Similarity of Triangles: Areas and Applications
Similarity of Triangles: Areas and Applications - Areas of Similar Triangles Theorem (Ratio of Areas = Square of Ratio of Sides)
We have learned about the definition of similar triangles and criteria to prove their similarity. A significant theorem related to similar triangles concerns the relationship between their areas. This theorem states that the ratio of the areas of two similar triangles is directly related to the square of the ratio of their corresponding sides. This property is very useful in solving problems involving areas and lengths of similar figures.
Theorem Statement
The theorem relating the areas of similar triangles is stated as follows:
Theorem: The ratio of the areas of two similar triangles is equal to the square of the ratio of any pair of their corresponding sides.
If $\triangle \text{ABC}$ is similar to $\triangle \text{PQR}$ (denoted as $\triangle \text{ABC} \sim \triangle \text{PQR}$), then:
$\frac{\text{Area}(\triangle \text{ABC})}{\text{Area}(\triangle \text{PQR})} = \left(\frac{\text{AB}}{\text{PQ}}\right)^2 = \left(\frac{\text{BC}}{\text{QR}}\right)^2 = \left(\frac{\text{AC}}{\text{PR}}\right)^2$
Here, $\text{AB}$ and $\text{PQ}$, $\text{BC}$ and $\text{QR}$, and $\text{AC}$ and $\text{PR}$ are pairs of corresponding sides. The scale factor $k = \frac{\text{AB}}{\text{PQ}} = \frac{\text{BC}}{\text{QR}} = \frac{\text{AC}}{\text{PR}}$. The theorem states that the ratio of areas is $k^2$.
Proof of the Areas of Similar Triangles Theorem
Given: Two triangles $\triangle \text{ABC}$ and $\triangle \text{PQR}$ such that $\triangle \text{ABC} \sim \triangle \text{PQR}$.
Since the triangles are similar, by definition, their corresponding angles are equal and their corresponding sides are in proportion:
$\angle \text{A} = \angle \text{P}, \quad \angle \text{B} = \angle \text{Q}, \quad \angle \text{C} = \angle \text{R}$
$\frac{\text{AB}}{\text{PQ}} = \frac{\text{BC}}{\text{QR}} = \frac{\text{AC}}{\text{PR}}$
... (i)
To Prove: $\frac{\text{Area}(\triangle \text{ABC})}{\text{Area}(\triangle \text{PQR})} = \left(\frac{\text{AB}}{\text{PQ}}\right)^2 = \left(\frac{\text{BC}}{\text{QR}}\right)^2 = \left(\frac{\text{AC}}{\text{PR}}\right)^2$.
Construction: Draw altitudes AM from vertex A to side BC in $\triangle \text{ABC}$, and PN from vertex P to side QR in $\triangle \text{PQR}$. M lies on BC and N lies on QR.
Proof:
We know that the area of a triangle is calculated using the formula: Area $= \frac{1}{2} \times \text{base} \times \text{height}$.
Area($\triangle \text{ABC}$) $= \frac{1}{2} \times \text{BC} \times \text{AM}$
Area($\triangle \text{PQR}$) $= \frac{1}{2} \times \text{QR} \times \text{PN}$
Taking the ratio of the areas of $\triangle \text{ABC}$ and $\triangle \text{PQR}$:
$\frac{\text{Area}(\triangle \text{ABC})}{\text{Area}(\triangle \text{PQR})} = \frac{\frac{1}{2} \times \text{BC} \times \text{AM}}{\frac{1}{2} \times \text{QR} \times \text{PN}}$
$\frac{\text{Area}(\triangle \text{ABC})}{\text{Area}(\triangle \text{PQR})} = \left(\frac{\text{BC}}{\text{QR}}\right) \times \left(\frac{\text{AM}}{\text{PN}}\right)$
... (ii)
Now, consider the right-angled triangles $\triangle \text{ABM}$ and $\triangle \text{PQN}$.
1. $\angle \text{B} = \angle \text{Q}$
(Corresponding angles of similar triangles $\triangle \text{ABC}$ and $\triangle \text{PQR}$)
2. $\angle \text{AMB} = 90^\circ$
(By Construction, AM $\perp$ BC)
3. $\angle \text{PNQ} = 90^\circ$
(By Construction, PN $\perp$ QR)
4. $\angle \text{AMB} = \angle \text{PNQ}$
(From statements 2 and 3, both are $90^\circ$)
Therefore, by the AA (Angle-Angle) similarity criterion:
$\triangle \text{ABM} \sim \triangle \text{PQN}$
(AA Similarity - from statements 1 and 4)
Since these triangles are similar, the ratio of their corresponding sides is equal.
$\frac{\text{AM}}{\text{PN}} = \frac{\text{AB}}{\text{PQ}}$
(Ratio of corresponding sides)
Let's call this equation (iii).
... (iii)
From equation (i), we know that the ratio of corresponding sides of the similar triangles $\triangle \text{ABC}$ and $\triangle \text{PQR}$ is constant, i.e., $\frac{\text{AB}}{\text{PQ}} = \frac{\text{BC}}{\text{QR}} = \frac{\text{AC}}{\text{PR}}$.
Substitute $\frac{\text{AB}}{\text{PQ}}$ with $\frac{\text{BC}}{\text{QR}}$ in equation (iii):
$\frac{\text{AM}}{\text{PN}} = \frac{\text{BC}}{\text{QR}}$
(Substituting from i into iii)
Let's call this equation (iv).
... (iv)
Now, substitute the result from (iv) back into equation (ii):
$\frac{\text{Area}(\triangle \text{ABC})}{\text{Area}(\triangle \text{PQR})} = \left(\frac{\text{BC}}{\text{QR}}\right) \times \left(\frac{\text{BC}}{\text{QR}}\right)$
(Substituting from iv into ii)
$\frac{\text{Area}(\triangle \text{ABC})}{\text{Area}(\triangle \text{PQR})} = \left(\frac{\text{BC}}{\text{QR}}\right)^2$
Using equation (i) again, since all the ratios of corresponding sides are equal, we can replace $\left(\frac{\text{BC}}{\text{QR}}\right)^2$ with the square of the ratio of any other pair of corresponding sides.
$\mathbf{\frac{\text{Area}(\triangle \text{ABC})}{\text{Area}(\triangle \text{PQR})} = \left(\frac{\text{AB}}{\text{PQ}}\right)^2 = \left(\frac{\text{BC}}{\text{QR}}\right)^2 = \left(\frac{\text{AC}}{\text{PR}}\right)^2}$
Hence, the theorem is proved: The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
Hence Proved.
Extension of the Theorem
The relationship between the ratio of areas and the ratio of corresponding sides extends to other corresponding linear elements in similar triangles. The ratio of the areas of two similar triangles is also equal to the square of the ratio of their corresponding altitudes, corresponding medians, and corresponding angle bisector segments.
If $\triangle \text{ABC} \sim \triangle \text{PQR}$, and AM and PN are corresponding altitudes, AD and PS are corresponding medians, and AX and PY are corresponding angle bisector segments, then:
$\frac{\text{Area}(\triangle \text{ABC})}{\text{Area}(\triangle \text{PQR})} = \left(\frac{\text{AB}}{\text{PQ}}\right)^2 = \left(\frac{\text{BC}}{\text{QR}}\right)^2 = \left(\frac{\text{AC}}{\text{PR}}\right)^2 = \left(\frac{\text{AM}}{\text{PN}}\right)^2 = \left(\frac{\text{AD}}{\text{PS}}\right)^2 = \left(\frac{\text{AX}}{\text{PY}}\right)^2$
This is because the ratio of any pair of corresponding linear elements in similar triangles is equal to the scale factor (the ratio of corresponding sides).
Example 1. Let $\triangle \text{ABC} \sim \triangle \text{DEF}$ and their areas be, respectively, 64 cm$^2$ and 121 cm$^2$. If $\text{EF} = 15.4$ cm, find the length of side $\text{BC}$.
Answer:
Given:
- $\triangle \text{ABC} \sim \triangle \text{DEF}$
- $\text{Area}(\triangle \text{ABC}) = 64 \text{ cm}^2$
- $\text{Area}(\triangle \text{DEF}) = 121 \text{ cm}^2$
- $\text{EF} = 15.4 \text{ cm}$
To Find: The length of side BC.
Solution:
Since $\triangle \text{ABC} \sim \triangle \text{DEF}$, the vertices correspond as A $\leftrightarrow$ D, B $\leftrightarrow$ E, and C $\leftrightarrow$ F. Therefore, BC and EF are corresponding sides.
According to the theorem on areas of similar triangles, the ratio of their areas is equal to the square of the ratio of their corresponding sides.
$\frac{\text{Area}(\triangle \text{ABC})}{\text{Area}(\triangle \text{DEF})} = \left(\frac{\text{BC}}{\text{EF}}\right)^2$
(Theorem on Areas of Similar Triangles)
Substitute the given values into the equation:
$\frac{64}{121} = \left(\frac{\text{BC}}{15.4}\right)^2$
To solve for BC, first take the square root of both sides of the equation:
$\sqrt{\frac{64}{121}} = \frac{\text{BC}}{15.4}$
$\frac{\sqrt{64}}{\sqrt{121}} = \frac{\text{BC}}{15.4}$
$\frac{8}{11} = \frac{\text{BC}}{15.4}$
Now, multiply both sides by 15.4 to isolate BC:
$\text{BC} = \frac{8}{11} \times 15.4$
To simplify the calculation, convert 15.4 to a fraction:
$\text{BC} = \frac{8}{11} \times \frac{154}{10}$
Cancel out common factors (11 divides 154):
$\text{BC} = \frac{8}{\cancel{11}_1} \times \frac{\cancel{154}^{14}}{10}$
$\text{BC} = \frac{8 \times 14}{10}$
$\text{BC} = \frac{112}{10}$
$\mathbf{\text{BC} = 11.2 \text{ cm}}$
Therefore, the length of side BC is 11.2 cm.
Similarity of Triangles: Areas and Applications - Similarity in Right Triangles (Altitude to Hypotenuse)
Right-angled triangles possess many unique properties. One particularly important configuration that leads to triangle similarity occurs when an altitude is drawn from the vertex of the right angle to the hypotenuse. This construction divides the original right triangle into two smaller triangles, and all three triangles in the figure are similar to each other.
Theorem: Similarity Induced by Altitude to Hypotenuse
Theorem Statement: If an altitude is drawn from the vertex of the right angle of a right triangle to its hypotenuse, then the two triangles formed on either side of the altitude are similar to the original triangle and are also similar to each other.
Given: A right-angled triangle $\triangle \text{ABC}$, with the right angle at vertex B ($\text{m}\angle \text{ABC} = 90^\circ$). BD is the altitude drawn from vertex B to the hypotenuse AC, where D lies on AC ($\text{BD} \perp \text{AC}$).
The construction creates two smaller triangles: $\triangle \text{ADB}$ and $\triangle \text{BDC}$. Both of these are right-angled triangles (at D).
To Prove:
- $\triangle \text{ADB} \sim \triangle \text{ABC}$ (The smaller triangle on one side is similar to the whole triangle)
- $\triangle \text{BDC} \sim \triangle \text{ABC}$ (The smaller triangle on the other side is similar to the whole triangle)
- $\triangle \text{ADB} \sim \triangle \text{BDC}$ (The two smaller triangles are similar to each other)
Proof:
We will use the AA (Angle-Angle) similarity criterion to prove the similarity of these triangles.
1. Comparing $\triangle \text{ADB}$ and $\triangle \text{ABC}$:
1. $\angle \text{DAB} = \angle \text{CAB}$
(Common angle, which is $\angle \text{A}$)
2. $\angle \text{ADB} = 90^\circ$
(By Construction, BD $\perp$ AC)
3. $\angle \text{ABC} = 90^\circ$
(Given)
4. $\angle \text{ADB} = \angle \text{ABC}$
(From statements 2 and 3, both are $90^\circ$)
By AA similarity criterion (using statements 1 and 4):
$\triangle \text{ADB} \sim \triangle \text{ABC}$
2. Comparing $\triangle \text{BDC}$ and $\triangle \text{ABC}$:
5. $\angle \text{BCD} = \angle \text{ACB}$
(Common angle, which is $\angle \text{C}$)
6. $\angle \text{BDC} = 90^\circ$
(By Construction, BD $\perp$ AC)
7. $\angle \text{ABC} = 90^\circ$
(Given)
8. $\angle \text{BDC} = \angle \text{ABC}$
(From statements 6 and 7, both are $90^\circ$)
By AA similarity criterion (using statements 5 and 8):
$\triangle \text{BDC} \sim \triangle \text{ABC}$
3. Comparing $\triangle \text{ADB}$ and $\triangle \text{BDC}$:
Since $\triangle \text{ADB} \sim \triangle \text{ABC}$ (from part 1) and $\triangle \text{BDC} \sim \triangle \text{ABC}$ (from part 2), and the similarity relation is transitive (if A is similar to B and B is similar to C, then A is similar to C), it follows that $\triangle \text{ADB}$ is similar to $\triangle \text{BDC}$.
$\triangle \text{ADB} \sim \triangle \text{BDC}$
Hence, the theorem is proved. All three triangles are similar to each other.
Alternative Proof for $\triangle \text{ADB} \sim \triangle \text{BDC}$:
Using angle properties:
In $\triangle \text{ABC}$, $\angle \text{A} + \angle \text{ABC} + \angle \text{C} = 180^\circ$. Since $\angle \text{ABC} = 90^\circ$, $\angle \text{A} + 90^\circ + \angle \text{C} = 180^\circ$, so $\angle \text{A} + \angle \text{C} = 90^\circ$.
In $\triangle \text{ADB}$, $\angle \text{DAB} + \angle \text{ADB} + \angle \text{ABD} = 180^\circ$. Since $\angle \text{ADB} = 90^\circ$, $\angle \text{A} + 90^\circ + \angle \text{ABD} = 180^\circ$, so $\angle \text{A} + \angle \text{ABD} = 90^\circ$.
Comparing $\angle \text{A} + \angle \text{C} = 90^\circ$ and $\angle \text{A} + \angle \text{ABD} = 90^\circ$, we get $\angle \text{C} = \angle \text{ABD}$.
In $\triangle \text{BDC}$, $\angle \text{DBC} + \angle \text{BDC} + \angle \text{BCD} = 180^\circ$. Since $\angle \text{BDC} = 90^\circ$, $\angle \text{DBC} + 90^\circ + \angle \text{C} = 180^\circ$, so $\angle \text{DBC} + \angle \text{C} = 90^\circ$.
Comparing $\angle \text{A} + \angle \text{C} = 90^\circ$ and $\angle \text{DBC} + \angle \text{C} = 90^\circ$, we get $\angle \text{A} = \angle \text{DBC}$.
Now consider $\triangle \text{ADB}$ and $\triangle \text{BDC}$:
$\angle \text{A} = \angle \text{DBC}$
(Proved above)
$\angle \text{ABD} = \angle \text{C}$
(Proved above)
$\angle \text{ADB} = \angle \text{BDC}$
(Each is $90^\circ$)
By AAA (or AA) similarity criterion:
$\triangle \text{ADB} \sim \triangle \text{BDC}$
Hence Proved.
Geometric Mean Theorems Derived from Similarity
The similarity relationships proven above lead directly to two important geometric mean theorems involving the altitude to the hypotenuse and the legs of the right triangle.
Let AD = $x$, CD = $y$, AC = $c = x+y$, BD = $h$ (altitude), AB = $a$, BC = $b$ (legs). The conventional labelling for sides opposite vertices is BC=a, AC=b, AB=c if the right angle is at B. Let's use the side names AB, BC, AC to avoid confusion with the previous section's labels.
1. Altitude Geometric Mean Theorem:
This theorem relates the altitude to the segments of the hypotenuse.
From the similarity $\triangle \text{ADB} \sim \triangle \text{BDC}$ (proven in part 3):
The ratio of corresponding sides is equal. Matching vertices (A $\leftrightarrow$ B, D $\leftrightarrow$ D, B $\leftrightarrow$ C) based on the angles $\angle \text{A} = \angle \text{DBC}$, $\angle \text{ABD} = \angle \text{C}$, $\angle \text{ADB} = \angle \text{BDC}$ ($90^\circ$).
$\frac{\text{AD}}{\text{BD}} = \frac{\text{BD}}{\text{CD}} = \frac{\text{AB}}{\text{BC}}$
(Ratio of corresponding sides from $\triangle \text{ADB} \sim \triangle \text{BDC}$)
Considering the first two ratios:
$\frac{\text{AD}}{\text{BD}} = \frac{\text{BD}}{\text{CD}}$
Cross-multiplying gives:
$\mathbf{\text{BD}^2 = \text{AD} \times \text{CD}}$
This states that the square of the altitude to the hypotenuse is equal to the product of the lengths of the two segments into which the hypotenuse is divided by the altitude. Equivalently, the altitude BD is the geometric mean of AD and CD ($\text{BD} = \sqrt{\text{AD} \times \text{CD}}$).
2. Leg Geometric Mean Theorems:
These theorems relate each leg to the hypotenuse and the adjacent segment of the hypotenuse.
From the similarity $\triangle \text{ADB} \sim \triangle \text{ABC}$ (proven in part 1):
$\frac{\text{AD}}{\text{AB}} = \frac{\text{AB}}{\text{AC}} = \frac{\text{BD}}{\text{BC}}$
(Ratio of corresponding sides from $\triangle \text{ADB} \sim \triangle \text{ABC}$)
Considering the first two ratios:
$\frac{\text{AD}}{\text{AB}} = \frac{\text{AB}}{\text{AC}}$
$\mathbf{\text{AB}^2 = \text{AD} \times \text{AC}}$
This states that the square of leg AB is equal to the product of the length of the hypotenuse AC and the length of the segment of the hypotenuse adjacent to AB (which is AD).
From the similarity $\triangle \text{BDC} \sim \triangle \text{ABC}$ (proven in part 2):
$\frac{\text{CD}}{\text{BC}} = \frac{\text{BC}}{\text{AC}} = \frac{\text{BD}}{\text{AB}}$
(Ratio of corresponding sides from $\triangle \text{BDC} \sim \triangle \text{ABC}$)
Considering the first two ratios:
$\frac{\text{CD}}{\text{BC}} = \frac{\text{BC}}{\text{AC}}$
$\mathbf{\text{BC}^2 = \text{CD} \times \text{AC}}$
This states that the square of leg BC is equal to the product of the length of the hypotenuse AC and the length of the segment of the hypotenuse adjacent to BC (which is CD).
Note: Relationship to Pythagorean Theorem
Adding the results of the two Leg Geometric Mean Theorems gives:
$\text{AB}^2 + \text{BC}^2 = (\text{AD} \times \text{AC}) + (\text{CD} \times \text{AC})$
$= \text{AC} \times (\text{AD} + \text{CD})$
Since D lies on AC, $\text{AD} + \text{CD} = \text{AC}$ (Segment Addition Postulate).
$\text{AB}^2 + \text{BC}^2 = \text{AC} \times \text{AC} = \text{AC}^2$
This provides another elegant proof of the Pythagorean Theorem using the similarity of triangles formed by the altitude to the hypotenuse.
Similarity of Triangles: Areas and Applications - Applications of Similarity in Problem Solving
The concepts and theorems related to similarity, especially triangle similarity, are not confined to theoretical geometry. They have wide-ranging practical applications in various fields and are powerful tools for solving real-world problems as well as proving complex geometric results. Understanding how to identify and utilize similar triangles can simplify many measurement and calculation tasks.
1. Indirect Measurement
One of the most classic applications of similarity is in finding measurements (lengths or distances) indirectly, i.e., without physically measuring them. This is particularly useful for objects that are too large, too distant, or otherwise inaccessible for direct measurement.
- Finding Heights of Tall Objects: By comparing the length of the shadow cast by a tall object (like a building, tree, or tower) to the length of the shadow cast by an object of known height (like a person or a pole) at the same time of day, we can find the height of the tall object. The sun's rays are considered parallel, forming similar right-angled triangles.
In the diagram, $\triangle \text{ABC}$ (formed by the pole and its shadow) is similar to $\triangle \text{PQR}$ (formed by the tower and its shadow) by AA similarity (right angles at B and Q, and equal angle of elevation of the sun at C and R). Thus, $\text{AB}/\text{PQ} = \text{BC}/\text{QR}$.
- Using Mirrors: Similar triangles can be formed by using a mirror on the ground. If you stand such that you can see the top of a tall object reflected in the mirror, the angles of incidence and reflection are equal. This creates two similar triangles (often right triangles, assuming the observer and object are vertical) that allow you to find the height of the object using proportions.
- Using Sighting Instruments: Tools like a clinometer or sextant (in a simplified application) can be used in conjunction with measuring distances on the ground to create similar triangles, enabling the calculation of distances across obstacles or heights of objects.
Example 1. A vertical pole of length 6 m casts a shadow 4 m long on the ground. At the same time, a tower casts a shadow 28 m long. Find the height of the tower.
Answer:
Given: Height of pole (AB) = 6 m, Shadow of pole (BC) = 4 m. Shadow of tower (QR) = 28 m.
To Find: Height of tower (PQ).
Solution:
At the same time of day, the angle of elevation of the sun is the same for both the pole and the tower. This means the angle formed by the top of the object, the base, and the end of the shadow is the same for both.
Let $\theta$ be the angle of elevation of the sun. In $\triangle \text{ABC}$ (for the pole) and $\triangle \text{PQR}$ (for the tower):
$\angle \text{ACB} = \angle \text{PRQ} = \theta$
(Angle of elevation of the sun)
Assuming both the pole and the tower are vertical and the ground is horizontal, the angles at the base are right angles.
$\angle \text{ABC} = 90^\circ$
(Pole is vertical)
$\angle \text{PQR} = 90^\circ$
(Tower is vertical)
Therefore, $\angle \text{ABC} = \angle \text{PQR}$.
Consider $\triangle \text{ABC}$ and $\triangle \text{PQR}$. We have two pairs of equal corresponding angles ($\angle \text{C} = \angle \text{R}$ and $\angle \text{B} = \angle \text{Q}$).
By the AA (Angle-Angle) similarity criterion:
$\triangle \text{ABC} \sim \triangle \text{PQR}$
(AA Similarity)
Since the triangles are similar, the ratio of their corresponding sides is equal.
$\frac{\text{AB}}{\text{PQ}} = \frac{\text{BC}}{\text{QR}} = \frac{\text{AC}}{\text{PR}}$
(Ratio of corresponding sides of similar triangles)
We are interested in the ratio of heights and shadows:
$\frac{\text{AB}}{\text{PQ}} = \frac{\text{BC}}{\text{QR}}$
Substitute the given values:
$\frac{6 \text{ m}}{h \text{ m}} = \frac{4 \text{ m}}{28 \text{ m}}$
Simplify the ratio on the right side:
$\frac{6}{h} = \frac{\cancel{4}^1}{\cancel{28}_7}$
$\frac{6}{h} = \frac{1}{7}$
Cross-multiply to solve for $h$:
$6 \times 7 = h \times 1$
$42 = h$
$\mathbf{h = 42 \text{ m}}$
The height of the tower is 42 meters.
2. Scale Drawings and Models
Scale drawings like maps, blueprints, architectural plans, and engineering diagrams, as well as physical scale models, are direct applications of the concept of similarity. They represent larger objects or areas reduced by a constant scale factor, preserving the shape and relative proportions.
- Maps: A map is a scaled-down representation of a geographical area. The map scale (e.g., 1:100000 or 1 cm = 10 km) is the ratio of any distance on the map to the corresponding actual distance on the ground. This is the scale factor of similarity.
- Blueprints and Plans: Architects, engineers, and designers use blueprints to represent buildings, machines, or objects to scale. The dimensions on the plan are proportional to the actual dimensions, allowing for precise construction.
- Scale Models: Models of vehicles, buildings, molecules, etc., are built to scale, maintaining similarity to the actual objects for study, visualization, or demonstration.
3. Geometric Proofs and Properties
Triangle similarity criteria are powerful tools for proving other geometric theorems and establishing relationships between lengths of segments in various figures.
- Proving Proportional Relationships: Similarity is often the key to proving that certain line segments in a figure are proportional, as shown in the proof of BPT and its converse, and in the geometric mean theorems for right triangles.
- Establishing Angle Equality: While equal angles are a criterion for similarity, similarity can also be used indirectly in proofs to show that certain angles are equal (if they are corresponding angles of similar triangles).
- Circle Theorems: Many theorems involving intersecting chords, secants, and tangents in circles are proven by identifying similar triangles within the figure.
4. Relationships between Perimeters and Areas of Similar Figures
The definition of similarity leads to specific and predictable relationships between the perimeters and areas of similar figures. These relationships apply to any pair of similar polygons, not just triangles.
- Ratio of Perimeters: If two polygons are similar with a scale factor $k$ (ratio of corresponding sides), then the ratio of their perimeters is equal to the scale factor $k$. If $\triangle \text{ABC} \sim \triangle \text{PQR}$ with $\frac{\text{AB}}{\text{PQ}} = \frac{\text{BC}}{\text{QR}} = \frac{\text{AC}}{\text{PR}} = k$, then the perimeter of $\triangle \text{ABC}$ is $\text{AB} + \text{BC} + \text{AC}$, and the perimeter of $\triangle \text{PQR}$ is $\text{PQ} + \text{QR} + \text{PR}$.
$\frac{\text{Perimeter}(\triangle \text{ABC})}{\text{Perimeter}(\triangle \text{PQR})} = \frac{\text{AB} + \text{BC} + \text{AC}}{\text{PQ} + \text{QR} + \text{PR}}$
Since $\text{AB} = k \cdot \text{PQ}$, $\text{BC} = k \cdot \text{QR}$, $\text{AC} = k \cdot \text{PR}$:
$= \frac{k \cdot \text{PQ} + k \cdot \text{QR} + k \cdot \text{PR}}{\text{PQ} + \text{QR} + \text{PR}} = \frac{k(\text{PQ} + \text{QR} + \text{PR})}{\text{PQ} + \text{QR} + \text{PR}}$
The ratio of perimeters is equal to the scale factor.$\mathbf{\frac{\text{Perimeter}(\triangle \text{ABC})}{\text{Perimeter}(\triangle \text{PQR})} = k = \frac{\text{AB}}{\text{PQ}}}$
- Ratio of Areas: As proved in Section I1, the ratio of the areas of two similar triangles is the square of the scale factor.
The ratio of areas is the square of the ratio of corresponding sides.
$\mathbf{\frac{\text{Area}(\triangle \text{ABC})}{\text{Area}(\triangle \text{PQR})} = k^2 = \left(\frac{\text{AB}}{\text{PQ}}\right)^2}$
These relationships allow us to find perimeters or areas of similar figures if we know the scale factor and the perimeter/area of one figure.
In summary, similarity is a powerful concept that provides a framework for understanding scaled shapes, solving indirect measurement problems, and proving various geometric theorems by establishing proportional relationships between lengths and equal relationships between angles.