It is well known that a triangle’s area is √(p(p-a)(p-b)(p-c)),
when p=(a+b+c)/2, such that a, b, c are the lengths of sides of the triangle. If the triangle has 360, 300, and 360 as the side’s lengths, what is the triangle’s area?  Answers.
A) 34,200  B) 36,200  C) 38,200  D) 49,089  E) 43,200
It is well known that a triangle’s area is √(p(p-a)(p-b)(p-c)),   when p=(a+b+c)/2,   -&gt; P=(360+300+360)/2= 
<gadget id="calculator">(360+300+360)/2</gadget>
<output>510</output>
510, area=√(510(510-360)(510-300)(510-360))=49,089. Therefore, the answer is 49,089.
<result>D</result>
It is well known that a triangle’s area is √(p(p-a)(p-b)(p-c)),
when p=(a+b+c)/2, such that a, b, c are the lengths of sides of the triangle. If the triangle has 360, 300, and 360 as the side’s lengths, what is the triangle’s area?