MathsICSE Class 10 · Similarity

Areas of Similar Triangles: Square of the Side Ratio, with Solved Questions (ICSE Class 10)

Updated 15 October 2026 · Fact-checked

The areas of two similar triangles are in the ratio of the squares of their corresponding sides: area ABC : area PQR = AB² : PQ². The same holds for corresponding altitudes, medians and perimeters. To go from an area ratio back to sides, take square roots.

Square the ratio of the whole sides AD : AB = 3 : 5 to get areas 9 : 25; the trapezium DBCE is the remaining 16 parts.

Illustration drawn for QuestionBuddy.

Understand Areas of Similar Triangles

The CISCE syllabus states the result directly: the areas of similar triangles are proportional to the squares of corresponding sides.

The rule. If $△ A B C \sim △ P QR$ with scale factor $k = \frac{A B}{P Q}$ , then

$\frac{area △ A B C}{area △ P QR} = \frac{A B ^{2}}{P Q ^{2}} = \frac{B C ^{2}}{Q R ^{2}} = \frac{C A ^{2}}{R P ^{2}} = k^{2} .$

Why squares. Area is $\frac{1}{2} \times base \times height$ . In similar triangles both the base and the height scale by $k$ , so the area scales by $k \times k = k^{2}$ .

Other lengths. Every corresponding length (altitude, median, angle bisector, perimeter) is in the ratio $k$ , so the area ratio is also the square of their ratio.

Two directions.

Sides to areas: square. Sides $2 : 5$ give areas $4 : 25$ .
Areas to sides: square-root. Areas $9 : 49$ give sides $3 : 7$ .

A line parallel to a side. If $D E ∥ B C$ with $D$ on $A B$ and $E$ on $A C$ , then $△ A D E \sim △ A B C$ and

$\frac{area △ A D E}{area △ A B C} = (\frac{A D}{A B})^{2} .$

Use the whole side $A B$ , not $D B$ . If $A D : D B = 3 : 2$ , then $A D : A B = 3 : 5$ and the areas are $9 : 25$ . The remaining trapezium $D B C E$ is the difference: $25 - 9 = 16$ parts.

Exam pattern. In 2024 Q6(iii)(d) the triangles $A P D$ and $B P C$ were similar with $A D = 6$ and $B C = 2.4$ , so the area ratio was $6^{2} : 2. 4^{2} = 25 : 4$ . The examiners reported candidates who did not use the square of the side ratio.

Areas of Similar Triangles: formulas to remember

Area ratio of similar triangles: $\frac{ar ( △ A B C )}{ar ( △ P QR )} = (\frac{A B}{P Q})^{2} = k^{2}$; Any pair of corresponding sides.
Side ratio from areas: $\frac{A B}{P Q} = \frac{ar ( △ A B C )}{ar ( △ P QR )}$; Square root of the area ratio.
Other corresponding lengths: $\frac{ar _{1}}{ar _{2}} = (\frac{h _{1}}{h _{2}})^{2} = (\frac{m _{1}}{m _{2}})^{2} = (\frac{P _{1}}{P _{2}})^{2}$; Altitudes, medians, perimeters.
Parallel line cutting a triangle: $D E ∥ B C \Rightarrow \frac{ar ( △ A D E )}{ar ( △ A B C )} = (\frac{A D}{A B})^{2}$; Use $A B$ , not $D B$ .

How to solve Areas of Similar Triangles questions

Establish similarity, find the side ratio k, then square it.

1Show or note that the triangles are similar, with vertices matched.
2Find the ratio $k$ of a pair of corresponding sides (or altitudes, medians, perimeters).
3Square it to get the area ratio.
4For a known area, multiply by $k^{2}$ (or divide).
5For a side from areas, take the square root of the area ratio.
6For a trapezium left after a parallel cut, subtract the small triangle's share from the whole.

Areas of Similar Triangles in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Material needed scales with the square

Suppose a boat-maker has a triangular sail 3 m high and wants a similar sail 4.5 m high.

How many times as much cloth will the larger sail need?

Try it first, then show the solutionSolution

Scale factor of heights $k = \frac{4.5}{3} = 1.5$ .
Area factor $k^{2} = 2.25$ .

Answer: 2.25 times as much cloth.

Memory hook: Lengths ×k, areas ×k².

Doubling lengths quadruples area

Imagine a triangular logo printed on a poster at twice its original width and height.

How does the amount of ink needed for the logo change?

Try it first, then show the solutionSolution

The enlarged logo is similar with $k = 2$ .
Area, and so ink, changes by $k^{2} = 4$ .

Answer: Four times as much ink.

Memory hook: Twice as wide and twice as tall is four times the area.

Tricks to remember Areas of Similar Triangles

Short memory aids. Say each one aloud once, then use it on a question the same day.

Square going up, root coming down
Sides → areas: square. Areas → sides: square root.
Use it when: Moving between side and area ratios.
Whole side, not the piece
With $D E ∥ B C$ , compare $A D$ with $A B$ (the whole), then square.
Use it when: Triangle-and-trapezium questions.
Every length, same k
Sides, heights, medians and perimeters all share the ratio $k$ ; only area uses $k^{2}$ .
Use it when: Questions that give altitudes or perimeters.

Areas of Similar Triangles: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: From sides to areas · 1 mark · MCQBoard level

Two similar triangles have corresponding sides in the ratio $2 : 5$ . The ratio of their areas is:
(a) $2 : 5$
(b) $4 : 25$
(c) $4 : 10$
(d) $8 : 125$

Try it first, then show the solutionSolution

Area ratio $= 2^{2} : 5^{2} = 4 : 25$ .

Answer: (b) $4 : 25$

Example 2: A side from areas · 3 marks · Short answerBoard level

$△ A B C \sim △ D E F$ . The areas of $△ A B C$ and $△ D E F$ are 36 cm² and 81 cm², and $A B = 8$ cm. Find $D E$ .

Try it first, then show the solutionSolution

$(\frac{A B}{D E})^{2} = \frac{36}{81} \Rightarrow \frac{A B}{D E} = \frac{6}{9} = \frac{2}{3}$ .
$D E = 8 \times \frac{3}{2} = 12$ cm.

Answer: $D E = 12$ cm

Example 3: Triangle and trapezium · 3 marks · Short answerHarder board level

In $△ A B C$ , $D E ∥ B C$ with $D$ on $A B$ , $E$ on $A C$ and $A D : D B = 3 : 2$ . The area of $△ A B C$ is 75 cm². Find the areas of $△ A D E$ and of trapezium $D B C E$ .

Try it first, then show the solutionSolution

$△ A D E \sim △ A B C$ (corresponding angles, as $D E ∥ B C$ ); $\frac{A D}{A B} = \frac{3}{5}$ .
Area $△ A D E = 75 \times \frac{9}{25} = 27$ cm².
Area $D B C E = 75 - 27 = 48$ cm².

Answer: 27 cm² and 48 cm²

Example 4: Areas from altitudes · 3 marks · Short answerHarder board level

Two similar triangles have corresponding altitudes 6 cm and 9 cm. The area of the larger triangle is 90 cm². Find the area of the smaller.

Try it first, then show the solutionSolution

Ratio of altitudes $= \frac{6}{9} = \frac{2}{3}$ , so the area ratio is $\frac{4}{9}$ .
Smaller area $= 90 \times \frac{4}{9} = 40$ cm².

Answer: 40 cm²

Example 5: Halving the area with a parallel line · 4 marks · Short answerHOTS / competency

A line $D E$ parallel to $B C$ cuts $A B$ at $D$ and $A C$ at $E$ so that $△ A D E$ and trapezium $D B C E$ have equal areas. Find $\frac{A D}{A B}$ and $\frac{A D}{D B}$ .

Try it first, then show the solutionSolution

Area $△ A D E = \frac{1}{2}$ area $△ A B C$ , so $(\frac{A D}{A B})^{2} = \frac{1}{2}$ and $\frac{A D}{A B} = \frac{1}{2}$ .
Let $A B = 2$ and $A D = 1$ ; then $D B = 2 - 1$ .
$\frac{A D}{D B} = \frac{1}{2 - 1} = 2 + 1$ (multiplying top and bottom by $2 + 1$ ).

Answer: $\frac{A D}{A B} = \frac{1}{2}$ ; $A D : D B = (2 + 1) : 1$

Example 6: Case study: a path across a garden · 4 marks · Case-basedHOTS / competency

A triangular garden $A B C$ has $∠ B = 90°$ , $A B = 30$ m and $B C = 40$ m. A straight path $D E$ runs parallel to $B C$ , with $D$ on $A B$ , $E$ on $A C$ and $A D = 12$ m.
(a) Find area $△ A D E$ : area $△ A B C$ .
(b) Find the area of the garden and of the part $D B C E$ beyond the path.
(c) Find the length of the path $D E$ .

Try it first, then show the solutionSolution

(a) $△ A D E \sim △ A B C$ ; $\frac{A D}{A B} = \frac{12}{30} = \frac{2}{5}$ , so the area ratio is $4 : 25$ .
(b) Area $A B C = \frac{1}{2} \times 30 \times 40 = 600$ m². Area $A D E = 600 \times \frac{4}{25} = 96$ m², so $D B C E = 504$ m².
(c) $D E = 40 \times \frac{2}{5} = 16$ m.

Answer: (a) $4 : 25$ (b) 600 m² and 504 m² (c) 16 m

Where students lose marks in Areas of Similar Triangles

The slips examiners see most often, and what to write instead.

Mistake: Using the side ratio for areas
Areas scale by the square of the side ratio.
Wrong working
Area $△ A P D$ : area $△ B P C = 6 : 2.4 = 5 : 2$ .
Right working
$6^{2} : 2. 4^{2} = 25 : 4$ .
Do this: Write 'ratio of areas = (ratio of sides)²' before substituting.
Usually costs: Reported in 2024 Q6(iii)(d).
Mistake: Using the part instead of the whole side
The similar triangles are $A D E$ and $A B C$ ; their sides are $A D$ and $A B$ .
Wrong working
$A D : D B = 3 : 2$ , so areas $9 : 4$ .
Right working
$A D : A B = 3 : 5$ , so areas $9 : 25$ .
Do this: Convert to a part-to-whole ratio first.
Usually costs: The final answer.
Mistake: Forgetting unit conversion in area scale
Area units are squared: $1 m^{2} = 10000 cm^{2}$ .
Wrong working
$576 m^{2} = 57600 cm^{2}$ .
Right working
$576 m^{2} = 5760000 cm^{2}$ .
Do this: Convert using the square of the length conversion.
Usually costs: Reported in 2025 Q2(iii).

Exam tips

2024 Q6(iii)(d) expected area $△ A P D$ : area $△ B P C = 25 : 4$ from $A D = 6$ and $B C = 2.4$ .
2025 Q2(iii) used an area ratio to find a scale factor ( $k^{2}$ first, then $k$ ).
State the similarity before using the area rule; it is the justification.
Keep ratios exact; simplify decimals such as $6 : 2.4$ to $5 : 2$ before squaring.

Practice questions on Areas of Similar Triangles

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The areas of two similar triangles are in the ratio $16 : 81$ . The ratio of their corresponding sides is:
(a) $16 : 81$
(b) $4 : 9$
(c) $8 : 9$
(d) $2 : 3$
Hint
Square roots.
Answer
(b) $4 : 9$
Q2 · 2 marks · Short answerBoard level
$△ A B C \sim △ P QR$ , $A B : P Q = 3 : 4$ and the area of $△ P QR$ is 48 cm². Find the area of $△ A B C$ .
Hint
Multiply by $\frac{9}{16}$ .
Answer
27 cm²
Q3 · 2 marks · Short answerHarder board level
Two similar triangles have perimeters 20 cm and 30 cm. The smaller has area 32 cm². Find the area of the larger.
Hint
Perimeter ratio $\frac{3}{2}$ .
Answer
$32 \times \frac{9}{4} = 72$ cm²
Q4 · 3 marks · Short answerHarder board level
In $△ A B C$ , $D$ and $E$ are the mid-points of $A B$ and $A C$ . If the area of $△ A B C$ is 64 cm², find the areas of $△ A D E$ and quadrilateral $D B C E$ .
Hint
$D E ∥ B C$ and $\frac{A D}{A B} = \frac{1}{2}$ .
Answer
16 cm² and 48 cm²
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): If every side of a triangle is doubled, its area is doubled.
Reason (R): The ratio of the areas of similar triangles equals the square of the ratio of corresponding sides.
(a) (A) is true and (R) is false.
(b) (A) is false and (R) is true.
(c) Both (A) and (R) are true, and (R) is the correct explanation for (A).
(d) Both (A) and (R) are true, but (R) is not the correct explanation for (A).
Hint
$2^{2} = 4$ .
Answer
(b). The area becomes four times as large.
Q6 · 4 marks · Short answerHarder board level
In $△ A B C$ , $∠ A = 90°$ , $A B = 6$ cm, $A C = 8$ cm and $A D ⊥ B C$ . Find area $△ A B D$ : area $△ A B C$ and area $△ A C D$ : area $△ A B C$ .
Hint
$△ D B A \sim △ A B C$ and $△ D A C \sim △ A B C$ ; $B C = 10$ .
Answer
$(\frac{6}{10})^{2} = \frac{9}{25}$ and $(\frac{8}{10})^{2} = \frac{16}{25}$ (they add to 1, as they should).
Q7 · 4 marks · Short answerHOTS / competency
In $△ P QR$ , $S T ∥ QR$ with $S$ on $P Q$ and $T$ on $P R$ . The area of $△ P S T$ is $\frac{1}{8}$ of the area of trapezium $S T R Q$ . Find $P S : S Q$ .
Hint
Area $P S T$ : area $P QR = 1 : 9$ .
Answer
$\frac{P S}{P Q} = \frac{1}{3}$ , so $P S : S Q = 1 : 2$ .

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.State the area rule for similar triangles.Show answer
Their areas are in the ratio of the squares of corresponding sides.
2.Sides $1 : 3$ . Areas?Show answer
$1 : 9$
3.Areas $25 : 64$ . Sides?Show answer
$5 : 8$
4.Altitudes $2 : 3$ . Areas?Show answer
$4 : 9$
5.Mid-point segment $D E$ : area $A D E$ is what fraction of area $A B C$ ?Show answer
$\frac{1}{4}$

Areas of Similar Triangles: frequently asked questions

What is the ratio of the areas of two similar triangles?

It equals the square of the ratio of any pair of corresponding sides. For sides in the ratio 2 : 3, the areas are in the ratio 4 : 9.

Does the area rule work for altitudes and medians too?

Yes. Corresponding altitudes, medians and perimeters of similar triangles are in the same ratio as the sides, so the area ratio is the square of their ratio as well.

How do I find the area of the trapezium left when a parallel line cuts a triangle?

Find the small triangle's area using (AD/AB)² times the whole area, then subtract it from the area of the whole triangle.