Show that \[\left( {{m^2} - 1} \right),\left( {2m} \right),\left( {{m^2} + 1} \right)\] always form a Pythagorean triplet.
Answer
328.5k+ views
Hint: When the length of the side of a right triangle satisfies the Pythagoras theorem, these three numbers are known as Pythagorean triplets or triples. According to Pythagoras theorem, ${a^2} + {b^2} = {c^2}$.
Complete step-by-step answer:
Let $x = {m^2} - 1$ ,$y = 2m$ , and $z = {m^2} + 1$
To show x, y and z are Pythagorean triplet .So, we have to prove ${x^2} + {y^2} = {z^2}$ .
Now, $LHS = {x^2} + {y^2}$
Put the value of x and y in LHS
$ \Rightarrow {x^2} + {y^2} = {\left( {{m^2} - 1} \right)^2} + {\left( {2m} \right)^2}$
Use identity, ${\left( {p - q} \right)^2} = {\left( p \right)^2} - 2pq + {\left( q \right)^2}$
$
\Rightarrow {x^2} + {y^2} = {\left( {{m^2}} \right)^2} - 2\left( {{m^2}} \right)\left( 1 \right) + {\left( 1 \right)^2} + 4{m^2} \\
\Rightarrow {x^2} + {y^2} = {m^4} - 2{m^2} + 1 + 4{m^2} \\
\Rightarrow {x^2} + {y^2} = {m^4} + 2{m^2} + 1 \\
$
Use identity, ${\left( p \right)^2} + 2pq + {\left( q \right)^2} = {\left( {p + q} \right)^2}$
$
\Rightarrow {x^2} + {y^2} = {\left( {{m^2}} \right)^2} + 2\left( {{m^2}} \right)\left( 1 \right) + {\left( 1 \right)^2} \\
\Rightarrow {x^2} + {y^2} = {\left( {{m^2} + 1} \right)^2} \\
LHS = {\left( {{m^2} + 1} \right)^2}............\left( 1 \right) \\
$
Now, $RHS = {z^2}$
Put value of z in RHS
$
\Rightarrow {z^2} = {\left( {m{}^2 + 1} \right)^2} \\
RHS = {\left( {m{}^2 + 1} \right)^2}.........\left( 2 \right) \\
$
From (1) and (2) equation, LHS=RHS
Now, it's proven ${x^2} + {y^2} = {z^2}$ .
So, we can say \[\left( {{m^2} - 1} \right),\left( {2m} \right),\left( {{m^2} + 1} \right)\] always form a Pythagorean triplet.
Note: Whenever we face such types of problems we use some important points. First we assume triplets are sides of the right triangle and apply Pythagoras theorem then after using some algebraic identities if Pythagoras theorem satisfies. So, we can assume triplets are Pythagorean triplet.
Complete step-by-step answer:
Let $x = {m^2} - 1$ ,$y = 2m$ , and $z = {m^2} + 1$
To show x, y and z are Pythagorean triplet .So, we have to prove ${x^2} + {y^2} = {z^2}$ .
Now, $LHS = {x^2} + {y^2}$
Put the value of x and y in LHS
$ \Rightarrow {x^2} + {y^2} = {\left( {{m^2} - 1} \right)^2} + {\left( {2m} \right)^2}$
Use identity, ${\left( {p - q} \right)^2} = {\left( p \right)^2} - 2pq + {\left( q \right)^2}$
$
\Rightarrow {x^2} + {y^2} = {\left( {{m^2}} \right)^2} - 2\left( {{m^2}} \right)\left( 1 \right) + {\left( 1 \right)^2} + 4{m^2} \\
\Rightarrow {x^2} + {y^2} = {m^4} - 2{m^2} + 1 + 4{m^2} \\
\Rightarrow {x^2} + {y^2} = {m^4} + 2{m^2} + 1 \\
$
Use identity, ${\left( p \right)^2} + 2pq + {\left( q \right)^2} = {\left( {p + q} \right)^2}$
$
\Rightarrow {x^2} + {y^2} = {\left( {{m^2}} \right)^2} + 2\left( {{m^2}} \right)\left( 1 \right) + {\left( 1 \right)^2} \\
\Rightarrow {x^2} + {y^2} = {\left( {{m^2} + 1} \right)^2} \\
LHS = {\left( {{m^2} + 1} \right)^2}............\left( 1 \right) \\
$
Now, $RHS = {z^2}$
Put value of z in RHS
$
\Rightarrow {z^2} = {\left( {m{}^2 + 1} \right)^2} \\
RHS = {\left( {m{}^2 + 1} \right)^2}.........\left( 2 \right) \\
$
From (1) and (2) equation, LHS=RHS
Now, it's proven ${x^2} + {y^2} = {z^2}$ .
So, we can say \[\left( {{m^2} - 1} \right),\left( {2m} \right),\left( {{m^2} + 1} \right)\] always form a Pythagorean triplet.
Note: Whenever we face such types of problems we use some important points. First we assume triplets are sides of the right triangle and apply Pythagoras theorem then after using some algebraic identities if Pythagoras theorem satisfies. So, we can assume triplets are Pythagorean triplet.
Last updated date: 05th Jun 2023
•
Total views: 328.5k
•
Views today: 5.84k
Recently Updated Pages
If a spring has a period T and is cut into the n equal class 11 physics CBSE

A planet moves around the sun in nearly circular orbit class 11 physics CBSE

In any triangle AB2 BC4 CA3 and D is the midpoint of class 11 maths JEE_Main

In a Delta ABC 2asin dfracAB+C2 is equal to IIT Screening class 11 maths JEE_Main

If in aDelta ABCangle A 45circ angle C 60circ then class 11 maths JEE_Main

If in a triangle rmABC side a sqrt 3 + 1rmcm and angle class 11 maths JEE_Main

Trending doubts
Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Epipetalous and syngenesious stamens occur in aSolanaceae class 11 biology CBSE
