Important Properties of Similarity
Two polygons of the same number of sides are similar if their corresponding angles are equal and their corresponding sides are in the same ratio (proportional). For triangles, the most important properties and criteria are:
- AA (Angle-Angle) Similarity: If two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar.
- SAS (Side-Angle-Side) Similarity: If one angle of a triangle is equal to one angle of another triangle and the sides including these angles are proportional, then the two triangles are similar.
- SSS (Side-Side-Side) Similarity: If in two triangles, the sides of one triangle are proportional to the sides of the other triangle, then their corresponding angles are equal and hence the two triangles are similar.
- Ratio of Areas Theorem: The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
Practice Problems for 10th Grade
Problem 1
In $\Delta ABC$ and $\Delta DEF$, it is given that $\angle A = 50^\circ$, $\angle B = 70^\circ$, $\angle D = 50^\circ$, and $\angle F = 60^\circ$. Are the two triangles similar? Why or why not?
Solution:
In $\Delta ABC$, the sum of the interior angles is $180^\circ$. Therefore, we can find $\angle C$:
$\angle C = 180^\circ - (50^\circ + 70^\circ) = 180^\circ - 120^\circ = 60^\circ$.Now, compare $\Delta ABC$ and $\Delta DEF$:
$\angle A = \angle D = 50^\circ$
$\angle C = \angle F = 60^\circ$Since two corresponding angles are equal, $\Delta ABC \sim \Delta DEF$ by the AA Similarity criterion.
Problem 2
Given that $\Delta ABC \sim \Delta PQR$. If $AB = 4 \text{ cm}$, $BC = 6 \text{ cm}$, $AC = 5 \text{ cm}$, and $PQ = 8 \text{ cm}$, find the lengths of the remaining sides of $\Delta PQR$.
Solution:
Since the triangles are similar, the ratio of their corresponding sides is equal:
$$ \frac{AB}{PQ} = \frac{BC}{QR} = \frac{AC}{PR} $$
Substitute the given values:
$$ \frac{4}{8} = \frac{6}{QR} = \frac{5}{PR} $$The scale factor is $\frac{4}{8} = \frac{1}{2}$. We can solve for $QR$ and $PR$:
$QR = 6 \times 2 = 12 \text{ cm}$
$PR = 5 \times 2 = 10 \text{ cm}$
Problem 3
The ratio of the corresponding sides of two similar triangles is $3:5$. If the area of the smaller triangle is $45 \text{ cm}^2$, find the area of the larger triangle.
Solution:
By the Ratio of Areas theorem, the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
$$ \frac{\text{Area of smaller triangle}}{\text{Area of larger triangle}} = \left( \frac{3}{5} \right)^2 = \frac{9}{25} $$
Substitute the area of the smaller triangle:
$$ \frac{45}{\text{Area of larger triangle}} = \frac{9}{25} $$Cross-multiplying yields:
$9 \times (\text{Area of larger triangle}) = 45 \times 25$
$\text{Area of larger triangle} = \frac{1125}{9} = 125 \text{ cm}^2$
Problem 4
A vertical pole of length $6 \text{ m}$ casts a shadow $4 \text{ m}$ long on the ground. At the same time, a nearby tower casts a shadow $28 \text{ m}$ long. Find the height of the tower.
Solution:
At the same time of day, the sun's rays fall at the same angle of elevation. Thus, the right-angled triangle formed by the pole and its shadow is similar to the right-angled triangle formed by the tower and its shadow (by AA similarity).
Let the height of the tower be $h$. Using the property of proportional sides in similar triangles:
$$ \frac{\text{Height of pole}}{\text{Shadow of pole}} = \frac{\text{Height of tower}}{\text{Shadow of tower}} $$
$$ \frac{6}{4} = \frac{h}{28} $$
Solving for $h$:
$h = \frac{6 \times 28}{4} = 6 \times 7 = 42 \text{ m}$.
The height of the tower is $42 \text{ m}$.
Problem 5
In a right-angled triangle $ABC$, right-angled at $B$, a perpendicular $BD$ is drawn from $B$ to the hypotenuse $AC$. If $AD = 4 \text{ cm}$ and $CD = 9 \text{ cm}$, find the length of the altitude $BD$.
Solution:
A fundamental property of right triangles states that the altitude drawn to the hypotenuse divides the triangle into two smaller triangles that are similar to each other and to the original triangle.
Therefore, $\Delta ADB \sim \Delta BDC$.
Because corresponding sides of similar triangles are proportional:
$$ \frac{AD}{BD} = \frac{BD}{CD} $$Cross-multiplying gives the geometric mean theorem:
$BD^2 = AD \times CD$Substitute the given values:
$BD^2 = 4 \times 9 = 36$
$BD = \sqrt{36} = 6 \text{ cm}$.