Geometry

Using Herons Formula,

sqrt(s(s-AB)(s-BC)(s-CA)) = sqrt(17(17-15)(17-9)(17-10)) = sqrt(17 * 2 * 8 * 7) = sqrt(1904) = 4√119

Then, you solve for the altitudes:

BC_a = (2 * A) / BC = (2 * 4√119) / 9 = (8√119) / 9

CA_a = (2 * A) / CA = (2 * 4√119) / 10 = (8√119) / 10 = (4√119) / 5

AB_a = (2 * A) / AB = (2 * 4√119) / 15 = (8√119) / 15

The shortest one is (8√119) / 15.

Find the length of the shortest altitude in this triangle.

general circle equation

$(x – x_P)² + (y – y_P)² = r²$

$y^2=10^2-x^2\\ y^2=9^2-(15-x)^2\\ y^2=81-225+30x-x^2\\ 100-x^2=-144+30x-x^2\\30x=244\\ x_C=8.1\overline{33}\\ \color{blue}y_C=h_C=5.818$

$y^2=15^2-x^2\\ y^2=10^2-(9-x)^2\\ y^2=100-x^2+18x-81\\ 225-x^2=19+18x-x^2\\ 18x=206\\ x_A=11.\overline{44}\\ y_A=h_A=9.697$

$y^2=9^2-x^2\\ y^2=15^2-(10-x)^2\\ y^2=225-100+20x-x^2\\ 81-x^2=125+20x-x^2\\20x=-44\\ x_B=-2.2\\ y_B=h_B=8.726$

$\color{blue}The\ length\ of\ the\ shortest\ altitude\ in\ this\ triangle\ is\ h_C=5.818$

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