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, 360, and 300 as the side’s lengths, what is the triangle’s area?
Pick: A) 34,200 B) 36,200 C) 38,200 D) 42,200 E) 49,089
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+360+300)/2= 
<gadget id="calculator">(360+360+300)/2</gadget>
<output>510</output>
510, area=√(510(510-360)(510-360)(510-300))=49,089. Therefore, the answer is 49,089.
<result>E</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, 360, and 300 as the side’s lengths, what is the triangle’s area?