Areas of similar triangles

Doubling a triangle

Take a triangle and double every side. The big triangle can be cut into $4$ copies of the small one. Treble every side, and it can be cut into $9$ copies.

So the area is not doubled. It is multiplied by $2 \times 2 = 4$. When the sides are trebled, it is multiplied by $3 \times 3 = 9$. This is the rule for all similar triangles, not only for these.

The theorem

Theorem. The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

If $\triangle ABC \sim \triangle PQR$, then

Here $\text{ar}$ stands for "the area of".

Why it is true

Draw the altitudes $AM \perp BC$ and $PN \perp QR$.

1. Write the areas. $\text{ar}(ABC) = \tfrac{1}{2} \times BC \times AM$ and $\text{ar}(PQR) = \tfrac{1}{2} \times QR \times PN$. So $\dfrac{\text{ar}(ABC)}{\text{ar}(PQR)} = \dfrac{BC \times AM}{QR \times PN} = \dfrac{BC}{QR} \times \dfrac{AM}{PN}$.
2. Find a second pair of similar triangles. In $\triangle ABM$ and $\triangle PQN$, $\angle B = \angle Q$, because the big triangles are similar, and $\angle M = \angle N = 90^\circ$. By AA, $\triangle ABM \sim \triangle PQN$. So $\dfrac{AM}{PN} = \dfrac{AB}{PQ}$.
3. Put them together. Since $\triangle ABC \sim \triangle PQR$, we have $\dfrac{AB}{PQ} = \dfrac{BC}{QR}$. So $\dfrac{AM}{PN} = \dfrac{BC}{QR}$, and the ratio of the areas is $\dfrac{BC}{QR} \times \dfrac{BC}{QR} = \left(\dfrac{BC}{QR}\right)^2$.

Step 2 also shows that corresponding altitudes are in the same ratio as corresponding sides. So the areas are also in the ratio of the squares of the corresponding altitudes.

Using the theorem

From sides to areas. The sides of two similar triangles are in the ratio $3 : 5$. The areas are in the ratio $3^2 : 5^2 = 9 : 25$.

From areas to sides. Two similar triangles have areas $36$ square cm and $81$ square cm. The ratio of the areas is $36 : 81$, so the ratio of the sides is $6 : 9$, which is $2 : 3$.

Using altitudes. The corresponding altitudes of two similar triangles are $10$ cm and $15$ cm. The altitudes are in the ratio $2 : 3$, so the areas are in the ratio $4 : 9$. If the smaller area is $40$ square cm, the larger is $40 \times \dfrac{9}{4} = 90$ square cm.

A line parallel to a side. In $\triangle ABC$, $DE \parallel BC$, with D on AB and E on AC. Let $AD = 2$ cm and $DB = 3$ cm.

1. $\angle A$ is common, and $\angle ADE = \angle B$. So $\triangle ADE \sim \triangle ABC$ by AA.
2. $\dfrac{AD}{AB} = \dfrac{2}{5}$, so $\dfrac{\text{ar}(ADE)}{\text{ar}(ABC)} = \dfrac{4}{25}$.
3. If $\text{ar}(ABC) = 75$ square cm, then $\text{ar}(ADE) = 75 \times \dfrac{4}{25} = 12$ square cm.
4. The trapezium DBCE has the rest: $75 - 12 = 63$ square cm.

Equilateral triangles. Every equilateral triangle is similar to every other. On sides of $4$ cm and $6$ cm, the equilateral triangles have areas in the ratio $4^2 : 6^2 = 16 : 36 = 4 : 9$.

Practise with new numbers

Try it in Python

The program takes a triangle with sides $5$, $6$ and $7$ cm, multiplies every side by $k$, and compares the areas using Heron's formula.

Common mistakes

• Using the ratio of the sides for the areas. The ratio of the areas is the square of the ratio of the sides.
• Squaring when going back. From areas to sides, take the square root of the ratio of the areas.
• Mixing up the order. If the sides are $2 : 5$, the areas are $4 : 25$, in the same order.
• Using the theorem on triangles that are not similar. Check similarity first.

Remember these

• Sides in the ratio $k$ give areas in the ratio $k^2$.
• Altitudes and sides are in the same ratio, so areas are in the ratio of the squares of the altitudes too.
• When $DE \parallel BC$, $\triangle ADE \sim \triangle ABC$, and $\dfrac{\text{ar}(ADE)}{\text{ar}(ABC)} = \left(\dfrac{AD}{AB}\right)^2$.
• Area of a trapezium DBCE $= \text{ar}(ABC) - \text{ar}(ADE)$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. The sides of two similar triangles are in the ratio $2 : 7$. In what ratio are their areas?

2. Two similar triangles have areas $25$ square cm and $64$ square cm. What is the ratio of corresponding sides?

3. A triangle has an area of $20$ square cm. Every side is doubled. What is the new area?

4. Corresponding altitudes of two similar triangles are $4$ cm and $12$ cm. What is the ratio of their areas?

5. Sunil says that when every side is trebled, the area is trebled. What is wrong?