Pythagoras Theorem: Solved Examples, Proof, Formula- Embibe
• Written By SHWETHA B.R
• Written By SHWETHA B.R

# Pythagoras Theorem: Check Formula Pythagoras Theorem: Pythagoras Theorem says that the square of the hypotenuse or longest side of a triangle is equal to the sum of squares of the other two sides of the triangle. The opposite side of the right-angle in a right-angled triangle is the hypotenuse. Pythagoras Theorem only applies to right-angled triangles. The theorem outlines the relationship between the base, perpendicular, and hypotenuse of a right-angled triangle. The converse of Pythagoras Theorem is also used for determining whether or not a triangle is right-angled.

Pythagoras Theorem is important since it is used for finding out the factual concept of whether the sum of other sides of a triangle is actually getting equal to the hypotenuse or not. In a similar manner, lengths of right-angled triangles can also be determined. There are a lot of applications of the Pythagoras Theorem and one of many is its usage in security cameras for face recognition. Continue reading to know more.

## Pythagoras Theorem Formula: Overview

Pythagoras theorem is a basic relation in Euclidean geometry. It is a study of plane and solid figures and the five most important theorem under Euclidean geometry are the sum of the angles in a triangle is 180 degrees, the Bridge of Asses, the fundamental theorem of similarity, the Pythagorean theorem, and the invariance of angles subtended by a chord in a circle.

Pythagoras’ Theorem talks about, the square of the hypotenuse equals the sum of the squares of the other two sides. Look at the triangle ABC below, where  BC2 = AB2 + AC2​​. The base is AB, the altitude (height) is AC, and the hypotenuse is BC. Thus, the formula goes like this:

$${\left( {{\rm{Hypotenuse}}} \right)^{\rm{2}}}{\rm{ = }}\,{\left( {{\rm{Base}}} \right)^{\rm{2}}}{\rm{ + }}{\left( {{\rm{Perpendicular}}} \right)^{\rm{2}}}$$

$${\left( {{\rm{Hypotenuse}}} \right)^{\rm{2}}}{\rm{ = }}\,{\left( {{\rm{Base}}} \right)^{\rm{2}}}{\rm{ + }}{\left( {{\rm{Perpendicular}}} \right)^{\rm{2}}}$$

Where, $${{b=}}$$ Base, $${{a=}}$$ Perpendicular, $${{c=}}$$ Hypotenuse

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### Basic Terms of Pythagoras Theorem

The basic terms used in the Pythagoras Theorem are base, perpendicular (altitude), and hypotenuse. Pythagoras Theorem in Hindi is पाइथागोरस प्रमेय. The two sides of a right-angled triangle that forms the right angle are known as the base and perpendicular, and the side which is opposite to the right angle is known as the hypotenuse.

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#### Base

Base is the side in a right-angled triangle which is opposite to the angle $$P.$$

#### Perpendicular (Altitude or Height)

Perpendicular is the side which is opposite to the angle $$R.$$

#### Hypotenuse

Hypotenuse is the side which is opposite to the right angle i.e., angle $$Q.$$

### Pythagorean Triplets

Three positive integers $${{a, b}}$$ and $${{c}}$$ are called Pythagorean triplets if $${c^2} = {a^2} + {b^2},$$ the triple is commonly written $${{(a, b, c)}}$$ i.e., in increasing order of $${{a, b}}$$ and $${{c}}{{.}}$$
Examples for Pythagorean triplets are $${{(3}},{{4}},{{5)}},{{(5}},{{12}},{{13)}},{{(7}},{{24}},{{25)}},{{(8}},{{15}},{{17)}},{{(9}},{{40}},{{41)}}{{.}}$$

### Proving Pythagoras Theorem

The proof of the Pythagoras Theorem is very interesting. It involves the concept of similarity of the triangle.

In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Given: A right-angled triangle $$PQR,$$ right angled at $${{Q}}{{.}}$$

To prove: $$P{R^2} = P{Q^2} + Q{R^2}$$

Construction: Draw a perpendicular line $${{QD}}$$ meeting $${{PR}}$$ at $${{D}}{{.}}$$

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Proof: we know that $$\Delta RDQ \sim \Delta RQP$$

So, $$\frac{{RD}}{{QR}} = \frac{{QR}}{{PR}}$$ (Corresponding sides of similar triangles)

$$\Rightarrow Q{R^2} = RD \times PR$$ —$$\left( {\rm{i}} \right)$$

Also, $$\Delta QDP \sim \Delta RQP$$

So, $$\frac{{PD}}{{PQ}} = \frac{{PQ}}{{PR}}$$ (Corresponding sides of similar triangles)

$$\Rightarrow P{Q^2} = PD \times RP$$ —$$\left( {{\rm{ii}}} \right)$$

Adding the equation $$\left( {\rm{i}} \right)$$ and $$\left( {{\rm{ii}}} \right)$$ we get,

$${{Q}}{{{R}}^{{2}}}{{ + P}}{{{Q}}^{{2}}}{{=RD \times PR + PD \times PR}}$$

$$\Rightarrow Q{R^2} + P{Q^2} = PR(RD + PD)$$

From the figure, $${{RD}} + {{PD}} = {{PR}}$$

From the figure, $${{P}}{{{R}}^2} = {{P}}{{{Q}}^2} + {{Q}}{{{R}}^2}$$

Hence, the Pythagoras Theorem is proved.

### Converse of Pythagoras Theorem and Its Proof

In a triangle, if the square of one side is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle.

Given: In $$\Delta XYZ,X{Y^2} + Y{Z^2} = X{Z^2}$$

$$\angle {\text{XYZ}} = 90^\circ$$

Construction: – A triangle $$PQR$$ is constructed such that

$$PQ = XY,\,QR = YZ,\,\angle PQR = 90^\circ$$

Proof:In $$\Delta PQR,\,\angle Q = 90^\circ$$

$$P{R^2} = P{Q^2} + Q{R^2}$$ [Pythagoras Theorem]

Or $$P{R^2} = X{Y^2} + Y{Z^2}$$ —$$\left( {\rm{i}} \right)$$

$$\left[ {PQ = XY,{\mkern 1mu} QR = YZ} \right]$$

Therefore, $$X{Z^2} = P{R^2}$$ [From equation $$\left( {\rm{i}} \right)$$ and $$\left( {{\rm{ii}}} \right)$$

Or $$XZ = PR$$

Or $$\Delta XYZ \cong \Delta PQR$$ [$$SSS$$ congruency rule]

Therefore $$\angle Y = \angle Q = 90^\circ$$ [CPCT]

Hence, $$\angle XYZ = 90^\circ$$

The converse of Pythagoras Theorem is proved.

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### Applications of Pythagoras Theorem

Some of the applications of the Pythagoras Theorem are

1. Pythagoras Theorem is commonly used to find the lengths of sides of a right-angled triangle.
2. The Theorem is used to find the length of the diagonal of a rectangle, square, etc.
3. Pythagoras Theorem is used in trigonometry to find the trigonometric ratios like $$\sin ,\,\cos ,\,\tan ,\,\cos ec,\,\sec ,\,\cot .$$.
4. Pythagoras Theorem is used in security cameras for face recognition.
5. The technique of the Pythagoras Theorem is used by architects in engineering and construction fields.
6. The Pythagoras Theorem is applied in surveying the mountains.
7. It is also used in navigation to find the shortest route.
8. By using the Pythagoras Theorem, we can derive the formula for base, perpendicular and hypotenuse.
9. Painters use ladders to paint on high buildings with the help of Pythagoras Theorem.
10. The Theorem is used to calculate the steepness of slopes of hills or mountains.
11. The converse of Pythagoras Theorem is used to check whether a triangle is a right triangle or not

### Pythagoras Theorem Examples

Question-1: Find the hypotenuse of a triangle whose lengths of two sides are $$4\,{\rm{cm}}\,\& \,3\,{\rm{cm}}.$$

Answer: Given: Base $$= 4\,{\rm{cm,}}$$ Perpendicular $$= 3\,{\rm{cm,}}$$ Hypotenuse $$= ?$$
$${\left( {{\rm{Hypotenuse}}} \right)^{\rm{2}}}{\rm{ = }}\,{\left( {{\rm{Base}}} \right)^{\rm{2}}}{\rm{ + }}{\left( {{\rm{Perpendicular}}} \right)^{\rm{2}}}$$
$$\Rightarrow {\left( {{\rm{Hypotenuse}}} \right)^{\rm{2}}}\,\,{\rm{ = }}\,\,{{\rm{4}}^{\rm{2}}}\,{\rm{ + }}\,\,{{\rm{3}}^{\rm{2}}}$$
$$\Rightarrow {\left( {{\rm{Hypotenuse}}} \right)^{\rm{2}}}\,\,{\rm{ = }}\,\,{\rm{16}}\,\,{\rm{ + }}\,\,{\rm{9}}$$
$$\Rightarrow {\rm{Hypotenuse}}\,\,{\rm{ = }}\,\,\sqrt {{\rm{25}}} \,\,{\rm{ = }}\,\,{\rm{5}}\,{\rm{cm}}$$
Therefore, the hypotenuse of a triangle is $${\rm{ = 5}}{\mkern 1mu} \,{\rm{cm}}{\rm{.}}$$

Question-2: If the hypotenuse of a right-angled triangle is $$5\,{\rm{cm}}$$ and one of the two sides is $$4\,{\rm{cm,}}$$ find the third side.

Answer: Given: Base $$= 4\,{\rm{cm,}}$$ Perpendicular $$= ?,$$ Hypotenuse $$= 5\,{\rm{cm,}}$$
$${\left( {{\rm{Hypotenuse}}} \right)^{\rm{2}}}\,\,{\rm{ = }}\,\,{\rm{Bas}}{{\rm{e}}^{\rm{2}}}\,\,{\rm{ + }}\,\,{\rm{Perpendicula}}{{\rm{r}}^{\rm{2}}}$$
$$\Rightarrow {\left( {\rm{5}} \right)^{\rm{2}}}\,\,{\rm{ = }}\,\,{{\rm{4}}^{\rm{2}}}\,{\rm{ + }}\,\,{\rm{Third\,sid}}{{\rm{e}}^{\rm{2}}}$$
$$\Rightarrow {\rm{Third}}{\mkern 1mu}\, {\rm{sid}}{{\rm{e}}^{\rm{2}}}{\rm{ = 25 – 16}}$$
$$\Rightarrow \,\,{\rm{Third\,side}}\,\,{\rm{ = }}\,\,\sqrt {\rm{9}} \,\,{\rm{ = }}\,\,{\rm{3}}\,{\rm{cm}}$$
Therefore, the third side is $${\rm{ = 3}}\,{\mkern 1mu} {\rm{cm}}{\rm{.}}$$

Question-3: Find the value of $$x.$$

Answer: Given: Base $$= 12\,{\rm{cm,}}$$ Perpendicular $$= 5\,{\rm{cm,}}$$ Hypotenuse $$= x$$
$${\left( {{\rm{Hypotenuse}}} \right)^{\rm{2}}}{\rm{ = Bas}}{{\rm{e}}^{\rm{2}}}{\rm{ + Perpendicula}}{{\rm{r}}^{\rm{2}}}$$
$$\Rightarrow {\left( {{\rm{Hypotenuse}}} \right)^{\rm{2}}}{\rm{ = 1}}{{\rm{2}}^{\rm{2}}}{\rm{ + }}{{\rm{5}}^{\rm{2}}}$$
$$\Rightarrow {\left( {{\rm{Hypotenuse}}} \right)^{\rm{2}}}{\rm{ = 144 + 25}}$$
$$\Rightarrow {\rm{Hypotenuse = }}\sqrt {{\rm{169}}} {\rm{ = 13 }}{\mkern 1mu} {\rm{cm}}$$
Therefore, the hypotenuse (value of $$x$$) of a triangle is $$= 13\,{\rm{cm}}{\rm{.}}$$

Question-4: In a right-angled triangle, the measure of its hypotenuse is $$= 20$$ inches. One of the sides of the triangle is $$= 16$$ inches. What would be the measure of the third side?

Answer: Given: Base $$= 16$$ inches, Perpendicular $$= ?$$, Hypotenuse $$= 20$$ inches
$${\left( {{\rm{Hypotenuse}}} \right)^{\rm{2}}}{\rm{ = Bas}}{{\rm{e}}^{\rm{2}}}{\rm{ + Perpendicula}}{{\rm{r}}^{\rm{2}}}$$
$$\Rightarrow {\left( {{\rm{20}}} \right)^{\rm{2}}}{\rm{ = 1}}{{\rm{6}}^{\rm{2}}}{\rm{ + Third}}{\mkern 1mu} {\rm{sid}}{{\rm{e}}^{\rm{2}}}$$
$$\Rightarrow {\rm{Third}}{\mkern 1mu} {\rm{sid}}{{\rm{e}}^{\rm{2}}}{\rm{ = 400 – 256}}$$
$${\rm{Third}}{\mkern 1mu} {\rm{side = }}\sqrt {{\rm{144}}} {\rm{ = 12}}\,{\rm{inches}}$$
Therefore, the third side is $$12$$ inches.

Question-5: In a right triangle, the lengths of the legs are $$6\,{\rm{cm}}\,\& \,8\,{\rm{cm}}.$$ Find the length of the hypotenuse.

Answer: Given: Base $$= 6\,{\rm{cm,}}$$ Perpendicular $$= 8\,{\rm{cm}}$$ Hypotenuse $$= ?$$
$${\left( {{\rm{Hypotenuse}}} \right)^{\rm{2}}}{\rm{ = Bas}}{{\rm{e}}^{\rm{2}}}{\rm{ + Perpendicula}}{{\rm{r}}^{\rm{2}}}$$
$$\Rightarrow {\left( {{\rm{Hypotenuse}}} \right)^{\rm{2}}}{\rm{ = }}{{\rm{6}}^{\rm{2}}}{\rm{ + }}{{\rm{8}}^{\rm{2}}}$$
$$\Rightarrow {\left( {{\rm{Hypotenuse}}} \right)^{\rm{2}}}{\rm{\; = \;36 + 64}}$$
$$\Rightarrow{\text{Hypotenuse}} = \sqrt{100} = 10\,{\text{cm}}$$
Therefore, the hypotenuse of a triangle is $$10\,{\rm{cm}}.$$

Question-6: In a right triangle, the lengths of the legs are $$8\,{\rm{cm}}\,\& \,15\,{\rm{cm}}.$$ Find the length of the hypotenuse.

Answer: Given: Base $$= 8\,{\rm{cm,}}$$ Perpendicular $$= 15\,{\rm{cm,}}$$ Hypotenuse $$= ?$$
$${\left( {{\rm{Hypotenuse}}} \right)^{\rm{2}}}{\rm{ = Bas}}{{\rm{e}}^{\rm{2}}}{\rm{ + Perpendicula}}{{\rm{r}}^{\rm{2}}}$$
$$\Rightarrow {\left( {{\rm{Hypotenuse}}} \right)^{\rm{2}}}{\rm{ = }}{{\rm{8}}^{\rm{2}}}{\rm{ + 1}}{{\rm{5}}^{\rm{2}}}$$
$$\Rightarrow {\left( {{\rm{Hypotenuse}}} \right)^{\rm{2}}}{\rm{ = 64 + 225}}$$
$$\Rightarrow{\text{Hypotenuse}} = \sqrt{289} = 17\,{\text{cm}}$$
Therefore, the hypotenuse of a triangle is $$= 17\,{\rm{cm}}{\rm{.}}$$

Question-7: In a right triangle, the lengths of the legs are $$2\,{\rm{cm}}\,\& \,1.5\,{\rm{cm}}.$$ Find the length of the hypotenuse.

Answer: Given: Base $$= 2\,{\rm{cm,}}$$ Perpendicular $$= 2\,{\rm{cm,}}$$ Hypotenuse $$= ?$$ $${\left( {{\rm{Hypotenuse}}} \right)^{\rm{2}}}{\rm{ = Bas}}{{\rm{e}}^{\rm{2}}}{\rm{ + Perpendicula}}{{\rm{r}}^{\rm{2}}}$$
$$\Rightarrow {\left( {{\rm{Hypotenuse}}} \right)^{\rm{2}}}{\rm{ = }}{{\rm{2}}^{\rm{2}}}{\rm{ + 1}}{\rm{.}}{{\rm{5}}^{\rm{2}}}$$
$$\Rightarrow {\left( {{\rm{Hypotenuse}}} \right)^{\rm{2}}}{\rm{ = 4 + 2}}{\rm{.25}}$$
$$\Rightarrow{\text{Hypotenuse}} = \sqrt{6.25} = 2.5\,{\text{cm}}$$
Therefore, the hypotenuse of a triangle is $$= 2.5\,{\rm{cm}}{\rm{.}}$$

Question-8: Find the hypotenuse of a triangle whose lengths of two sides are $$9\,{\rm{cm}}\,\& \,40\,{\rm{cm}}{\rm{.}}$$

Answer: Given: Base $$= 9\,{\rm{cm,}}$$ Perpendicular $$= 40\,{\rm{cm,}}$$ Hypotenuse =?
$${\left( {{\rm{Hypotenuse}}} \right)^{\rm{2}}}{\rm{ = Bas}}{{\rm{e}}^{\rm{2}}}{\rm{ + Perpendicula}}{{\rm{r}}^{\rm{2}}}$$
$$\Rightarrow {\left( {{\rm{Hypotenuse}}} \right)^{\rm{2}}}{\rm{ = }}{{\rm{9}}^{\rm{2}}}{\rm{ + 4}}{{\rm{0}}^{\rm{2}}}$$
$$\Rightarrow {\left( {{\rm{Hypotenuse}}} \right)^{\rm{2}}}{\rm{ = 81 + 1600}}$$
$$\Rightarrow{\text{Hypotenuse}} = \sqrt{1681} = 41\,{\text{cm}}$$
therefore, the hypotenuse of a triangle is $$= 41\,{\rm{cm}}{\rm{.}}$$

Question-9: Find the value of $$x.$$

Answer: Base $$= 8\,ft,$$ Perpendicular $$= ?,$$ Hypotenuse $$= 10\,ft$$
$${\left( {{\rm{Hypotenuse}}} \right)^{\rm{2}}}{\rm{ = Bas}}{{\rm{e}}^{\rm{2}}}{\rm{ + Perpendicula}}{{\rm{r}}^{\rm{2}}}$$
$$\Rightarrow {\left( {{\rm{10}}} \right)^{\rm{2}}}{\rm{ = }}{{\rm{8}}^{\rm{2}}}{\rm{ + Third}}{\mkern 1mu} {\rm{sid}}{{\rm{e}}^{\rm{2}}}$$
$$\Rightarrow {\rm{Third}}{\mkern 1mu} {\rm{sid}}{{\rm{e}}^{\rm{2}}}{\rm{ = 100 – 64}}$$
$$\Rightarrow {\rm{Third}}{\mkern 1mu} {\rm{side = }}\sqrt {{\rm{36}}} {\rm{ = 6}}{\mkern 1mu} \,ft$$
Therefore, the value of $$x$$ is $$6\,ft.$$

Question-10: Find the hypotenuse of a triangle whose lengths of two sides are $$7$$ inches & $$24$$ inches.

Answer: Given: Base $$= 7$$ inches, Perpendicular $$= 24$$ inches, Hypotenuse $$= ?$$
$${\left( {{\rm{Hypotenuse}}} \right)^{\rm{2}}}{\rm{ = Bas}}{{\rm{e}}^{\rm{2}}}{\rm{ + Perpendicula}}{{\rm{r}}^{\rm{2}}}$$
$$\Rightarrow {\left( {{\rm{Hypotenuse}}} \right)^{\rm{2}}}{\rm{ = }}{{\rm{7}}^{\rm{2}}}{\rm{ + 2}}{{\rm{4}}^{\rm{2}}}$$
$$\Rightarrow {\left( {{\rm{Hypotenuse}}} \right)^{\rm{2}}}{\rm{ = 49 + 576}}$$
$$\Rightarrow {\rm{Hypotenuse = }}\sqrt {{\rm{625}}} {\rm{ = 25 }}\,{\rm{inches}}$$
Therefore, the hypotenuse of a triangle is $$25$$ inches.

### Summary

In this article, we have learned that the Pythagoras Theorem and its converse are the basic and fundamental theories of mathematics. A right-angled triangle can be identified given the length of the longest side squared is equal to the sum of the squares of the other two sides. This article is useful to learn about the Pythagoras Theorem and its converse in detail with the formula, proof, and its applications. Some solved examples and FAQs are also included at the end for a better understanding of the topic.

### FAQs on Pythagoras Theorem

We have provided some frequently asked questions about Pythogarus Theorem here:

Q.1. Where did Pythagorean Theorem come from?
Ans: The Pythagorean Theorem was first discovered by Babylonian mathematician 100100 years before Pythagoras was born.

Q.2. How do we use Pythagorean Theorem in daily life?
AnsThe Pythagorean theorem applications in daily life

1. Pythagoras Theorem is commonly used to find the sides of a right-angled triangle.
2. Pythagoras Theorem is used in trigonometry to find the trigonometric ratios like sin,cos,tan,cosec,sec,cot.sin,cos,tan,cos⁡ec,sec,cot..
3. Pythagoras Theorem is used in security cameras for face recognition.
4. The technique of the Pythagoras Theorem is used by architects in engineering and construction fields.
5. The Pythagoras Theorem is applied in surveying the mountains.
6. It is useful in the navigation to find the shortest route.

Q.3. What is meant by Pythagoras Theorem?
Ans: The Pythagoras Theorem states that in a right-angled triangle the sum of the square of the base and the square of the perpendicular is equal to the square of the hypotenuse. (Hypotenuse)2=Base2+Perpendicular2

Q.4. What is the significance of the Pythagorean Theorem?
Ans: Discovering Pythagorean Theorem led the Greeks to prove the existence of numbers that could not be expressed as rational numbers. For example, if a right-angled triangle with a base and perpendicular 11 units each is constructed then hypotenuse will be 2–√2 units which is not a rational number.

Q.5. Can we apply the Pythagoras Theorem for any triangle?
Ans
:
No, the Pythagoras Theorem is applicable for only the right-angled triangle.

Q.6. What is the use of the Pythagoras Theorem?
Ans: Pythagoras Theorem is commonly used to find the sides of a right-angled triangle and used in trigonometry to find the trigonometric ratios.

Q.7. What is the formula for Pythagoras Theorem?
AnsThe formula of the Pythagoras Theorem is given by (Hypotenuse)2=Base2+Perpendicular2(Hypotenuse)2=Base2+Perpendicular2

Q.8. Is Pythagorean Theorem only for right-angled triangles?
Ans: Yes, the Pythagoras Theorem is applicable for only the right-angled triangle.

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