Question

i have drawn a triangle PQR whose angleQ is right angle. if S is any point on QR , then let us prove that, PS^2+QR^2 = PR^2+ QS^2

Answer 1

Restatement :

• In the adjoining figure ∆PQR is a right angled triangle where angle Q is 90° . S is a point on side QR.

To Prove :

• PS² + QR² = PR² + QS²

Proof :

Since angle Q is right so applying Pythagorean theorem in ∆PQS we have

$\sf{ {PQ}^{2} + QS {}^{2} =PS {}^{2} } \\ \sf\implies {PQ}^{2} = PS {}^{2} - QS {}^{2} .......(1)$

Now applying the theorem in ∆PQR we have

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

Substituting the value of PQ² from (1) we have

$\implies \sf{PS {}^{2} - QS {}^{2} + QR {}^{2} = PR {}^{2} } \\ \implies\sf{PS {}^{2} + QR {}^{2} =PR {}^{2} + QS {}^{2}}$

$\sf{Hence \: \: Shown}$

Formula Used :

Pythagorean theorem : $\sf{Perpendicular {}^{2} \: + Base {}^{2} = Hypetenus {}^{2} }$

Answer 2

$\tt{\huge{\purple{ ............. }}}$

QUESTION :

I have drawn a triangle PQR whose angleQ is right angle. if S is any point on QR , then let us prove that, PS^2+QR^2 = PR^2+ QS^2.

SOLUTION :

See the attachment above .

Now, from the figure, we can see that ∆ PQS and ∆ PQR are right angled Triangle.

From the Pythagoras Theorem, we can state that :

{ Base } ^ 2 + { Perpendicular } ^ 2 = { Hypotenuse } ^ 2

Using the above relation,

[ PS ] ^ 2 = [ PQ ] ^ 2 + [ QS ] ^ 2 ..... ( 1 )

[ PR ] ^ 2 = [ PQ ] ^ 2 + [ QR ] ^ 2 ..... ( 2 )

Now, substitute the value of [ PQ ] ^ 2 from Equation ( 1 ) in Equation ( 2 )

=> [ PR ] ^ 2 = { [ PS ] ^ 2 - [ QS ] ^ 2 } + [ QR ] ^ 2

Shifting the negative terms to LHS :

=> [ PS ] ^ 2 + [ QR ] ^ 2 = [ PR ] ^ 2 + [ QS ] ^ 2

HENCE PROVED