Answer:
[tex]x = 5[/tex]
Step-by-step explanation:
Given
[tex]\triangle JDC[/tex] similar to [tex]\triangle JKL[/tex]
Required
Find x
To do this, we make use of the following equivalent ratios
[tex]JK:JL = JD:JC[/tex]
This gives
[tex]98 : 63 = 7x + 7:27[/tex]
Express as fraction
[tex]\frac{98 }{ 63} =\frac{ 7x + 7}{27}[/tex]
Multiply both sides by 27
[tex]27 * \frac{98 }{ 63} =\frac{ 7x + 7}{27} * 27[/tex]
[tex]27 * \frac{98 }{ 63} =7x + 7[/tex]
[tex]3* \frac{98 }{ 7} =7x + 7[/tex]
[tex]3* 14 =7x + 7[/tex]
[tex]42 =7x + 7[/tex]
Collect like terms
[tex]7x = 42 - 7[/tex]
[tex]7x = 35[/tex]
Divide both sides by 7
[tex]x = 5[/tex]
