[tex]\bold{\huge{\underline{ Solution\: C }}}[/tex]
Part 5 :-
Here,
- Two non parallel lines are intersected by three parallel lines .
- These three parallel lines are also parallel to each other.
By using basic proportionality theorem
- It states that the line which is parallel to one side of triangle intersects the other two sides then the other two sides are divided in the same ratios.
That is,
[tex]\sf{\dfrac{ x }{ x + 16 }}{\sf{=}}{\sf{\dfrac{9}{12 + 9}}}[/tex]
By cross multiplication ,
[tex]\sf{ 21x = 9( x + 16)}[/tex]
[tex]\sf{ 21x = 9x + 144}[/tex]
[tex]\sf{ 21x - 9x = 144}[/tex]
[tex]\sf{ 12x = 144}[/tex]
[tex]\sf{ x = }{\sf{\dfrac{ 144}{12}}}[/tex]
[tex]\bold{ x = 12}[/tex]
Hence, The value of x is 12
Part 6 :-
Here also ,
- Two non parallel lines are intersected by three parallel lines .
- These three parallel lines are also parallel to each other.
By using basic proportionality theorem :-
[tex]\sf{\dfrac{ 4 }{ 4 + 10 }}{\sf{=}}{\sf{\dfrac{6}{x +6}}}[/tex]
[tex]\sf{\dfrac{ 4 }{ 14 }}{\sf{=}}{\sf{\dfrac{6}{x +6}}}[/tex]
By cross multiplication ,
[tex]\sf{ 4(x + 6) = 6 {\times} 14}[/tex]
[tex]\sf{ 4x + 24 = 84}[/tex]
[tex]\sf{ 4x = 84 - 24}[/tex]
[tex]\sf{ x = }{\sf{\dfrac{ 60}{4}}}[/tex]
[tex]\bold{ x = 15}[/tex]
Hence, The value of x is 15 .
[tex]\bold{\huge{\underline{ Solution\: D }}}[/tex]
Part 7 :-
Here,
- Here, JL bisects Angle J in ΔHJK
- It means Angle JHL = Angle JKL
- The length of HJ = 12 , JK = 15 , HL = 18
Therefore,
In triangle JHL and In triangle JKL
By using similarity theorem
- It states that if the two triangles are similar then the corresponding sides of triangle are in proportion and corresponding angles are equal
That is,
[tex]\sf{\dfrac{ JH }{ JK}}{\sf{=}}{\sf{\dfrac{HL}{KL}}}[/tex]
Subsitute the required values,
[tex]\sf{\dfrac{ 12 }{ 15 }}{\sf{=}}{\sf{\dfrac{18}{KL}}}[/tex]
By cross multiplication
[tex]\sf{ 12( KL) = 18 {\times} 15}[/tex]
[tex]\sf{ 12(KL) = 270}[/tex]
[tex]\sf{ KL = }{\sf{\dfrac{ 270}{12}}}[/tex]
[tex]\bold{ KL = 22.5 }[/tex]
Hence, The length of KL is 22.5
Part 8 :-
Here also,
- Here, JL bisects Angle J in ΔHJK
- It means Angle JHL = Angle JKL
- The length of KL = 6 , JK = 20 , HJ = 14
Therefore,
In triangle JHL and In triangle JKL
By using similarity theorem :-
[tex]\sf{\dfrac{ JH }{ JK}}{\sf{=}}{\sf{\dfrac{HL}{KL}}}[/tex]
Subsitute the required values,
[tex]\sf{\dfrac{ 14 }{ 20}}{\sf{=}}{\sf{\dfrac{HL}{6}}}[/tex]
[tex]\sf{\dfrac{ 7 }{ 10}}{\sf{=}}{\sf{\dfrac{HL}{6}}}[/tex]
[tex]\sf{\dfrac{ 7 }{ 10}}{\sf{{\times} 6= HL }}[/tex]
[tex]\sf{\dfrac{ 7 }{ 5 }}{\sf{{\times} 3 = HL }}[/tex]
[tex]\sf{\dfrac{ 21 }{ 5}}{\sf{ = HL }}[/tex]
[tex]\bold{ HL = 4.2 }[/tex]
Hence, The length of HL is 4.2 .
