefully and use the linked foundation lessons when a term or operation is unfamiliar.

Shared PSAT / SAT foundations. The complementary-angle sine/cosine exercises are labeled here as SAT extension practice.

The angle decides which side is opposite

Reference angle

For $\displaystyle \theta$ : opposite = 3, adjacent = 4, hypotenuse = 5. Sine = $\displaystyle \frac{3}{5}$ , cosine = $\displaystyle \frac{4}{5}$ , tangent = $\displaystyle \frac{3}{4}$ .

$\displaystyle \theta + \phi = 90^{\circ}$ . Switching angles reveals why $\displaystyle \operatorname{sin}\, \theta = \operatorname{cos}\, \phi$ . The hypotenuse always stays opposite the right angle.

Make sense of the idea

In a right triangle the hypotenuse is opposite the $\displaystyle 90^{\circ}$ angle. Relative to an acute angle $\displaystyle \theta$ , label the other two sides opposite and adjacent. Moving $\displaystyle \theta$ to the other acute corner swaps these two roles.

Choose a method

SOHCAHTOA: $\displaystyle \operatorname{sin}\, \theta$ = opposite/hypotenuse, $\displaystyle \operatorname{cos}\, \theta$ = adjacent/hypotenuse, $\displaystyle \operatorname{tan}\, \theta$ = opposite/adjacent. Choose the ratio containing the known and wanted sides.

Check and interpret

The two acute angles sum to $\displaystyle 90^{\circ}$ , so $\displaystyle \operatorname{sin}\, \theta = \operatorname{cos}\, \left(90^{\circ} - \theta\right)$ . Use inverse sine, cosine or tangent to recover an angle, with the calculator in degree mode when the answer is requested in degrees.

WORKED EXAMPLE 1

A right triangle has, relative to $\displaystyle \theta$ , opposite side 12, adjacent side 16, hypotenuse 20. Find $\displaystyle \operatorname{sin}\, \theta$ as a fraction.

$\displaystyle \operatorname{sin}\, \theta = \frac{12}{20}$ .
Cancel the common scale factor to get $\displaystyle \frac{3}{5}$ .

$\displaystyle \frac{3}{5}$

WORKED EXAMPLE 2

A triangular frame has acute angles $\displaystyle \theta$ and $\displaystyle \phi$ . If $\displaystyle \operatorname{sin}\, \theta = \frac{5}{8}$ , find $\displaystyle \operatorname{cos}\, \phi$ .

Since $\displaystyle \phi = 90^{\circ} - \theta$ , $\displaystyle \operatorname{cos}\, \phi = \operatorname{sin}\, \theta$ .
Thus $\displaystyle \operatorname{cos}\, \phi = \frac{5}{8}$ .

$\displaystyle \frac{5}{8}$

Common pitfalls

Possible mix-up: The opposite side never changes.

It changes when the reference angle changes. The hypotenuse stays fixed.

Possible mix-up: $\displaystyle \operatorname{cos}\, \theta$ is opposite divided by hypotenuse.

That is sine. Cosine uses adjacent/hypotenuse.

Explain it to yourself

In right-triangle sine, cosine & tangent, what information determines your method, and how can you check whether your result fits the original conditions?

Preview the eight practice prompts

Problem 1
A right triangle has, relative to $\displaystyle \theta$ , opposite side 6, adjacent side 8, hypotenuse 10. Find $\displaystyle \operatorname{sin}\, \theta$ as a fraction.
Practise this exact problem →
Problem 2
A right-triangle hypotenuse has length 15, and $\displaystyle \operatorname{sin}\, \theta = \frac{3}{5}$ . Find the side opposite $\displaystyle \theta$ .
Practise this exact problem →
Problem 3
A right triangle has opposite leg 2 and adjacent leg 2, relative to acute angle $\displaystyle \theta$ . Find $\displaystyle \theta$ in degrees.
Practise this exact problem →
Problem 4
In a right triangle, acute angles $\displaystyle \theta$ and $\displaystyle \phi$ . If $\displaystyle \operatorname{sin}\, \theta = \frac{2}{5}$ , find $\displaystyle \operatorname{cos}\, \phi$ .
Practise this exact problem →
Problem 5New context
A right-triangle ramp has, relative to $\displaystyle \theta$ , opposite side 9, adjacent side 12, hypotenuse 15. Find $\displaystyle \operatorname{sin}\, \theta$ as a fraction.
Practise this exact problem →
Problem 6New context
A ladder has length 20, and $\displaystyle \operatorname{sin}\, \theta = \frac{3}{5}$ . Find the side opposite $\displaystyle \theta$ .
Practise this exact problem →
Problem 7New context
A ramp rises 3 units over a horizontal run of 3, relative to acute angle $\displaystyle \theta$ . Find $\displaystyle \theta$ in degrees.
Practise this exact problem →
Problem 8New context
A triangular frame has acute angles $\displaystyle \theta$ and $\displaystyle \phi$ . If $\displaystyle \operatorname{sin}\, \theta = \frac{3}{6}$ , find $\displaystyle \operatorname{cos}\, \phi$ .
Practise this exact problem →

Open stage PSAT 4.4 in the student workspace →

Browse another level →

Geometry & Trigonometry: Applications

Surface Area, Volume & Missing Dimensions

Rectangular prism · 3D explorer

From the model to the calculation

Make sense of the idea

Choose a method

Check and interpret

A rectangular prism has dimensions 5, 4, 2 cm. Find its total surface area in cm².

A spherical ball has radius 5. Its volume is c π\displaystyle c\, \picπ. Find c as a fraction if needed.

Common pitfalls

Explain it to yourself

Similar Solids & Scale Factors

Rectangular prism · 3D explorer

From the model to the calculation

Make sense of the idea

Choose a method

Check and interpret

A figure is enlarged so every length is multiplied by 4. Its original area is 9. Find its new area.

Two similar storage tanks have volume ratio large:small = 125:1\displaystyle 125 : 1125:1. What is their surface-area ratio large:small? Enter the number before :1.

Common pitfalls

Explain it to yourself

Triangles, Similarity & Sufficient Information

Same shape, different size

Make sense of the idea

Choose a method

Check and interpret

A triangle has two interior angles 32∘\displaystyle 32^{\circ}32∘ and 52∘\displaystyle 52^{\circ}52∘. Find the exterior angle at the third vertex.

Parallel rails are cut by a transversal. One acute angle is 43∘\displaystyle 43^{\circ}43∘. Find the obtuse angle supplementary to a corresponding acute angle.

Common pitfalls

Explain it to yourself

Right-Triangle Sine, Cosine & Tangent

The angle decides which side is opposite

Make sense of the idea

Choose a method

Check and interpret

A right triangle has, relative to θ\displaystyle \thetaθ, opposite side 12, adjacent side 16, hypotenuse 20. Find sin⁡ θ\displaystyle \operatorname{sin}\, \thetasinθ as a fraction.

A triangular frame has acute angles θ\displaystyle \thetaθ and ϕ\displaystyle \phiϕ. If sin⁡ θ=58\displaystyle \operatorname{sin}\, \theta = \frac{5}{8}sinθ=85​, find cos⁡ ϕ\displaystyle \operatorname{cos}\, \phicosϕ.

Common pitfalls

Explain it to yourself

A spherical ball has radius 5. Its volume is $\displaystyle c\, \pi$ . Find c as a fraction if needed.

Two similar storage tanks have volume ratio large:small = $\displaystyle 125 : 1$ . What is their surface-area ratio large:small? Enter the number before :1.

A triangle has two interior angles $\displaystyle 32^{\circ}$ and $\displaystyle 52^{\circ}$ . Find the exterior angle at the third vertex.

Parallel rails are cut by a transversal. One acute angle is $\displaystyle 43^{\circ}$ . Find the obtuse angle supplementary to a corresponding acute angle.

A right triangle has, relative to $\displaystyle \theta$ , opposite side 12, adjacent side 16, hypotenuse 20. Find $\displaystyle \operatorname{sin}\, \theta$ as a fraction.

A triangular frame has acute angles $\displaystyle \theta$ and $\displaystyle \phi$ . If $\displaystyle \operatorname{sin}\, \theta = \frac{5}{8}$ , find $\displaystyle \operatorname{cos}\, \phi$ .