Similarity (With Applications to Maps and Models)

319. Area Ratio of Similar Triangles · Areas scale with the square of the sides

The ratio of the areas always equals the square of the ratio of corresponding sides.

ABCDEFMNBC = 400BC = 400EF = 240EF = 240AM = 300AM = 300DN = 180DN = 180Side ratio k = EF/BC = 0.6Side ratio k = EF/BC = 0.6Area ABC = ½·BC·AM = 60000Area ABC = ½·BC·AM = 60000Area DEF = ½·EF·DN = 21600Area DEF = ½·EF·DN = 21600Area ratio = DEF/ABC = 0.36 = k² = 0.36Area ratio = DEF/ABC = 0.36 = k² = 0.36K
For similar triangles, the ratio of areas equals the square of the ratio of corresponding sides: if the sides are in ratio k, the areas are in ratio k². Doubling every length quadruples the area.

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Selina ICSE: Similarity (With Applications to Maps and Models)

What this lesson covers

Try to break it

Drag K to resize △DEF. The side ratio DEF/ABC reads k, and the area ratio always reads k². When the sides scale by 2, the area scales by 4; when by 3, by 9. Try to find a scale factor where (area ratio) ≠ (side ratio)²; impossible.

How you build it

Build a triangle and its half-scale copy, and see the area shrink by k².

  • Point tool: mark A near the top — the apex of the triangle.
  • Point tool: mark B at the lower-left.
  • Point tool: mark C at the lower-right, roughly level with B so BC is a horizontal base.
  • Triangle tool: click your three points A, B, and C to draw the triangle.
  • Perpendicular tool: click vertex A, then click side BC — it drops the altitude (the height line) from A straight down to BC.
  • Point tool: mark M where the altitude meets BC. AM is the height of triangle ABC, so its area = ½·BC·AM.
  • Midpoint tool: click A then B to drop E exactly halfway — AE is half of AB. No ruler needed.
  • Midpoint tool: click A then C to drop F halfway — AF is half of AC.
  • Segment tool: join E to F. By the midpoint theorem EF is parallel to BC and exactly half its length, so triangle AEF is similar to ABC with ratio k = ½.
  • Point tool: mark N where the altitude AM crosses EF. AN is the height of the small triangle and equals ½·AM. So area AEF = ½·EF·AN = ½·(½BC)·(½AM) = ¼ of area ABC — the base and height each scale by k = ½, so the AREA scales by k² = ¼.

The proof, step by step

Prove that the ratio of areas of similar triangles equals the square of the ratio of their sides.

  • Since ΔABC ~ ΔDEF, the ratio of corresponding sides is constant: AB/DE = BC/EF = AC/DF = k.
  • Area of a triangle = ½ × base × height. So, Area(ABC) = ½ × BC × AM and Area(DEF) = ½ × EF × DN.
  • Because the triangles are similar, the ratio of their altitudes equals the ratio of their sides: AM/DN = AB/DE = k.
  • Dividing the area formulas: Area(ABC)/Area(DEF) = (BC/EF) × (AM/DN) = k × k = k².
  • Therefore, the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

Worked example

In ΔPQR ~ ΔXYZ, PQ = 6 cm, XY = 9 cm. If Area(ΔPQR) = 24 cm², find Area(ΔXYZ).

Ratio of sides = XY/PQ = 9/6 = 1.5. Ratio of areas = (1.5)² = 2.25. Area(ΔXYZ) = 24 × 2.25 = 54 cm².

  • 36 cm²
  • 48 cm²
  • 54 cm² — correct
  • 72 cm²
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