NCEA Level 1Properties of similar shapesAS 91947

Similar shapes: 7 exam questions solved step by step

What this lesson covers

  • Similar shapes have equal corresponding angles and corresponding sides in the same ratio.
  • If lengths scale by k, areas scale by k squared and volumes by k cubed.
Key idea

7 questions, one method

  • 1. Show they are similar, and why.
  • 2. Match corresponding sides.
  • 3. Find k from a known pair.
  • 4. Lengths × k, areas × k², volumes × k³.
Your turn

Question 1 · Achieved

  • Rectangle A: 4 cm by 10 cm
  • Rectangle B: 6 cm by 15 cm
  • Rectangle C: 7 cm by 13 cm
  • Which is similar to A? Why?
Worked example

Q1: which rectangle is similar?

  • Angles match already.
  • Short ÷ short, long ÷ long.
  1. all angles 90°rectangles
  2. B: 6 ÷ 4 = 1.5, 15 ÷ 10 = 1.5
  3. B is similar, k = 1.5
  4. C: 7 ÷ 4 = 1.75, 13 ÷ 10 = 1.3
  5. C is not similarratios differ
  6. B is similar to A; C is not.
Your turn

Question 2 · Achieved

  • Two triangles are similar.
  • Small: 6 cm, 8 cm, y
  • Large, same order: 9 cm, x, 15 cm
  • Find x and y.
Worked example

Q2: find x and y

  • k from the known pair.
  • Up × k, down ÷ k.
  1. 6 with 9, 8 with x, y with 15
  2. k = 9 ÷ 6 = 1.5small to large
  3. x = 8 × 1.5 = 12 cmsmall to large: × k
  4. y = 15 ÷ 1.5 = 10 cmlarge to small: ÷ k
  5. 12 > 8, 10 < 15
  6. x = 12 cm, y = 10 cm
Your turn

Question 3 · Merit

  • Triangle PQR: angles 48° and 62°
  • Triangle XYZ: angles 62° and 70°
  • Similar? Give your reasoning.
Question 3 · Merit
Worked example

Q3: are they similar?

  • Angles add to 180°.
  • Equal angles: similar.
  1. PQR: 180 − 48 − 62 = 70°angle sum 180°
  2. PQR: 48°, 62°, 70°
  3. XYZ: 180 − 62 − 70 = 48°angle sum 180°
  4. XYZ: 48°, 62°, 70°
  5. all three angles equal
  6. Similar: all angles equal
Your turn

Question 4 · Merit

  • DE is parallel to BC.
  • AD = 6 cm, DB = 4 cm, DE = 9 cm
  • Find BC.
  • Careful: which side of the big one?
Question 4 · Merit
Worked example

Q4: DE parallel to BC

  • Prove similar first.
  • Use the whole side.
  1. angle A sharedcommon angle
  2. angle D = angle Bcorresponding, DE ∥ BC
  3. triangles similar
  4. AB = 6 + 4 = 10 cmthe whole side
  5. k = 10 ÷ 6 = 5/3AB ÷ AD
  6. BC = 9 × 5/3 = 15 cm
  7. 9 × 4 ÷ 6 = 6 cmused the piece DB
Your turn

Question 5 · Merit, in context

  • A student 160 cm tall
  • casts a shadow 2 m long.
  • The flagpole's shadow is 12.5 m.
  • How tall is the flagpole?
Worked example

Q5: the flagpole

  • Reason, then units.
  • Shadows give k.
  1. same sun angle, both 90°two equal angles
  2. triangles similar
  3. 160 cm = 1.6 msame units
  4. k = 12.5 ÷ 2 = 6.25shadow ÷ shadow
  5. h = 1.6 × 6.25 = 10 mheight × k
  6. The flagpole is 10 m tall.
Your turn

Question 6 · Excellence

  • Two similar triangular sails.
  • Small: area 16 m², height 4.8 m
  • Large: area 25 m²
  • How tall is the large sail?
Worked example

Q6: from areas to a height

  • Area ratio is k².
  • Square root for k.
  1. 4.8 × 25 ÷ 16 = 7.5 mareas scale by k²
  2. k² = 25 ÷ 16area ratio
  3. k = √25 ÷ √16 = 5/4 = 1.25square root
  4. h = 4.8 × 1.25 = 6 mlengths scale by k
  5. 16 × 1.25² = 16 × 1.5625 = 25
  6. The large sail is 6 m tall.
Your turn

Question 7 · Excellence

  • Two similar paint tins.
  • Surface areas: 225 cm² and 576 cm²
  • The small tin holds 250 mL.
  • How much does the large tin hold?
Worked example

Q7: from areas to a volume

  • Areas back to k.
  • Then k³ for volume.
  1. k² = 576 ÷ 225 = 2.56areas scale by k²
  2. k = √2.56 = 1.6square root
  3. k³ = 1.6³ = 4.096volumes scale by k³
  4. V = 250 × 4.096small volume × k³
  5. = 1024 mL
  6. 250 × 2.56 = 640 mLk² is for areas
  7. The large tin holds 1024 mL.
Summary

The method that solved all seven

  • Prove similar, find k, then use k, k² or k³.
StepHow
Similar?equal angles, sides in 1 ratio
Trianglestwo equal angles are enough
Match sidesin order, by their angles
Scale factorfrom a pair known on both
Lengths× k, or ÷ k to go back
Whole sidenever just the piece
Area factork², so k = √(area ratio)
Volumesscale by k³, never k²

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