Question

The sides of a triangle are $25,$ $39,$ and $40.$ The diameter of the circumscribed circle is:

$\dfrac{133}{3}$

$\dfrac{125}{3}$

$42$

$41$

$40$

Answer: B

Concepts:Heron’s Formula circumcircle, circumcenter, and circumradius triangle area

Difficulty rating: 1750

Small Hint:

Use Heron’s formula with semiperimeter $52$

Big Hint:

After finding the area $K,$ use $abc=4KR$ and double the circumradius

Solution:

The semiperimeter is $52,$ so Heron’s formula gives $K=\sqrt{52\cdot27\cdot13\cdot12}=468.$ If $R$ is the circumradius, then $R=\frac{25\cdot39\cdot40}{4\cdot468}=\frac{125}{6}.$ Hence the diameter is $2R=\frac{125}{3}.$

Thus, the correct answer is B.

1959 AMC 12 Problem 43

Answer: B

Difficulty rating: 1750

Small Hint:

Big Hint:

Solution:

Problem 43 in Other Years