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Introduction to Geometry / LEVEL 2 · DIFFICULTY 2/5

Similarity and Right Triangles

Connect scale, area and distance.

3 stages · 24 practice problems · two 6-question assessment forms

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MINI QUEST GEO 2.1Similarity and ScaleRead the lesson
MINI QUEST GEO 2.2Area under ScalingRead the lesson
MINI QUEST GEO 2.3Right-Triangle DistanceRead the lesson
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STAGE GEO 2.1 · 8 PRACTICE PROBLEMS · 2 NEW-CONTEXT APPLICATIONS

Similarity and Scale

Useful preparation: Corresponding Parts

Goal: Understand and apply similarity and scale.

Before you begin: Corresponding Parts

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.

Understand the idea

Similar figures have equal corresponding angles and proportional lengths. A scale factor is a directed comparison: the new length divided by the original length.

new length=k·old length

Choose and carry out a method

Match a known side pair, form the new-to-old ratio, then multiply the requested original length by it. Fractions are valid scale factors.

Check the reasoning

Every corresponding side pair should give the same ratio. A factor above one enlarges the figure; a factor below one shrinks it.

WORKED EXAMPLE 1

Two similar triangles have corresponding sides 4 and 7. A second side in the first triangle is 8. Find its corresponding side in the second triangle.

Every pair of corresponding lengths has the same scale factor.
The scale factor is $\displaystyle \frac{7}{4}$ ; multiply 8 by that ratio.
The new length is 14. The ratio must point from the first figure to the second.

WORKED EXAMPLE 2

Two similar triangles have corresponding sides 5 and 8. A second side in the first triangle is 10. Find its corresponding side in the second triangle.

Every pair of corresponding lengths has the same scale factor.
The scale factor is $\displaystyle \frac{8}{5}$ ; multiply 10 by that ratio.
The new length is 16. The ratio must point from the first figure to the second.

Common pitfalls

Possible mix-up: Use the same difference for every side.

Similarity preserves ratios, not differences.

Possible mix-up: A correct numerical answer alone explains the method.

State the governing relationship and check the conditions described above.

Explain it to yourself

Why must all corresponding sides use the same scale factor?

Preview the eight practice prompts

Problem 1
Two similar triangles have corresponding sides 7 and 10. A second side in the first triangle is 14. Find its corresponding side in the second triangle.
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Problem 2
Two similar triangles have corresponding sides 8 and 11. A second side in the first triangle is 16. Find its corresponding side in the second triangle.
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Problem 3
Two similar triangles have corresponding sides 9 and 12. A second side in the first triangle is 18. Find its corresponding side in the second triangle.
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Problem 4
Two similar triangles have corresponding sides 10 and 13. A second side in the first triangle is 20. Find its corresponding side in the second triangle.
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Problem 5
Two similar triangles have corresponding sides 11 and 14. A second side in the first triangle is 22. Find its corresponding side in the second triangle.
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Problem 6
Two similar triangles have corresponding sides 12 and 15. A second side in the first triangle is 24. Find its corresponding side in the second triangle.
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Problem 7New context
A scale model has a 13 cm edge representing a 16 m edge. Another model edge is 26 cm. What real length in meters does it represent?
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Problem 8New context
A scale model has a 14 cm edge representing a 17 m edge. Another model edge is 28 cm. What real length in meters does it represent?
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Open stage GEO 2.1 in the student workspace →

STAGE GEO 2.2 · 8 PRACTICE PROBLEMS · 2 NEW-CONTEXT APPLICATIONS

Area under Scaling

Useful preparation: Similarity and Scale

Goal: Understand and apply area under scaling.

Before you begin: Similarity and Scale

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.

Understand the idea

Scaling every length by k multiplies a rectangle’s two dimensions by k. Decomposing other shapes into small rectangles or triangles explains why their areas also scale by $\displaystyle k^{2}$ .

new area=

\displaystyle k^{2}

·old area

Choose and carry out a method

Find the linear scale factor before squaring it. Multiply the original area by that square and retain square units.

Check the reasoning

Compare a square with side one to a square with side two. Its area becomes four, illustrating the second power.

WORKED EXAMPLE 1

A triangle of area 10 is enlarged by a length scale factor of 3. What is the new area?

Area scales as the square of the length scale factor.
Both a base and its perpendicular height gain a factor of 3. Multiply area by 9.
The new area is $\displaystyle 10 \cdot 9 = 90$ .

WORKED EXAMPLE 2

A triangle of area 11 is enlarged by a length scale factor of 4. What is the new area?

Area scales as the square of the length scale factor.
Both a base and its perpendicular height gain a factor of 4. Multiply area by 16.
The new area is $\displaystyle 11 \cdot 16 = 176$ .

176

Common pitfalls

Possible mix-up: Area scales by k.

Both independent length dimensions scale, giving $\displaystyle k^{2}$ .

Possible mix-up: A correct numerical answer alone explains the method.

State the governing relationship and check the conditions described above.

Explain it to yourself

Predict the volume scale factor for a similar solid and explain it.

Preview the eight practice prompts

Problem 1
A triangle of area 13 is enlarged by a length scale factor of 2. What is the new area?
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Problem 2
A triangle of area 14 is enlarged by a length scale factor of 3. What is the new area?
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Problem 3
A triangle of area 15 is enlarged by a length scale factor of 4. What is the new area?
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Problem 4
A triangle of area 16 is enlarged by a length scale factor of 5. What is the new area?
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Problem 5
A triangle of area 17 is enlarged by a length scale factor of 2. What is the new area?
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Problem 6
A triangle of area 18 is enlarged by a length scale factor of 3. What is the new area?
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Problem 7New context
A square logo uses 19 square centimeters of ink. Every length in a poster version is 4 times as large. How many square centimeters of ink does the new logo cover?
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Problem 8New context
A square logo uses 20 square centimeters of ink. Every length in a poster version is 5 times as large. How many square centimeters of ink does the new logo cover?
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Open stage GEO 2.2 in the student workspace →

STAGE GEO 2.3 · 8 PRACTICE PROBLEMS · 2 NEW-CONTEXT APPLICATIONS

Right-Triangle Distance

Useful preparation: Area under Scaling

Goal: Understand and apply right-triangle distance.

Before you begin: Area under Scaling

NOTICE → MARK → REASON

Right triangle: find the useful sides

Mark the two legs, the right angle and the hypotenuse. 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.

Go further: a diagonal inside a 3D box

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.
First find the base diagonal: $\displaystyle d^{2} = 4^{2} + 3^{2} = 25$ . It is perpendicular to the vertical edge. Then $\displaystyle D^{2} = d^{2} + 6^{2} = 61$ , so $\displaystyle D \approx 7.81$ . The space diagonal must be longer than any edge.
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.

Understand the idea

In a right triangle, the square on the hypotenuse has the combined area of the squares on the two legs. The hypotenuse is opposite the right angle and is the longest side.

\displaystyle a^{2} + b^{2} = c^{2}

Choose and carry out a method

Identify the right angle first. Add the leg squares when finding the hypotenuse; subtract when finding a leg. Take the nonnegative square root.

Check the reasoning

The longest side must be the hypotenuse. Substitution of the three lengths should satisfy the squared equation exactly.

WORKED EXAMPLE 1

A right triangle has legs 9 and 12. Find its hypotenuse.

For a right triangle, the hypotenuse squared is the sum of the leg squares.
$\displaystyle c^{2} = 81 + 144 = 225$ .
The positive length is 15, which exceeds either leg.

WORKED EXAMPLE 2

A right triangle has legs 12 and 16. Find its hypotenuse.

For a right triangle, the hypotenuse squared is the sum of the leg squares.
$\displaystyle c^{2} = 144 + 256 = 400$ .
The positive length is 20, which exceeds either leg.

Common pitfalls

Possible mix-up: Any triangle satisfies this equation.

The equation requires a right angle.

Possible mix-up: A correct numerical answer alone explains the method.

State the governing relationship and check the conditions described above.

Explain it to yourself

How can a diagonal across a rectangular floor become a right-triangle problem?

Preview the eight practice prompts

Problem 1
A right triangle has legs 18 and 24. Find its hypotenuse.
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Problem 2
A right triangle has legs 21 and 28. Find its hypotenuse.
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Problem 3
A right triangle has legs 24 and 32. Find its hypotenuse.
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Problem 4
A right triangle has legs 27 and 36. Find its hypotenuse.
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Problem 5
A right triangle has legs 30 and 40. Find its hypotenuse.
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Problem 6
A right triangle has legs 33 and 44. Find its hypotenuse.
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Problem 7New context
A park is a rectangle 36 m by 48 m. A straight path joins opposite corners. How long is it, in meters?
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Problem 8New context
A park is a rectangle 39 m by 52 m. A straight path joins opposite corners. How long is it, in meters?
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Open stage GEO 2.3 in the student workspace →

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