In ABC, L and M are the points on the
sides CA and CB such that LM is parallel
to AB. IF AL=X-3, AC= 2x, BM = x-2
BC2x +3 , find the value of x.
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In ABC, L and M are the points on the
sides CA and CB such that LM is parallel
to AB. IF AL=X-3, AC= 2x, BM = x-2
BC2x +3 , find the value of x.
Answer this. ...☝
Wanted verified answers ✅
No spamm ❎
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Correct Question:-
In ∆ABC, L and M are the points on the sides CA and CB such that LM is parallel to AB. If AL = x - 3, AC = 2x, BM = x - 2 and BC = 2x + 3, find the value of x.
Solution:-
Given:-
To find:-
The value of x.
Note:-
Refer to the attachment for a clear concept.
In triangle:-
As the L and M are points on AC and BC respectively,
LC + AL = AC
[tex]\implies[/tex] LC = AC - AL [tex]\longrightarrow[/tex] (1)
MC + BM = BC
[tex]\implies[/tex] MC = BC - BM [tex]\longrightarrow[/tex] (2)
Solution:-
In ∆ABC,
LM || AB
According to Thale's Theorem,
[tex]\sf{\dfrac{AL}{LC} = \dfrac{BM}{MC}}[/tex]
Now Substituting the values of LC and MC from (1) and (2)
= [tex]\sf{\dfrac{AL}{AC - AL} = \dfrac{BM}{BC - BM}}[/tex]
Now, Substituting the values from given,
[tex]\sf{\dfrac{x-3}{2x - (x-3)} = \dfrac{x-2}{(2x+3)-(x-2)}}[/tex]
= [tex]\sf{\dfrac{x-3}{2x-x+3} = \dfrac{x-2}{2x+3 -x+2}}[/tex]
= [tex]\sf{\dfrac{x-3}{x+3} = \dfrac{x-2}{x+5}}[/tex]
By Cross-multiplication,
[tex]\sf{(x-3)(x+5) = (x+3)(x-2)}[/tex]
= [tex]\sf{x(x+5) -3(x+5) = x(x-2)+3(x-2)}[/tex]
= [tex]\sf{x^2 + 5x - 3x -15 = x^2 - 2x +3x - 6}[/tex]
= [tex]\sf{x^2 + 2x -15 = x^2 + x - 6}[/tex]
Taking all the variables on LHS,
The signs of the variables will change according to:-
(-) into (+)
(+) into (-)
= [tex]\sf{x^2 + 2x -15 - x^2 - x + 6 = 0}[/tex]
Taking like terms together.
= [tex]\sf{x^2 - x^2 + 2x - x - 15 + 6 = 0}[/tex]
= [tex]\sf{\cancel{x^2} - \cancel{x^2} + x -9 = 0}[/tex]
= [tex]\sf{x - 9 = 0}[/tex]
= [tex]\sf{ x = 9}[/tex]
Therefore the value of x is 9
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Additional Information:-
→ What is Thale's Theorem?
✓ Thale's Theorem states that if a line which is parallel to one side of the triangle intersecting the other two at distinct points then the line divides the two sides in proportion.
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