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The segments are tangents to the circle. Find the perimeter of JLNQ.

The perimeter of the polygon is

User Thecheech
by
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1 Answer

5 votes

Answer:

36 Units.

Explanation:

In the diagram attached, K, M, P and R are points of tangency.

Theorem: Tangents to a circle from the same point are equal.

By the theorem above,

LK=LM=6 Units

NM=NP=2 Units

QP=QR=7 Units

JR=JK=3 Units

Therefore:

JL=JK+KL=3+6=9 Units

LN=LM+LN=6+2=8 Units

NQ=NP+PQ=2+7=9 Units

QJ=QR+RJ=7+3=10 Units

Therefore, the perimeter of the polygon JLNQ =9+8+9+10

=36 Units.

Note: The second diagram shows the theorem already applied.

The segments are tangents to the circle. Find the perimeter of JLNQ. The perimeter-example-1
The segments are tangents to the circle. Find the perimeter of JLNQ. The perimeter-example-2
User Xiaochen
by
9.8k points

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