Choose the correct statement:
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Of all the line segments that can be drawn from a point outside a line, the perpendicular is the shortest.
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The difference of two sides of a triangle is equal to the third side.
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The sum of the three sides of a triangle is less than the sum of its three medians.
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If two sides of a triangle are unequal then the larger side has the smaller angle opposite to it.
Reveal answer
Fill a bubble to check yourself
A
Correct answer
Explanation
XY is a fixed line.
$LX=MO=NT=FP=a$
$In\triangle PXL$
we can see that
${ PX }^{ 2 }={ PL }^{ 2 }+{ LX }^{ 2 }$
$\implies\quad { PX }^{ 2 }={ PL }^{ 2 }+{ a }^{ 2 }\quad $-(1)
$ In\quad \triangle PMO:$
$ { PO }^{ 2 }+{ OM }^{ 2 }={ PM }^{ 2 }$
$ \implies\quad { PM }^{ 2 }={ PO }^{ 2 }+{ a }^{ 2 }\quad$ -(2)
$ In\quad \triangle PNT:$
$ { PT }^{ 2 }+{ TN }^{ 2 }={ PN }^{ 2 }$
$\implies\quad { PN }^{ 2 }={ PT }^{ 2 }+{ a }^{ 2 }\quad$ -(3)
$ In\quad \triangle PFN:$
$ { PN }^{ 2 }={ PF }^{ 2 }+{ FN }^{ 2 }$
$\implies\quad { PF }^{ 2 }={ PN }^{ 2 }-{ FN }^{ 2 }$
$ \because In\quad \triangle PFN,PN\quad is\quad $hypotenuse,
$ \therefore PN>PF\quad -(4)$
comparing$(1)(2)(3)& (4)$
$ PL>PO>PT$
$\implies\quad PX>PM>PN$
$comparing \quad with(4)$
$PN>PF$
$\implies\quad PX>PM>PN>PF$
Similarly,we can prove for right side of $PF.$
$\therefore PF$ is the shortest distance on $XY$ from $P.$
$\therefore A)$is correct.