Given:
Bl parallel to RA.
To find:
The value of x.
Solution:
In triangle ATR and ITB,
[tex]\angle ATR\cong \angle ITB[/tex] [Common angles]
[tex]\angle ART\cong \angle IBT[/tex] [Corresponding angle]
[tex]\triangle ATR\sim \triangle ITB[/tex] [AA property of similarity]
We know that the corresponding sides of a similar triangle are proportional. So,
[tex]\dfrac{AT}{IT}=\dfrac{AR}{IB}[/tex]
[tex]\dfrac{x}{x+8}=\dfrac{12}{18}[/tex]
[tex]\dfrac{x}{x+8}=\dfrac{2}{3}[/tex]
On cross multiplication, we get
[tex]3(x)=2(x+8)[/tex]
[tex]3x=2x+16[/tex]
[tex]3x-2x=16[/tex]
[tex]x=16[/tex]
Therefore, the correct option is C.
