Areas of Similar Triangles: Ratio Property

Similar triangles are triangles that have the same shape but not necessarily the same size. This means their corresponding angles are equal, and their corresponding sides are proportional. A fundamental relationship exists between the areas of similar triangles and the lengths of their corresponding sides. The key concept here is that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

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Formulae

Let’s consider two similar triangles, $\triangle ABC$ and $\triangle DEF$. Let $A(\triangle ABC)$ and $A(\triangle DEF)$ represent their respective areas. Let $a, b, c$ be the sides of $\triangle ABC$, and $d, e, f$ be the corresponding sides of $\triangle DEF$. Then:

• $\frac{A(\triangle ABC)}{A(\triangle DEF)} = \left(\frac{AB}{DE}\right)^2 = \left(\frac{BC}{EF}\right)^2 = \left(\frac{CA}{FD}\right)^2$
• Or, if the scale factor (ratio of corresponding sides) is $k$, then $\frac{A(\triangle ABC)}{A(\triangle DEF)} = k^2$.

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Examples

Example-1:

Suppose two similar triangles have corresponding sides with lengths 4 cm and 6 cm. If the area of the smaller triangle is 10 cm², what is the area of the larger triangle?

Let $A_1$ and $A_2$ be the areas of the smaller and larger triangles, respectively. Let $s_1$ and $s_2$ be the lengths of the corresponding sides. We have:

• $s_1 = 4$ cm, $s_2 = 6$ cm, and $A_1 = 10$ cm²
• $\frac{A_1}{A_2} = \left(\frac{s_1}{s_2}\right)^2$
• $\frac{10}{A_2} = \left(\frac{4}{6}\right)^2 = \left(\frac{2}{3}\right)^2 = \frac{4}{9}$
• $A_2 = \frac{10 \cdot 9}{4} = 22.5$ cm²

Therefore, the area of the larger triangle is 22.5 cm².

Example-2:

Two similar triangles have areas of 36 cm² and 81 cm². If the shortest side of the smaller triangle is 6 cm, what is the length of the shortest side of the larger triangle?

Let $A_1 = 36$ cm², $A_2 = 81$ cm², $s_1 = 6$ cm (shortest side of smaller triangle), and $s_2$ be the corresponding side of the larger triangle.

• $\frac{A_1}{A_2} = \left(\frac{s_1}{s_2}\right)^2$
• $\frac{36}{81} = \left(\frac{6}{s_2}\right)^2$
• $\frac{4}{9} = \frac{36}{s_2^2}$
• $s_2^2 = \frac{36 \cdot 9}{4} = 81$
• $s_2 = \sqrt{81} = 9$ cm

Therefore, the length of the shortest side of the larger triangle is 9 cm.

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Theorem with Proof

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

Proof:

Consider two similar triangles, $\triangle ABC$ and $\triangle DEF$. Let $AB$ and $DE$ be corresponding sides. Since the triangles are similar, $\angle A = \angle D$, $\angle B = \angle E$, and $\angle C = \angle F$. Also, $\frac{AB}{DE} = \frac{BC}{EF} = \frac{CA}{FD} = k$ (where $k$ is the scale factor).

Let $h_1$ be the height of $\triangle ABC$ from vertex $C$ to side $AB$, and $h_2$ be the height of $\triangle DEF$ from vertex $F$ to side $DE$.

The area of $\triangle ABC$, $A(\triangle ABC) = \frac{1}{2} \cdot AB \cdot h_1$.

The area of $\triangle DEF$, $A(\triangle DEF) = \frac{1}{2} \cdot DE \cdot h_2$.

Since the triangles are similar, the ratio of corresponding heights is also equal to the ratio of corresponding sides, i.e., $\frac{h_1}{h_2} = k$ or $h_1=k\cdot h_2$.

Now, $\frac{A(\triangle ABC)}{A(\triangle DEF)} = \frac{\frac{1}{2} \cdot AB \cdot h_1}{\frac{1}{2} \cdot DE \cdot h_2} = \frac{AB}{DE} \cdot \frac{h_1}{h_2} = k \cdot k = k^2$.

Since $k = \frac{AB}{DE}$, we have $\frac{A(\triangle ABC)}{A(\triangle DEF)} = \left(\frac{AB}{DE}\right)^2$. This holds for any pair of corresponding sides.

Therefore, the ratio of the areas of similar triangles is the square of the ratio of their corresponding sides.

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Common mistakes by students

• Forgetting to square the ratio: Students often calculate the ratio of the sides but forget to square it when finding the ratio of the areas.
• Using incorrect corresponding sides: Students might mistakenly use non-corresponding sides in the ratio, leading to incorrect results. Careful identification of corresponding sides is crucial.
• Confusing area ratios with side ratios: Students sometimes confuse the relationship, using the area ratio directly as the side ratio or vice versa. Remember that side ratios are linear, while area ratios are squared.

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Real Life Application

This concept is used in various real-life applications:

• Scale drawings and maps: Architects and cartographers use this principle extensively. The ratio of the areas in a scaled model or map relates directly to the square of the scale factor of the lengths.
• Photography: When enlarging or reducing a photo, the ratio of the areas changes according to the square of the scale factor.
• Engineering: Designing similar structures or components (e.g., bridges, buildings, or electronic components). The concept is fundamental to maintaining structural integrity.
• Computer Graphics: Scaling objects involves area scaling based on the side length scaling, often employed in rendering 3D models.

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Fun Fact

The relationship between areas and side ratios in similar shapes isn’t limited to triangles. It extends to all similar figures, such as squares, circles, and other polygons. For instance, the ratio of the areas of two similar squares is also the square of the ratio of their side lengths.

Recommended YouTube Videos for Deeper Understanding

Q.1 If the ratio of the corresponding sides of two similar triangles is $2:3$, what is the ratio of their areas?

Check Solution

Ans: B

The ratio of areas is the square of the ratio of corresponding sides. Therefore, the ratio is $(2/3)^2 = 4/9$.

Q.2 The area of a triangle is $25 cm^2$. If a similar triangle has sides that are $1.5$ times the length of the original triangle’s sides, what is the area of the larger triangle?

Check Solution

Ans: B

The ratio of the sides is $1.5$, so the ratio of the areas is $1.5^2 = 2.25$. Thus, the new area is $25 * 2.25 = 56.25 cm^2$.

Q.3 Two similar triangles have areas of $36 cm^2$ and $81 cm^2$. What is the ratio of the lengths of their corresponding sides?

Check Solution

Ans: D

The ratio of areas is $36/81 = 4/9$. The ratio of sides is the square root of the ratio of areas, which is $\sqrt{4/9} = 2/3$. However, since the sides ratio asks for a pair, and $6/9$ simplifies to $2/3$, then $6:9$ is a possible answer.

Q.4 Triangle ABC is similar to triangle DEF. If AB = $6 cm$, DE = $9 cm$, and the area of triangle ABC is $24 cm^2$, find the area of triangle DEF.

Check Solution

Ans: B

The ratio of sides is $6/9 = 2/3$. The ratio of areas is $(2/3)^2 = 4/9$. Let the area of DEF be $x$. Then $24/x = 4/9$, so $x = (24*9)/4 = 54 cm^2$.

Q.5 The ratio of the areas of two similar triangles is $16:25$. If the perimeter of the smaller triangle is $16 cm$, what is the perimeter of the larger triangle?

Check Solution

Ans: A

The ratio of the sides is $\sqrt{16/25} = 4/5$. Since the perimeter is proportional to the side length, the ratio of the perimeters is also $4/5$. Let the perimeter of the larger triangle be $x$. Then $16/x = 4/5$, so $x = (16*5)/4 = 20 cm$.

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