if M and N are points on the sides of QR and PQ respectively of a triangle PQR right angled at Q. Prove that PM^2 + RN^2 = PR^2 + MN^2
Answers
Step-by-step explanation:
Given:-
M and N are points on the sides of QR and PQ respectively of a triangle PQR right angled at Q.
To find:-
If M and N are points on the sides of QR and PQ respectively of a triangle PQR right angled at Q. Prove that PM^2 + RN^2 = PR^2 + MN^2.
Solution :-
Given that
∆PQR is a right angled triangle and right angle at Q
By Pythagoras theorem
"In a right angled triangle ,The square of the hypotenuse is equal to the sum of the squares of the other two sides"
PR^2 = PQ^2+QR^2 ------------(1)
M is the point on the QR
N is the point on the PQ
Join M and N ,
P and M
N and R
we get 3 another right angles
From ∆PQM ,by Pythagoras theorem,
PM^2 = PQ^2 + QM^2 ------------(2)
From ∆NQM ,by Pythagoras theorem
MN^2 = NQ^2+QM^2 ----------(3)
From ∆NQR ,by Pythagoras theorem
RN^2 = QN^2+QR^2--------------(4)
on adding (2)&(4) then
PM^2+RN^2 = PQ^2+QM^2+QN^2+QR^2
=>PM^2+RN^2 = (PQ^2+QR^2)+(QM^2+QN^2)
=>PM^2+RN^2 = PR^2+MN^2
(From (1) and (3))
Answer:-
if M and N are points on the sides of QR and PQ respectively of a triangle PQR right angled at Q. Prove that PM^2 + RN^2 = PR^2 + MN^2
Used formula:-
Pythagoras theorem:-
- In a right angled triangle,The square of the hypotenuse is equal to the sum of the squares of the other two sides
