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Answer:
c) 4cm
Step-by-step explanation:
From the figure, we observe that
∠LOP = ∠MOP ,
OL = 3cm , OP = 5cm ,
∠PLO = ∠PMO [ as right angles ]
Let us prove that ΔLOP ≅Δ MOP
From ΔLOP and Δ MOP we note that
∠LOP = ∠MOP
OP = OP [common]
∠PLO = ∠PMO [ right angles ]
Hence , ΔLOP ≅ΔMOP
PM = PL [ by CPCT ]—-(i)
By Pythagoras theorem we can determine the value of PL
In ΔPLO ,
OP²= LP²+OL²
[5]²= [3]²+LP²
25= 9 +LP²
LP²= 25- 9
LP= √16 = 4cm
As from above equation (i) we know that PM= PL
So , PL = PM = 4cm
Answer:
c) 4cm
Step-by-step explanation:
In ∆OPL and ∆OPM
∠LOP = ∠MOP (Common angle O)
∠OLP = ∠OMP (both 90°)
OP = OP (common)
∆OPL ≈ ∆OPM
Therefore,
PL = PM ----------- (eq 1)
(By CPCT)
Now, In ∆PLO apply pythagoras theorem. Where OP is hypotenuse having value 5 cm, LP is base and OL is perpendicular having value 3cm.
(Hypotenuse)² = (Base)² + (Perpendicular)²
(OP)² = (LP)² + (OL)²
Substitute the values,
→ (5)² = (LP)² + (3)²
→ 25 = (LP)² + 9
→ 25 - 9 = (LP)²
→ 16 = (LP)²
→ (4)² = (LP)²
→ LP = 4
As, LP = PM from equation 1). So, LP = PM = 4 cm
Therefore, the value of PM is 4 cm.