Asked by Edward

A triangle has side lengths 10,17 and 21cm. Find the length of the shortest altitude.

And please show working out to help me understand.

Answers

Answered by Steve
if a,b,c = 10,17,21,

a^2 = b^2+c^2-2bc cos C, so
cosC = (17^2 + 21^2 - 10^2)/(2*17*21) = 15/17
so, sinC = 0.47

figure sinA and sinB the same way

If the altitudes from A,B,C are Ha,Hb,Hc, then

Hc = a*sinB = b*sinA

Figure the other altitudes in like wise.

OR--------
figure the area using Heron's formula.
Then since 1/2 c*Hc = area
Hc = 2*area/c

Since the area does not change, the shortest altitude is on the longest side.
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