Menu Top
Complete Course of Mathematics
Topic 1: Numbers & Numerical Applications Topic 2: Algebra Topic 3: Quantitative Aptitude
Topic 4: Geometry Topic 5: Construction Topic 6: Coordinate Geometry
Topic 7: Mensuration Topic 8: Trigonometry Topic 9: Sets, Relations & Functions
Topic 10: Calculus Topic 11: Mathematical Reasoning Topic 12: Vectors & Three-Dimensional Geometry
Topic 13: Linear Programming Topic 14: Index Numbers & Time-Based Data Topic 15: Financial Mathematics
Topic 16: Statistics & Probability


Content On This Page
Pythagoras Property of a Right Angled Triangle: Statement and Proof Converse of the Pythagorean Theorem Applications of the Pythagorean Theorem


Pythagorean Theorem



Pythagorean Theorem - Statement and Proof

The Pythagorean Theorem is undoubtedly one of the most famous and fundamental theorems in all of mathematics, particularly in geometry. It reveals a crucial relationship that exists exclusively in right-angled triangles. Named after the ancient Greek mathematician Pythagoras, who is traditionally credited with its first proof (though the relationship was known in other ancient civilizations like Babylon and India centuries earlier), this theorem is indispensable for solving problems involving lengths in right triangles.


Theorem Statement

The Pythagorean theorem states the following property for any right-angled triangle:

Theorem: In a right-angled triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides (which are called the legs or catheti).

Consider a triangle $\triangle \text{ABC}$ that is right-angled at vertex B. Let the lengths of the sides opposite to vertices A, B, and C be denoted by the corresponding lowercase letters $a, b$, and $c$ respectively. In this case, the side opposite the right angle (at B) is AC, which is the hypotenuse. The sides opposite the other two angles, AB and BC, are the legs.

The theorem can be expressed as an equation:

$(\text{length of hypotenuse})^2 = (\text{length of one leg})^2 + (\text{length of other leg})^2$

Using the standard notation for $\triangle \text{ABC}$ right-angled at B:

$\text{AC}^2 = \text{AB}^2 + \text{BC}^2$

Or, using side lengths $b$ (hypotenuse opposite B), $c$ (opposite C), and $a$ (opposite A) where the right angle is at A:

$a^2 = b^2 + c^2$

(If the right angle is at A)

Let's stick to the notation where the right angle is at B, and the sides are AB, BC, and AC.

Right-angled triangle ABC, with the right angle at vertex B. The hypotenuse AC is labelled c, leg BC is labelled a, and leg AB is labelled b.

In this figure, $\triangle \text{ABC}$ is right-angled at B. The side opposite the right angle is AC (hypotenuse). The sides AB and BC are the legs. The theorem states $\text{AC}^2 = \text{AB}^2 + \text{BC}^2$. If we label the lengths $AC=c$, $BC=a$, $AB=b$, then the theorem is $c^2 = a^2 + b^2$. However, the image provided labels the sides opposite vertices A, B, C as a, b, c respectively. If the right angle is at B, then the side opposite B is b (AC, the hypotenuse), opposite A is a (BC, a leg), and opposite C is c (AB, a leg). So, for the given image notation, the theorem is $b^2 = a^2 + c^2$. Let's clarify and use the notation from the proof description later (AB, BC, AC as side names).

Let's re-state using vertex names:

In $\triangle \text{ABC}$, if $\angle \text{B}$ is the right angle ($90^\circ$), then $(\text{length of AC})^2 = (\text{length of AB})^2 + (\text{length of BC})^2$.


Proof using Similar Triangles

There are numerous proofs of the Pythagorean theorem. One elegant method involves using the properties of similar triangles.

Given: A right-angled triangle $\triangle \text{ABC}$, right-angled at vertex B.

To Prove: $\text{AC}^2 = \text{AB}^2 + \text{BC}^2$

Construction: Draw a perpendicular line segment from the vertex B to the hypotenuse AC. Let this perpendicular intersect AC at point D.

Right-angled triangle ABC, right-angled at B. A perpendicular line segment BD is drawn from B to the hypotenuse AC, intersecting AC at point D.

This construction creates two smaller right-angled triangles, $\triangle \text{ADB}$ and $\triangle \text{BDC}$, and the original triangle $\triangle \text{ABC}$. All three of these triangles are similar to each other.

Proof:

Consider $\triangle \text{ADB}$ and $\triangle \text{ABC}$.

1. $\angle \text{DAB} = \angle \text{CAB}$

(Common angle - it is the angle $\angle \text{A}$)

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 $90^\circ$)

Therefore, by the Angle-Angle (AA) similarity criterion, if two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar.

$\triangle \text{ADB} \sim \triangle \text{ABC}$

(AA Similarity - from statements 1 and 4)

When two triangles are similar, the ratio of their corresponding sides is equal.

The side opposite $\angle \text{ADB}$ in $\triangle \text{ADB}$ is AB. The side opposite $\angle \text{ABC}$ in $\triangle \text{ABC}$ is AC.

The side opposite $\angle \text{DAB}$ in $\triangle \text{ADB}$ is BD. The side opposite $\angle \text{CAB}$ in $\triangle \text{ABC}$ is BC.

The side opposite $\angle \text{ABD}$ in $\triangle \text{ADB}$ is AD. The side opposite $\angle \text{ACB}$ in $\triangle \text{ABC}$ is AB.

So, from $\triangle \text{ADB} \sim \triangle \text{ABC}$, we have the proportion of corresponding sides:

$\frac{\text{AD}}{\text{AB}} = \frac{\text{AB}}{\text{AC}}$

(Ratio of corresponding sides AD/AB = AB/AC)

Cross-multiplying gives:

$\text{AD} \times \text{AC} = \text{AB}^2$

... (i)


Now, consider $\triangle \text{BDC}$ and $\triangle \text{ABC}$.

5. $\angle \text{BCD} = \angle \text{ACB}$

(Common angle - it is the angle $\angle \text{C}$)

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 $90^\circ$)

Therefore, by the Angle-Angle (AA) similarity criterion:

$\triangle \text{BDC} \sim \triangle \text{ABC}$

(AA Similarity - from statements 5 and 8)

From $\triangle \text{BDC} \sim \triangle \text{ABC}$, we have the proportion of corresponding sides:

$\frac{\text{CD}}{\text{BC}} = \frac{\text{BC}}{\text{AC}}$

(Ratio of corresponding sides CD/BC = BC/AC)

Cross-multiplying gives:

$\text{CD} \times \text{AC} = \text{BC}^2$

... (ii)


Now, add equation (i) and equation (ii):

$\text{AB}^2 + \text{BC}^2 = (\text{AD} \times \text{AC}) + (\text{CD} \times \text{AC})$

(Adding corresponding sides of equations i and ii)

Factor out AC on the right side:

$\text{AB}^2 + \text{BC}^2 = \text{AC} \times (\text{AD} + \text{CD})$

(Distributive property)

From the figure, point D lies on the line segment AC. By the Segment Addition Postulate, the length of AC is the sum of the lengths of AD and CD.

$\text{AD} + \text{CD} = \text{AC}$

(Segment Addition Postulate)

Substitute $(\text{AD} + \text{CD})$ with $\text{AC}$ in the equation for $\text{AB}^2 + \text{BC}^2$:

$\text{AB}^2 + \text{BC}^2 = \text{AC} \times \text{AC}$

(Substitution)

$\text{AB}^2 + \text{BC}^2 = \text{AC}^2$

Thus, the square of the hypotenuse (AC) is equal to the sum of the squares of the other two sides (AB and BC).

Hence Proved.

$\mathbf{\text{AC}^2 = \text{AB}^2 + \text{BC}^2}$



Pythagorean Theorem - Converse of the Pythagorean Theorem

The Pythagorean Theorem provides a property that is true for all right-angled triangles. The converse of this theorem is equally important and useful. It allows us to determine if a triangle is a right-angled triangle just by knowing the lengths of its three sides. This is a powerful tool for classifying triangles based on their side lengths.


Statement of the Converse of the Pythagorean Theorem

The converse of the Pythagorean theorem states:

Theorem: If in a triangle, the square of the length of one side is equal to the sum of the squares of the lengths of the other two sides, then the angle opposite the first side is a right angle ($90^\circ$), and the triangle is a right-angled triangle.

Consider a triangle $\triangle \text{ABC}$ with side lengths $a, b,$ and $c$, where $c$ is the length of the longest side. If you find that the relationship $c^2 = a^2 + b^2$ holds true for these side lengths, then the angle opposite the side with length $c$ (which is angle C, $\angle \text{C}$) must be a right angle ($90^\circ$).

Alternatively, using vertex names for the sides:

If in $\triangle \text{ABC}$, the square of the length of side AC is equal to the sum of the squares of the lengths of sides AB and BC (i.e., $\text{AC}^2 = \text{AB}^2 + \text{BC}^2$), then the angle opposite side AC, which is $\angle \text{B}$, is a right angle ($90^\circ$).


Proof of the Converse of the Pythagorean Theorem

We can prove the converse of the Pythagorean theorem by constructing a right-angled triangle with the given side lengths and showing that it is congruent to the original triangle.

Given: A triangle $\triangle \text{ABC}$ such that the square of the length of side AC is equal to the sum of the squares of the lengths of sides AB and BC ($\text{AC}^2 = \text{AB}^2 + \text{BC}^2$).

Two triangles shown side by side: Triangle ABC with sides AB, BC, AC. Triangle PQR right-angled at Q with PQ = AB and QR = BC.

To Prove: $\triangle \text{ABC}$ is a right-angled triangle, specifically right-angled at vertex B (i.e., $\text{m}\angle \text{B} = 90^\circ$).

Construction: Construct a new triangle $\triangle \text{PQR}$ that is right-angled at vertex Q. Ensure that the lengths of the legs of $\triangle \text{PQR}$ are equal to the lengths of two sides of $\triangle \text{ABC}$, specifically construct it such that $\text{PQ} = \text{AB}$ and $\text{QR} = \text{BC}$.

Proof:

Consider the constructed triangle $\triangle \text{PQR}$. It is right-angled at Q.

(By Construction)

We can apply the Pythagorean theorem to $\triangle \text{PQR}$.

$\text{PR}^2 = \text{PQ}^2 + \text{QR}^2$

(Pythagorean Theorem applied to $\triangle \text{PQR}$)

By construction, $\text{PQ} = \text{AB}$ and $\text{QR} = \text{BC}$. Substitute these equalities into the equation for $\text{PR}^2$.

$\text{PR}^2 = \text{AB}^2 + \text{BC}^2$

... (i)

We are given the relationship between the sides of $\triangle \text{ABC}$.

$\text{AC}^2 = \text{AB}^2 + \text{BC}^2$

(Given)

Comparing equation (i) and the given equation, we see that both $\text{PR}^2$ and $\text{AC}^2$ are equal to the same quantity ($\text{AB}^2 + \text{BC}^2$).

$\text{PR}^2 = \text{AC}^2$

(From equation (i) and Given - Common Notion 1)

Take the square root of both sides. Since side lengths must be positive:

$\text{PR} = \text{AC}$

(Taking positive square root)


Now, consider $\triangle \text{ABC}$ and $\triangle \text{PQR}$. We have the following equalities between their sides:

$\text{AB} = \text{PQ}$

(By Construction)

$\text{BC} = \text{QR}$

(By Construction)

$\text{AC} = \text{PR}$

(Proved above)

Since all three sides of $\triangle \text{ABC}$ are equal in length to the corresponding three sides of $\triangle \text{PQR}$, the two triangles are congruent by the Side-Side-Side (SSS) congruence criterion.

$\triangle \text{ABC} \cong \triangle \text{PQR}$

(SSS Congruence Criteria)

When two triangles are congruent, their corresponding parts (sides and angles) are equal. Thus, the corresponding angles are equal.

The angle opposite side AC in $\triangle \text{ABC}$ is $\angle \text{B}$. The angle opposite side PR in $\triangle \text{PQR}$ is $\angle \text{Q}$.

$\angle \text{B} = \angle \text{Q}$

(Corresponding Parts of Congruent Triangles - CPCT)

By construction, $\triangle \text{PQR}$ is a right-angled triangle at Q, so the measure of $\angle \text{Q}$ is $90^\circ$.

$\text{m}\angle \text{Q} = 90^\circ$

(By Construction)

Therefore, substituting the value of $\text{m}\angle \text{Q}$:

$\mathbf{\text{m}\angle \text{B} = 90^\circ}$

This proves that the angle at vertex B in $\triangle \text{ABC}$ is a right angle ($90^\circ$). Consequently, $\triangle \text{ABC}$ is a right-angled triangle.

Hence Proved.


Application of the Converse

The converse of the Pythagorean theorem is widely used to check if a triangle is right-angled, given its side lengths. For example, a triangle with side lengths 3, 4, and 5 units is a right-angled triangle because $3^2 + 4^2 = 9 + 16 = 25 = 5^2$. The angle opposite the side of length 5 is the right angle.


Example 1. The sides of a triangle are 6 cm, 8 cm, and 10 cm. Determine if it is a right-angled triangle.

Answer:

Given: Side lengths of a triangle are 6 cm, 8 cm, and 10 cm.

To Determine: If the triangle is right-angled.

Solution:

To check if a triangle is right-angled using the converse of the Pythagorean theorem, we need to see if the square of the longest side is equal to the sum of the squares of the other two sides.

The longest side is 10 cm.

Let's square the lengths of all three sides:

$(6 \text{ cm})^2 = 36 \text{ cm}^2$

$(8 \text{ cm})^2 = 64 \text{ cm}^2$

$(10 \text{ cm})^2 = 100 \text{ cm}^2$

Now, check if the sum of the squares of the two shorter sides equals the square of the longest side:

Sum of squares of shorter sides $= 36 \text{ cm}^2 + 64 \text{ cm}^2$

$= 100 \text{ cm}^2$

The square of the longest side is $100 \text{ cm}^2$.

Since the sum of the squares of the two shorter sides is equal to the square of the longest side ($36 + 64 = 100$), the condition of the converse of the Pythagorean theorem is met.

Therefore, the triangle with side lengths 6 cm, 8 cm, and 10 cm is a right-angled triangle. The right angle is opposite the side of length 10 cm.



Pythagorean Theorem - Applications of the Pythagorean Theorem

The Pythagorean Theorem is not just a fundamental result in geometry; it is also one of the most widely applicable mathematical theorems across numerous fields. Its power lies in providing a direct relationship between lengths in right-angled triangles, which appear frequently in various contexts, whether physical or abstract.


Calculating Side Lengths in Right-Angled Triangles

The most immediate application of the Pythagorean theorem is to find the length of an unknown side of a right-angled triangle when the lengths of the other two sides are known. Let $a$ and $b$ be the lengths of the legs, and $c$ be the length of the hypotenuse in a right-angled triangle.


Coordinate Geometry (Distance Formula)

One of the most significant applications of the Pythagorean theorem is in establishing the distance formula in a Cartesian coordinate system (analytic geometry). The distance between any two points in a 2D plane can be found by considering them as vertices of a right-angled triangle.

Let $P_1 = (x_1, y_1)$ and $P_2 = (x_2, y_2)$ be two points in the Cartesian plane. We can form a right-angled triangle by drawing a horizontal line through one point and a vertical line through the other, meeting at a third point, say $P_3 = (x_2, y_1)$. The horizontal leg of this triangle has length $|x_2 - x_1|$, and the vertical leg has length $|y_2 - y_1|$. The distance between $P_1$ and $P_2$ is the hypotenuse of this right triangle.

Points (x1, y1) and (x2, y2) connected by a line segment, forming the hypotenuse of a right triangle with horizontal leg (x2-x1) and vertical leg (y2-y1)

Applying the Pythagorean theorem, if $d$ is the distance between $(x_1, y_1)$ and $(x_2, y_2)$:

$d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2$

Taking the square root gives the distance formula:

$\mathbf{d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}}$

This formula is widely used in coordinate geometry and other areas of mathematics and science.


Architecture and Construction

The Pythagorean theorem is invaluable in architecture and construction for ensuring square corners and checking the alignment of structural elements. Builders use it frequently, especially the converse theorem and Pythagorean triples.

For example, to verify that a wall is exactly perpendicular to the floor (forms a right angle), a builder can measure a distance up the wall (say 4 units, like 4 feet or 40 cm) and a distance out along the floor from the base of the wall (say 3 units). If the diagonal distance measured between these two points is exactly 5 units, then the corner is indeed a right angle ($3^2 + 4^2 = 9 + 16 = 25 = 5^2$). This simple check using the (3, 4, 5) Pythagorean triple ensures structural integrity where right angles are required.


Navigation

The theorem is used in navigation to calculate the shortest distance between two points or to determine a displacement. If a ship or aircraft travels a certain distance east and then a certain distance north, the direct distance from the starting point to the end point is the hypotenuse of a right triangle formed by the eastward and northward legs.


Applications in Various Geometry Problems


Pythagorean Triples

A set of three positive integers, traditionally denoted as $(a, b, c)$, is called a Pythagorean triple if they satisfy the equation $a^2 + b^2 = c^2$. These triples represent the side lengths of right-angled triangles where all sides have integer lengths. Pythagorean triples are useful in problems and constructions as they simplify calculations.

If $(a, b, c)$ is a Pythagorean triple, then $(ka, kb, kc)$ is also a Pythagorean triple for any positive integer $k$. This means any multiple of a Pythagorean triple also forms the sides of a right-angled triangle. A triple is considered primitive if the three integers $a, b,$ and $c$ have no common divisor greater than 1.

Some common primitive Pythagorean triples include:

Recognizing these triples can sometimes provide a shortcut in solving problems involving right triangles.


Example 1. A ladder 10 m long reaches a window 8 m above the ground. Assuming the wall is perpendicular to the ground, find the distance of the foot of the ladder from the base of the wall.

Answer:

Given: Length of ladder (hypotenuse, $c$) = 10 m. Height of window from ground (one leg, $a$) = 8 m.

To Find: Distance of the foot of the ladder from the wall (other leg, $b$).

Solution:

The situation forms a right-angled triangle where the ladder is the hypotenuse, the height up the wall is one leg, and the distance from the wall on the ground is the other leg. The right angle is formed between the wall and the ground.

Diagram showing a vertical wall, horizontal ground, and a ladder leaning against the wall. This forms a right triangle. Ladder length is 10m, height on wall is 8m. The unknown distance on the ground is labelled.

Using the Pythagorean theorem, $a^2 + b^2 = c^2$:

$8^2 + b^2 = 10^2$

$64 + b^2 = 100$

Subtract 64 from both sides:

$b^2 = 100 - 64$

$b^2 = 36$

Take the square root of both sides. Since $b$ represents a length, it must be positive.

$b = \sqrt{36}$

$\mathbf{b = 6 \text{ m}}$

Therefore, the distance of the foot of the ladder from the base of the wall is 6 meters.

Note: This is a $(6, 8, 10)$ triangle, which is a multiple of the primitive Pythagorean triple $(3, 4, 5)$ since $6 = 2 \times 3$, $8 = 2 \times 4$, and $10 = 2 \times 5$.


Example 2. Check if a triangle with sides 7 cm, 24 cm, and 25 cm is a right-angled triangle.

Answer:

Given: Side lengths of a triangle are 7 cm, 24 cm, and 25 cm.

To Determine: If the triangle is right-angled.

Solution:

We use the converse of the Pythagorean theorem. If the square of the longest side is equal to the sum of the squares of the other two sides, the triangle is right-angled.

The longest side is 25 cm.

Let's square the lengths of the two shorter sides and find their sum:

$7^2 + 24^2$

$= 49 + 576$

$= 625$

Now, let's square the length of the longest side:

$25^2 = 625$

We observe that the sum of the squares of the two shorter sides ($7^2 + 24^2 = 625$) is equal to the square of the longest side ($25^2 = 625$).

$7^2 + 24^2 = 25^2$

Since $a^2 + b^2 = c^2$ holds for the side lengths, by the converse of the Pythagorean theorem, the triangle is right-angled.

Therefore, the triangle with side lengths 7 cm, 24 cm, and 25 cm is a right-angled triangle. The angle opposite the side of length 25 cm is the right angle.

Note: $(7, 24, 25)$ is a primitive Pythagorean triple.