pls answer the first part of the question
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LovableLiana:
Which question
first part
U mean question no 1
you mean we have to prove that PA*PB=PT^2
I THINK SO
yh
then what do u mean by first part
OK SORRY
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given,
OL is perpendicular to AB
AL=BL [ perpendicular from the centre to the chord bisects the chord]..................(1)
LHS=PA×PB
= (PL-AL)(PL+BL)
=(PL-AL)(PL+AL). [from (1)]
=PL²-AL²
In right ∆OLP,
OP²=OL ²+PL²
=>PL²=OP ²-OL²..................(2)
LHS=PL²-AL²
=OP²-OL²-AL²
=OP²-(OL²+AL²). [from (2)]
In right ∆OAL,
OA²=OL²+AL²...............(3)
LHS=OP²-(OL²+AL²)
=OP²-OA². [from (3)]
=OP²-OT² [OA=OT because radii of the same circle are equal]
In right ∆OPT,
OP²=OT²+PT²
=>PT²=OP²-OT²...............(4)
LHS=OP²-OT²
=PT². [from (4)]
Hence, LHS=PA×PB=PT²=RHS
Hence proved
HOPE U CAN UNDERSTAND
pls mark it as brainliest
OL is perpendicular to AB
AL=BL [ perpendicular from the centre to the chord bisects the chord]..................(1)
LHS=PA×PB
= (PL-AL)(PL+BL)
=(PL-AL)(PL+AL). [from (1)]
=PL²-AL²
In right ∆OLP,
OP²=OL ²+PL²
=>PL²=OP ²-OL²..................(2)
LHS=PL²-AL²
=OP²-OL²-AL²
=OP²-(OL²+AL²). [from (2)]
In right ∆OAL,
OA²=OL²+AL²...............(3)
LHS=OP²-(OL²+AL²)
=OP²-OA². [from (3)]
=OP²-OT² [OA=OT because radii of the same circle are equal]
In right ∆OPT,
OP²=OT²+PT²
=>PT²=OP²-OT²...............(4)
LHS=OP²-OT²
=PT². [from (4)]
Hence, LHS=PA×PB=PT²=RHS
Hence proved
HOPE U CAN UNDERSTAND
pls mark it as brainliest
actually only 2 steps were needed for this answer is wanted to check if anyone could answer like me!
I mean like 3 steps. .
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