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TTriangle H L I is shown. Line segment J K is drawn near point L to create triangle J L K. If △HLI ~ △JLK by the SSS similarity theorem, then StartFraction H L Over J L EndFraction = StartFraction I L Over K L EndFraction is also equal to which ratio? StartFraction H I Over J K EndFraction StartFraction H J Over J L EndFraction StartFraction I K Over K L EndFraction StartFraction J K Over H I EndFraction

Respuesta :

MrRoyal

Question:

[tex]\triangle[/tex]HLI is shown. Line segment JK is drawn near point L to create [tex]\triangle[/tex]JLK. If [tex]\triangle[/tex]HLI ~ [tex]\triangle[/tex]JLK by the SSS similarity theorem, then [tex]\frac{HL}{JL} =\frac{IL}{KL}[/tex] is also equal to which ratio?

[tex]\frac{H I}{J K}[/tex]      [tex]\frac{H J}{J L}[/tex]     [tex]\frac{I K}{K L}[/tex]     [tex]\frac{J K}{H I}[/tex]      

Answer:

[tex]\frac{H I}{J K}[/tex]

Step-by-step explanation:

Given

HLI ~ [tex]\triangle[/tex]JLK

[tex]\frac{HL}{JL} =\frac{IL}{KL}[/tex]

Required

Which other ratio equals [tex]\frac{HL}{JL} =\frac{IL}{KL}[/tex]

[tex]\triangle[/tex]HLI ~ [tex]\triangle[/tex]JLK  implies that:

HL, IL and HI corresponds to JL, KL and JK respectively.

So, the possible ratios are:

HL : IL : HI = JL : KL : JK

Convert to fractions

[tex]\frac{HL}{JL} = \frac{IL}{KL} = \frac{HI}{JK}[/tex]

So, from the list of options

[tex]\frac{H I}{J K}[/tex] is equivalent to [tex]\frac{HL}{JL} =\frac{IL}{KL}[/tex]

Answer:

its A

Step-by-step explanation:

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