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Application of Pythagoras Theorem: For two-dimensional navigation, the Pythagorean theorem is useful. A right-angled triangle is the subject of this theorem. The length of the third side of a right-triangled triangle can be calculated using the lengths of any two sides. Students must explore several parts of the Pythagoras theorem, including its statement, proof, the converse of Pythagoras theorem and proof, and real-life applications of the Pythagoras theorem.
Students must understand the application of the Pythagoras theorem by heart to score high in the exam. This article explains the real-life application of the Pythagoras theorem, the practical application of the Pythagoras theorem, the application of the Pythagoras theorem in daily life, and the model of Pythagoras theorem with application and more. Read on to know more.
Pythagoras theorem is a basic relation in Euclidean geometry among the sides of a right-angled triangle. It is commonly used to find the length of an unknown side in a right-angled triangle. It describes the interrelationship between a right-angled triangle’s base, perpendicular and hypotenuse. Using the Pythagoras theorem, we can derive base, perpendicular and hypotenuse formulas. The converse of the Pythagoras theorem is used to check whether a triangle is a right triangle or not.
The Pythagoras theorem states that in a right-angled triangle, the sum of squares of the length of base and perpendicular is equal to the square of the length of the hypotenuse.
\({\rm{(Hypotenuse)}}{{\rm{ }}^{\rm{2}}}{\rm{ = Base}}{{\rm{ }}^{\rm{2}}}{\rm{ + Perpendicular}}{{\rm{ }}^{\rm{2}}}\)
\({\rm{(Hypotenuse)}}{{\rm{ }}^{\rm{2}}}{\rm{ = Base}}{{\rm{ }}^{\rm{2}}}{\rm{ + Perpendicular}}{{\rm{ }}^{\rm{2}}}\)
where, \(b=\) base, \(a=\) perpendicular, \(c=\) hypotenuse
The proof of the Pythagoras theorem is very important in mathematics. In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
We were 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 \(D.\)
Proof: we know that \(\Delta RDQ \sim \Delta RQP\)
So, \(\frac{R D}{Q R}=\frac{Q R}{P R}\) (Corresponding sides of similar triangles)
\(\Longrightarrow Q R^{2}=R D \times R P\)…..(i)
Also, \(\Delta QDP \sim \Delta RQP\)
So \(\frac{P D}{P Q}=\frac{P Q}{P R}\) (Corresponding sides of similar triangles)
\(\Longrightarrow P Q^{2}=P D \times R P\) —-(ii)
Adding the equation (i) and (ii) we get,
\(Q R^{2}+P Q^{2}=R D \times P R+P D \times P R\)
\(\Rightarrow Q R^{2}+P Q^{2}=P R(R D+P D)\)
From the figure, \(RD+PD=PR\)
Therefore, \(P R^{2}=P Q^{2}+Q R^{2}\)
Hence, the Pythagoras theorem is proved.
In a triangle, if the square of one side is equal to the sum of the other two sides, then the angle opposite the first side is a right angle.
Given: In \(\Delta X Y Z, X Y^{2}+Y Z^{2}=X Z^{2}\) To prove \(\angle XYZ = {90^ \circ }\)
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 \(P R^{2}=X Y^{2}+Y Z^{2} \ldots \ldots\) (i) \([P Q=X Y, Q R=Y Z]\)
But we know, \(X Z^{2}=X Y^{2}+Y Z^{2} \ldots \ldots\) (ii) (Given)
Therefore, \(X Z^{2}=P R^{2}\) [From equation (i) and (ii)]
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.
Some of the applications of the Pythagoras theorem are
When the length of the three sides of the triangle is provided, then to identify the triangle, whether it is a right-angled triangle or not, we have to use the Pythagoras theorem.
Example: Assume the triangle with the sides of \(10, 24\) and \(26\) are given.
We can see that \(26\) is the longest side.
So, it also satisfies the condition, \(10+24>26\)
We know that,
\(c^{2}=a^{2}+b^{2}\) (1)
Now, let \(a=10, b=24\) and \(c=26\)
Here, we will solve the RHS of the equation \((1)\)
\(a^{2}+b^{2}=10^{2}+24^{2}=100+576=676\)
So, take LHS, then we get,
\(c^{2}=26^{2}=676\)
We can see that,
\(LHS=RHS\)
Hence, the given triangle is the right-angled triangle as it satisfies the theorem.
Q.1. Find the hypotenuse of a triangle whose lengths of two sides are 4 cm & 3 cm.
Ans: Given: Base \(=4 \mathrm{~cm}\) Perpendicular \(=3 \mathrm{~cm}\) Hypotenuse =?
\({{\rm{(Hypotenuse)}}^{\rm{2}}}{\rm{ = Bas}}{{\rm{e}}^{\rm{2}}}{\rm{ + Perpendicula}}{{\rm{r}}^{\rm{2}}}\)
\( \Rightarrow {({\rm{ Hypotenuse }})^2} = {4^2} + {3^2}\)
\(\Rightarrow {({\rm{ Hypotenuse }})^2} = 16 + 9\)
\( \Rightarrow {\rm{Hypotenuse}} = \sqrt {25} = 5\;{\rm{cm}}\)
Therefore, the hypotenuse of a triangle is \(5\;{\rm{cm}}.\)
Q.2. If the hypotenuse of a right-angled triangle is 5 cm and one of the two sides is 4 cm, find the third side.
Ans: Given: Base \( = 4\;{\rm{cm}}\) Perpendicular = ? \({\rm{cm}}\) Hypotenuse \(= 5\;{\rm{cm}}\)
\({{\rm{(Hypotenuse)}}^{\rm{2}}}{\rm{ = Bas}}{{\rm{e}}^{\rm{2}}}{\rm{ + Perpendicula}}{{\rm{r}}^{\rm{2}}}\)
\(\Rightarrow {(5)^2} = {4^2} + {\rm{Third\,sid}}{{\rm{e}}^{\rm{2}}}\)
\(\Rightarrow {\rm{Third\,sid}}{{\rm{e}}^2} = 25 – 16\)
\(\Rightarrow {\rm{Third\,side}} = \sqrt 9 = 3\;{\rm{cm}}\)
Therefore, the third side is \(3\;{\rm{cm}}.\)
Q.3. Find the value of x.
Ans: Given: Base \(\Rightarrow x=13 \mathrm{~cm}\), Perpendicular \(= 5\;{\rm{cm}}\) Hypotenuse =?
\({{\rm{(Hypotenuse)}}^{\rm{2}}}{\rm{ = Bas}}{{\rm{e}}^{\rm{2}}}{\rm{ + Perpendicula}}{{\rm{r}}^{\rm{2}}}\)
\(\Rightarrow {({\rm{ Hypotenuse }})^2} = {12^2} + {5^2}\)
\(\Rightarrow {({\rm{ Hypotenuse }})^2} = 144 + 25\)
\(\Rightarrow {\rm{Hypotenuse}} = \sqrt {169} = 13\;{\rm{cm}}\)
\(\Rightarrow x = 13\;{\rm{cm}}\)
Therefore, the value of \(x\) is \(13\;{\rm{cm}}\)
Q.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?
Ans: Given: Base \(=16\) inches, Perpendicular = ?, Hypotenuse \(=20\) inches
\({{\rm{(Hypotenuse)}}^{\rm{2}}}{\rm{ = Bas}}{{\rm{e}}^{\rm{2}}}{\rm{ + Perpendicula}}{{\rm{r}}^{\rm{2}}}\)
\(\Rightarrow {(20)^2} = {16^2} + {\rm{Third\,sid}}{{\rm{e}}^{\rm{2}}}\)
\({\rm{Third}}\,{\rm{side}} = \sqrt {144} = 12\)
\({\rm{Third\,side}} = \sqrt {144} = 12\) inches
Therefore, the third side is \(12\) inches.
Q.5. In a right triangle ABC, the lengths of the legs are 6 cm & 8 cm. Find the length of the hypotenuse.
Ans: Given: Base \({\rm{ = 6cm}}\) Perpendicular \({\rm{ = 8cm}}\) Hypotenuse =?
\({{\rm{(Hypotenuse)}}^{\rm{2}}}{\rm{ = Bas}}{{\rm{e}}^{\rm{2}}}{\rm{ + Perpendicula}}{{\rm{r}}^2}\)
\(\Rightarrow {({\rm{ Hypotenuse }})^2} = {6^2} + {8^2}\)
\(\Rightarrow {({\rm{ Hypotenuse }})^2} = 36 + 64\)
\(\Rightarrow {\rm{Hypotenuse}} = \sqrt {100} = 10\;{\rm{cm}}\)
Therefore, the hypotenuse of a triangle is \(10\;{\rm{cm}}\)
Q.6. In a right triangle ABC, the lengths of the legs are 8 cm & 15 cm. Find the lengths of the hypotenuse.
Ans: Given: Base \({\rm{ = 8cm}}\), Perpendicular \({\rm{ = 15cm}}\) Hypotenuse =?
\({{\rm{(Hypotenuse)}}^{\rm{2}}}{\rm{ = Bas}}{{\rm{e}}^{\rm{2}}}{\rm{ + Perpendicula}}{{\rm{r}}^2}\)
\(\Rightarrow {({\rm{ Hypotenuse }})^2} = {8^2} + {15^2}\)
\(\Rightarrow {({\rm{ Hypotenuse }})^2} = 64 + 225\)
\(\Rightarrow {\rm{Hypotenuse}} = \sqrt {289} = 17\;{\rm{cm}}\)
Therefore, the hypotenuse of a triangle is \(17\;{\rm{cm}}\)
Q.7. In a right triangle ABC, the lengths of the legs are 2 cm & 1.5 cm. Find the lengths of the hypotenuse.
Ans: Given: Base \({\rm{ = 2cm}}\) Perpendicular \({\rm{ = 1}}{\rm{.5cm}}\) Hypotenuse=?
\({{\rm{(Hypotenuse)}}^{\rm{2}}}{\rm{ = Bas}}{{\rm{e}}^{\rm{2}}}{\rm{ + Perpendicula}}{{\rm{r}}^2}\)
\(\Rightarrow {({\rm{ Hypotenuse }})^2} = {2^2} + {1.5^2}\)
\(\Rightarrow {({\rm{ Hypotenuse }})^2} = 4 + 2.25\)
\(\Rightarrow {\rm{Hypotenuse}} = \sqrt {6.25} = 2.5\;{\rm{cm}}\)
Therefore, the hypotenuse of a triangle is \(2.5\;{\rm{cm}}\)
In the given article, we have discussed the statement, proof of the Pythagoras theorem and its converse. We discussed the various applications of the Pythagoras theorem. We have provided solved examples along with a few FAQs.
Learn About Pythagorean Triplets
Q.1. What are the applications of the Pythagoras theorem in daily life?
Ans: The Pythagorean theorem applications in daily life are
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 \(\sin , \cos , \tan , \operatorname{cosec}, \sec , \cot .\)
3. The Pythagoras theorem is used in security cameras for face recognition.
4. Architects use the technique of the Pythagoras theorem for engineering and construction fields.
5. The Pythagoras theorem is applied in surveying the mountains.
6. It is useful in navigation to find the shortest route.
Q.2. What is the converse of the Pythagoras theorem and its proof?
Ans:
In a triangle, if the square of one side is equal to the sum of the other two sides, then the angle opposite the first side is a right angle.
Given: In \(\Delta X Y Z, X Y^{2}+Y Z^{2}=X Z^{2}\) To prove \(\angle XYZ = {90^ \circ }\)
Construction: – A triangle PQR is constructed such that
\(PO = XY,OR = YZ\angle POR = {90^ \circ }\)
Proof: In \(\Delta PQR,\angle Q = {90^ \circ }\)
\(P R^{2}=P Q^{2}+Q R^{2}\) [Pythagoras theorem]
Or \(P R^{2}=X Y^{2}+Y Z^{2} \ldots \ldots .\) (i) (\(P Q=X Y, Q R=Y Z\))
But we know, \(X Z^{2}=X Y^{2}+Y Z^{2}\)(ii) (Given)
Therefore, \(X Z^{2}=P R^{2}\) [From equation (i) and (ii)]
Or \(XZ=PR\)
Or \(X Z^{2}=P R^{2}\) [SSS congruency rule]
Therefore, \(\angle Y = \angle Q = {90^ \circ }\) (CPCT)
Hence, \(\angle XYZ = {90^ \circ }\)
The converse of Pythagoras theorem is proved.
Q.3. Why is the Pythagoras theorem important?
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.4. Is the Pythagorean theorem only for right triangles?
Ans: The hypotenuse is the longest side, and it is always opposite the right angle. Pythagoras’ theorem only works for the right-angled triangles, so we can use it to test whether the triangle has a right angle or not.
Q.5. How did the Pythagorean theorem change the world?
Ans: The Pythagoras theorem has indeed changed the world with its huge applications in mathematics, science and technology. For the past \(2500\) years, the Pythagoras’ theorem, arguably the most well-known theorem globally, has greatly been useful to mankind. Its useful right angles are everywhere, whether it is a building, a table, a graph with axes, or the atomic structure.
Q.6. What is the significance of the Pythagorean theorem?
Ans: The discovery of the Pythagorean theorem led the Greeks to prove the existence of numbers that could not be expressed as rational numbers.
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