Question

In Δ PQR , PM is perpendicular to QR. Prove that PR2=PQ2+QR2−2QM.QR

Answer 1

Answer:

$PR^2=PQ^2+QR^2-2\:QR.QM$

Step-by-step explanation:

Concept used:

Pythagors theorem:

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

In right Δ PMR and Δ PMQ

By Pythagors theorem, we have

$PR^2=PM^2+MR^2$.......(1)

$PQ^2=PM^2+QM^2$........(2)

Now,

$PR^2=PM^2+MR^2$

$PR^2=(PQ^2-QM^2)+MR^2$ (using (2))

$PR^2=(PQ^2-QM^2)+(QR-QM)^2$

$PR^2=PQ^2-QM^2+QR^2+QM^2-2\:QR.QM$

$PR^2=PQ^2+QR^2-2\:QR.QM$

Answer 2

Answer :

PR² = PQ² + QR² - 2×QR×QM

Explanation :

In the right angle ∆PMR & ∆PMQ ,

Using Pythagoras theorem :-

(i) PR² = PM² + MR²

(ii) PQ² = PM² + QM²

Now ,

=> PR² = PM² + MR²

=> PR² = PQ² - QM² + MR² ( using (ii) )

=> PR² = PQ² - QM² + ( QR - QM )²

=> PR² = PQ² - QM² + QR² + QM² - 2QR×M

=> PR² = PQ² + QR² - 2×QR×QM