Perimeter of polygons with an inscribed circle

🟢 What is an Inscribed Circle (Incircle)?

  • A circle drawn inside a polygon that touches all its sides.
  • The point where the circle touches a side is called the point of tangency.
  • The center of the circle is called the incenter (🔺 for triangles).

🟣 Key Property of Tangents from a Point

👉 If two tangents are drawn from the same vertex to the circle,
then they are equal in length.

Example in a quadrilateral:

  • From vertex A, the two tangent lengths are AP = AS.
  • From vertex B, the two tangent lengths are BP = BQ, and so on.

This property is the secret behind finding the perimeter! 🔑✨


🔵 How Perimeter Is Found Using Tangent Lengths

Each side of the polygon is formed by adding two tangent segments,
one from each of its endpoints.

Example for a quadrilateral ABCD:

  • Side AB = AP + BP
  • Side BC = BQ + CQ
  • Side CD = CR + DR
  • Side DA = DS + AS

🟠 Amazing Result: Perimeter = Sum of All Tangent Lengths × 2

Since every tangent length repeats twice (once for each side),
the perimeter = (AP + BP + CQ + DR + …) × 2.


🟢 Special Case — Triangle

For a triangle, if tangent lengths are x, y, z at three vertices,
then:

  • Perimeter = x + y + z + x + y + z = 2(x + y + z)
  • Or simpler:
    Perimeter = sum of the three sides (as usual!)
    because each side is formed by adding tangent lengths.

🔴 Formula Summary

Perimeter of polygon with an inscribed circle

P = 2(x1 + x2 + x3 + …)

where x1 , x2 , x3… are tangent lengths from each vertex.


🟣 Visual Shortcut

  • Tangent lengths from each vertex are equal.
  • Add all tangent lengths.
  • Multiply by 2.
  • 🎉 You get the perimeter!

🟡 Why Useful?

This method is useful when:
✔ Side lengths are not given
Tangent lengths (from vertices to points of tangency) are provided

FG= _______

Look at the diagram:

Find the unknown segment lengths.

JK and IJ are tangents to the inscribed circle from J. So , JK is congruent to IJ.
JK=IJ=8.

You know JK and FJ. Use the Additive Property of Length to write an equation and find FK.

FK+JK= FJ Additive Property of Length

FK + 8 = 13 Plug in JK=8 and FJ=13

FK = 5 Subtract 8 from both sides

So, FK is 5.

FG and FK are tangents to the inscribed circle from F. So , FG is congruent to FK.
FG=FK=5.

What is TU?

TU=______

Look at the diagram:

Find the unknown segment lengths.

SX and WX are tangents to the inscribed circle from X. So , SX is congruent to WX .

SX=WX=1.

You know SX and TX. Use the Additive Property of Length to write an equation and find ST.

SX + ST = TX Additive Property of Length

1 + ST = 2 Plug in SX=1 and TX=2

ST= 1 Subtract 1 from both sides

So, ST is 1.

TU and ST are tangents to the inscribed circle from T. So , TU is congruent to ST.
TU=ST=1.

let’s practice!