Hola Brainlians !!! 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 . About the author Madeline
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 Reply
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 Reply
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:–
” The square of the hypotenuse is equal to the sum of the squares of the other two sides”.
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