Triangles • Topic 2 of 3

Area of similar triangles

What is the Area of Similar Triangles Theorem? When two triangles are similar, their sides are scaled by a specific factor. But what happens to their area? The Area of Similar Triangles Theorem provides the exact rule: the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

Think of it like painting a square wall. If you double the length and width of the wall, you do not just need twice as much paint—you need four times as much paint! This is because area is a two-dimensional measure (length multiplied by width).

Mathematical Rule: If Triangle ABC is similar to Triangle PQR, then: (Area of Triangle ABC) / (Area of Triangle PQR) = (AB / PQ) squared = (BC / QR) squared = (AC / PR) squared.

This same rule also applies to other corresponding linear measurements of the triangles, such as:

  • The ratio of their corresponding altitudes (heights).
  • The ratio of their corresponding medians.
  • The ratio of their corresponding angle bisectors.

---

DIAGRAM 1: VISUALISING THE SQUARED SCALE FACTOR

    Triangle 1 (Side = 1 unit)     Triangle 2 (Side = 2 units)
          /\                                   /\
         /__\                                 /__\
      Area = 1 unit                       /\  /\  /\
                                         /__\/__\/__\
                                         Area = 4 units
    
    When side length doubles (x2), the area quadruples (x4 or 2 squared).

DIAGRAM 2: THEOREM RELATIONSHIP

         Triangle ABC                     Triangle PQR
              A                                P
             /\                               /\
            /  \                             /  \
           /____\                           /____\
          B      C                         Q      R
          
    Area(ABC)     / AB \ 2     / BC \ 2     / AC \ 2
    ---------  =  | -- |   =   | -- |   =   | -- |
    Area(PQR)     \ PQ /       \ QR /       \ PR /

DIAGRAM 3: ALTITUDE RATIO APPLICATION
              A                                P
             /\                               /\
            / | \                            / | \
           /__|__\                          /__|__\
          B   M   C                        Q   N  R
          
    Area(ABC) / Area(PQR) = (AM / PN) squared, where AM and PN are heights.
Solved Examples
4
Worked Example
Triangle ABC is similar to Triangle DEF. The length of side BC is 3 cm and the length of side EF is 5 cm. If the area of Triangle ABC is 18 square cm, find the area of Triangle DEF.
Solution
  1. Step 1: Apply the Area of Similar Triangles Theorem.*
  2. (Area of Triangle ABC) / (Area of Triangle DEF) = (BC / EF) squared*
  3. Step 2: Plug in the given values.*
  4. 18 / (Area of Triangle DEF) = (3 / 5) squared*
  5. 18 / (Area of Triangle DEF) = 9 / 25*
  6. Step 3: Cross-multiply to solve for the unknown area.*
  7. 9 (Area of Triangle DEF) = 18 25*
  8. 9 (Area of Triangle DEF) = 450
  9. Area of Triangle DEF = 450 / 9 = 50 square cm.*
  10. Answer: The area of Triangle DEF is 50 square cm.
5
Worked Example
The areas of two similar triangles are 49 square cm and 81 square cm. If the longest side of the smaller triangle is 14 cm, find the longest side of the larger triangle.
Solution
  1. Step 1: State the theorem relation.*
  2. (Area of Small Triangle) / (Area of Large Triangle) = (Side of Small Triangle / Side of Large Triangle) squared*
  3. Step 2: Substitute the known areas.*
  4. 49 / 81 = (14 / Side of Large Triangle) squared*
  5. Step 3: Take the square root of both sides to simplify.*
  6. Square root of (49 / 81) = 14 / Side of Large Triangle*
  7. 7 / 9 = 14 / Side of Large Triangle*
  8. Step 4: Solve for the unknown side.*
  9. 7 (Side of Large Triangle) = 9 14*
  10. 7 (Side of Large Triangle) = 126
  11. Side of Large Triangle = 126 / 7 = 18 cm.*
  12. Answer: The longest side of the larger triangle is 18 cm.
6
Worked Example
Triangle ABC is similar to Triangle PQR. AM is the median drawn from vertex A to side BC, and PN is the median drawn from vertex P to side QR. If Area(ABC) / Area(PQR) = 4 / 9, and PN = 12 cm, find the length of AM.
Solution
  1. Step 1: Understand the extended rule of the theorem.*
  2. The ratio of the areas of similar triangles is also equal to the square of the ratio of their corresponding medians.*
  3. Area(ABC) / Area(PQR) = (AM / PN) squared*
  4. Step 2: Substitute the known values.*
  5. 4 / 9 = (AM / 12) squared*
  6. Step 3: Take the square root of both sides.*
  7. 2 / 3 = AM / 12*
  8. Step 4: Solve for AM.*
  9. 3 AM = 2 12*
  10. 3 AM = 24
  11. AM = 24 / 3 = 8 cm.*
  12. Answer: The length of the median AM is 8 cm.
  13. --

Key Points

  • The area ratio of two similar triangles equals the square of the ratio of their corresponding sides.
  • If the sides are scaled by a factor of $k$, the area scales by a factor of $k^2$.
  • This square relationship holds true for all corresponding line segments, including altitudes, medians, and perpendicular bisectors.
  • To find side ratios when areas are given, always take the square root of the area ratio.
Quick Check — 4 MCQs
Tap an option to check your answer0 / 4
Q1.The ratio of areas of two similar triangles equals the ratio of the:
Explanation: Area ratio $=$ (side ratio)$^2$.
Q2.If the sides are in ratio $2:3$, the areas are in ratio:
Explanation: Square the side ratio.
Q3.If areas are in ratio $9:16$, the sides are in ratio:
Explanation: Take square roots.
Q4.Similar triangles with equal areas are:
Explanation: Equal area $+$ similar $\Rightarrow$ congruent.
Practice more MCQs → 📝 Assignment · 35 marks