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
28.27 is the area of the circle with a diameter of 6
Let ∠RTS=∠RST = a (say)
∠QUA=∠QSU= b(say)
then we know , at point S, a+40+b=180. so, a+b=140 we'll use this later.
consider trianglePQR, ∠P+∠Q+∠R=180
i.e.P+(180-2b)+(180-2a)=180
P+180+180-2(a+b)=180 ⇒P+180-2(a+b)=0 ⇒P=2(140)- 180=280-180=100
hence,answer is E
i cant tell you that it is from a test
Answer:
3. 16
Explanation:
Factor the numerator and denominator and cancel the common factors.
