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:
The area of the triangle formed by origin, and the points (4,5) and (-2,5) will be 15 sq. units.
Step-by-step explanation:
The two parabolas are and y = x² - 2x - 3.
Now, solving those two equations we will get the points of intersection.
So,
⇒ - (x - 1)² + 24 = 3x² - 6x - 9
⇒ -x² + 2x - 1 + 24 = 3x² - 6x - 9
⇒ 4x² - 8x - 32 = 0
⇒ x² - 2x - 8 = 0
⇒ x² - 4x + 2x - 8 = 0
⇒ (x - 4)(x + 2) = 0
So, x = 4 or - 2.
Now, for x = 4 , y = 4² - 2(4) - 3 = 5 and the point of intersection is (4,5).
Or, for x = - 2, y = (- 2)² - 2(- 2) - 3 = 5 and the point of intersection is (-2,5).
Now, points (4,5) and (-2,5) make a straight line parallel to the x-axis at a perpendicular distance of 5 units from origin and its length is (4 - (- 2)) = 6 units.
So, the area of the triangle formed by origin, and the points (4,5) and (-2,5) will be = 0.5 × 5 × 6 = 15 sq. units. (Answer)