# How Right Triangles Demonstrate Similarity

Right triangles, with their perfect $$90^\circ$$ angles, hold a unique position in the world of triangles. They serve as the backbone for many mathematical concepts and real-world applications. One of the captivating aspects of right triangles is their inherent property of similarity under specific conditions. Ever wondered how the shadow of a tall building can reveal its height? Or how surveyors determine the distance across a river? The answers lie in the similarity of right triangles. Let's delve into this engaging topic!

## Step-by-step Guide: Right Triangles and Similarity

Understanding Right Triangles:
A right triangle is defined by one angle measuring $$90^\circ$$. The side opposite this angle is the hypotenuse, the longest side, while the other two sides are the legs.

Altitude and the Two Formed Triangles:
Dropping an altitude from the right angle to the hypotenuse creates two new right triangles. The intriguing part? These two triangles are similar not only to each other but also to the original right triangle!

Criteria for Similarity:

• AA (Angle-Angle) Criterion: If two angles of one triangle are congruent to two angles of another triangle, then the triangles are similar. Since right triangles formed by an altitude share a common angle and both have a $$90^\circ$$ angle, they’re similar by the AA criterion.

Properties of Similar Right Triangles:
The ratios of the corresponding sides of similar right triangles are equal. In the case of an altitude dropped onto the hypotenuse:

• The altitude is geometrically mean between the two segments it divides the hypotenuse into.
• Each leg of the large triangle is the geometric mean of the hypotenuse and the segment of the hypotenuse adjacent to that leg.

### Examples

Example 1:
In right triangle $$ABC$$, where $$AC$$ is the hypotenuse and $$BD$$ is an altitude drawn to $$AC$$, if $$AB = 6 \text{ cm}$$, $$BC = 8 \text{ cm}$$, find the lengths of $$AD$$, $$CD$$, and $$BD$$.

Solution:
Using the property of geometric means:
$$AB^2 = AD \times AC$$
$$AD = \frac{AB^2}{AC} = \frac{36 \text{ cm^2}}{10 \text{ cm}} = 3.6 \text{ cm}$$ Similarly,
$$BC^2 = CD \times AC$$
$$CD = \frac{BC^2}{AC} = \frac{64 \text{ cm^2}}{10 \text{ cm}} = 6.4 \text{ cm}$$ Now, using the Pythagorean theorem for triangle $$ABD$$:
$$BD^2 = AB^2 – AD^2 = 7.2 \text{ cm^2}$$
$$BD = 2.68 \text{ cm}$$ (approx)

Example 2:
In a right triangle $$XYZ$$ with $$XZ$$ as the hypotenuse, an altitude $$WM$$ is drawn. If $$XM = 5 \text{ cm}$$ and $$YZ = 15 \text{ cm}$$, determine $$YM$$.

Solution:
Using the properties of geometric means:
$$XY^2 = XM \times XZ$$
$$XY^2 = 5 \times (5 + YM)$$ Also, using the Pythagorean theorem:
$$XY^2 + YZ^2 = XZ^2$$
$$XY^2 + 225 \text{ cm^2} = (5 + YM)^2$$ Equating the two expressions for $$XY^2$$, we can determine $$YM = 20 \text{ cm}$$.

### Practice Questions:

1. In right triangle $$PQR$$, if $$PR$$ is the hypotenuse and $$PS$$ is an altitude dividing $$PR$$ into segments of $$3 \text{ cm}$$ and $$9 \text{ cm}$$, find the length of $$PS$$.
2. Given a right triangle $$LMN$$ with $$LN$$ as the hypotenuse of length $$17 \text{ cm}$$ and altitude $$LO$$ dividing it into segments of $$8 \text{ cm}$$ and $$15 \text{ cm}$$, find the lengths of $$LO$$ and $$MO$$.

1. $$PS = 6 \text{ cm}$$
2. $$LO = 7.2 \text{ cm}$$ and $$MO = 9.6 \text{ cm}$$

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