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PQR is a right angled at P and M is a point on QR such that PM Perpendicular to QR. Show that PM^2 = QM.MR .
Answers
Step-by-step explanation:
In PQR, By pythagoras theorem
QR^2 = PR^2+PQ^2 ----- 1
Similarly, In PMQ & PMR By pythagoras theorem
PQ^2 = QM^2 + PM^2 ------2
PR^2 = PM^2+MR^2 ---------3
Putting the value of 2&3 in 1
QR^2 = QM^2+ PM^2 + PM^2+ MR^2
QR^2 = 2PM^2 + (QM^2+MR^2)
QR = QM+MR
(QM+MR)^2 = 2PM^2 + (QM^2+MR^2)
(QM^2+MR^2)+2QM*MR = 2PM^2 + (QM^2+MR^2)
(QM^2+MR^2) - (QM^2+MR^2) +2QM*MR = 2PM^2
2QM*MR = 2PM^2
2 will cancel 2
PM^2 = QM*MR
Hence, proved
Step-by-step explanation:
Given:-
PQR is a right angled at P and M is a point on QR such that PM Perpendicular to QR.
To show:-
Show that PM^2 = QM.MR .
Construction:-
Join P and M
Proof :-
See the above attachment
∆PQR is a right angled triangle and right angle at P
We know that Pythagoras theorem,
" The square of the hypotenuse is equal to the sum of the squares of the other two sides".
QR^2=PQ^2+PR^2 -----------------(1)
M is the point on QR such that PM is a perpendicular to QR
QR = QM +MR -----------------------(2)
and ∆ PMR and ∆PMQ are the two right angled triangles
In ∆ PMR, by Pythagoras theorem
PR^2=PM^2+MR^2------------------(3)
In ∆PMQ ,by Pythagoras theorem
PQ^2=PM^2+QM^2 ----------------(4)
On adding (3)&(4) equations then we get
PR^2+PQ^2=PM^2+MR^2+PM^2+QM^2
From (1)
=>QR^2 = 2 PM^2+QM^2+MR^2
We know that (a+b)^2 = a^2+2ab+b^2
=>a^2+b^2 = (a+b)^2-ab
=>QR^2 = 2PM^2+(QM+MR)^2 -2 QM.MR
From (2)
=>QR^2=2PM^2 +QR^2 -2 QM.MR
On cancelling QR^2 both sides
=>2PM^2-2QM.MR = 0
=>2(PM^2-QM.MR) = 0
=>PM^2-QM.MR = 0/2
=>PM^2-QM.MR = 0
Therefore,PM^2 = QM.MR
Hence, Proved.
Used formulae:-
- Pythagoras theorem:-
" The square of the hypotenuse is equal to the sum of the squares of the other two sides".
- (a+b)^2 = a^2+2ab+b^2
- a^2+b^2 = (a+b)^2-ab
