Problem
A circle is inscribed in an isosceles triangle with two equal sides of $10 \ inches$ and a base of $12 \ inches$.
What is the diameter of the inscribed circle?
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Solution
If we draw a perpendicular line $h$ from the vertex of this triangle to the base,
$$h^2=b^2-\left( \dfrac{a}{2} \right)^2,$$
$$∴ \ h=\sqrt{b^2-\left( \dfrac{a}{2} \right)^2 }=\sqrt{10^2-6^2}=8.$$
If the area of this isosceles triangle is $S$, then
$$S=\frac{1}{2}×a×h=\frac{ah}{2}. \qquad [1]$$
Also, by the Heron’s formula,
$$S=\frac{1}{2}×(a+b+c)×\frac{d}{2}=\frac{d(a+b+c)}{4}. \qquad [2]$$
From $[1]$ and $[2]$,
$$\frac{ah}{2}=\frac{d(a+b+c)}{4},$$
$$∴ \ d=\frac{2ah}{a+b+c}. \qquad [3]$$
Substituting $a=12, \ b=c=10$, and $h=8$ into $[3]$, we get
About the “Jutsu” of This Problem
The “jutsu” (general solution) to this problem states:
“[To find the diameter of the inscribed circle] using the Pythagorean theorem, find the length of the perpendicular line from the vertex to the base. Moreover, multiply this by $12 inches$ (the length of the base), double it, and divide by the sum of the three sides.”
Reference
Yoshikazu Yamakawa, ed. (1997) Okayama ken no Sangaku (Sangaku in Okayama Prefecture), p.41; p.343.