Jk,kl, and lj are all tangent to circle o, ja=13,al=7, and ck=10 what is the perimeter of angle JKL
1 answer:
see the picture attached to better understand the problem
we know that
two tangent segments drawn from the same exterior point are congruent
so
JA=JB ,
LA=LC,
KC=KB
JA=13 units
LA=7 units
kC=10 units
hence
perimeter = JA+JB+LA+LC+KC+KB------> 13+13+7+7+10+10------> =60 units
therefore
the answer is
the perimeter of triangle JKL is 60 units
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Answer:
An equation in point-slope form of the line that passes through (-4,1) and (4,3) will be:
Step-by-step explanation:
Given the points
- (-4,1)
- (4,3)
Finding the slope between the points (-4,1) and (4,3)
Refine
Point slope form:
where
- m is the slope of the line
- (x₁, y₁) is the point
in our case,
- m = 1/4
- (x₁, y₁) = (-4,1)
substituting the values m = 1/4 and the point (-4,1) in the point slope form of line equation.
Thus, an equation in point-slope form of the line that passes through (-4,1) and (4,3) will be: