SME
Grade 10 Math · Problem type
The incircle of a right triangle
Use this for “A right triangle has legs 9 and 12. Find the radius of the circle inside it.”
Diagram
Each angle bisector splits its angle into two equal parts (equal arcs).
They meet at the incenter I, the center of the circle that touches all three sides.
r = 3
This is an example. Type your own values to start a new problem.
r
=
+
−
2
Formulas
c
=
√
a
2
+
b
2
15
=
√
9
2
+
12
2
r
=
a
+
b
−
c
2
3
=
9 + 12 − 15
2
Assumptions
•
The angle bisectors meet at the incenter I, the same distance r from all three sides.
•
The two tangent lengths from each corner are equal, and at the right angle they are both r.
•
So the legs share the hypotenuse plus 2r: r = (a + b − c) ÷ 2.
•
Lengths are in any one unit.
Step-by-step
WE KNOW
a = 9, b = 12
FIND
hypotenuse AB (c), inradius r (r)
Step 1 · Find hypotenuse AB (c)
Use c = √(a² + b²)
The Pythagorean theorem gives the hypotenuse.
c
=
√
9
2
+
12
2
c
=
√
81 + 144
c
=
√
225
c
=
15
Step 2 · Find inradius r (r)
Use r = (a + b − c) ÷ 2
The
two
legs
add
to
the
hypotenuse
plus
2
r
:
take
c
away
and
halve.
r
=
9 + 12 − 15
2
r
=
6
2
r
=
3
Check
Put the numbers back in. Both sides should match.
15
=
√
9
2
+
12
2
✓
3
=
9 + 12 − 15
2
✓
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Medians and the centroid
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The circle through the corners of a right triangle
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Relationships within triangles: bisectors, medians, midsegments and inequalities
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