Question

1. \\(\overline{jk}\\), \\(\overline{kl}\\), and \\(\overline{lj}\\) are all tangent to circle o. \\(ja = 13\\), \\(al = 7\\), and \\(ck = 10\\). what is the perimeter of \\(\triangle jkl\\)? \\(\bigcirc\\) 40 units \\(\bigcirc\\) 60 units

Explanation:

Step1: Recall Tangent Segment Theorem

Tangent segments from a common external point to a circle are congruent. So, \( JA = JB = 13 \), \( AL = CL = 7 \), and \( CK = BK = 10 \).

Step2: Calculate Sides of Triangle

• Side \( JL \): \( JA + AL = 13 + 7 = 20 \)
• Side \( JK \): \( JB + BK = 13 + 10 = 23 \) (Wait, no, wait. Wait, perimeter is \( JL + LK + JK \). Let's re - express:
• \( JL=JA + AL = 13 + 7=20 \)
• \( LK=CL + CK = 7+10 = 17 \)
• \( JK=JB + BK \), and since \( JB = JA = 13 \) and \( BK = CK = 10 \), \( JK = 13+10 = 23 \)
• Wait, another way: Perimeter \( P=JL + LK+JK=(JA + AL)+(CL + CK)+(JB + BK) \). But since \( JA = JB \), \( AL = CL \), \( CK = BK \), we can rewrite as \( P = 2(JA + AL+CK) \)
• Substitute \( JA = 13 \), \( AL = 7 \), \( CK = 10 \)
• \( P=2(13 + 7+10)=2\times(30)=60 \)

Answer:

60 units