the area of a triangle can be determined by A=Square Root S(s-a)(s-b)(s-c).in the formula A is the area;a,b and c are the side lengths; and s is half the perimeter or a+b+c divided by two. determine the area of a triangle with 15, 12, and 10 as the side lengths.express each area to the nearest tenth of a square meter.
+130477 S = the semi-perimeter = (15 + 12+ 10) / 2 = 18.5
So we have
A = √[(18.5)*(18.5-15)*(18.5-12)*(18.5-10)]= √[18.5 * 3.5 * 6.5 * 8.5]= 59.8 m2
BTW......this "formula" is known as "Hero's Formula" or "Heron's Formula".....however, many mathematicians believe that it bears the distinct "pawprint" of Archimedes...the proof of it, geometrically, is somewhat involved......something at which Archimedes excelled !!!
+23254 Since a = 15 b = 12 and c = 10
then s = ( 15 + 12 + 10 ) / 2 ---> s = 37/2 = 18.5
Area = √( 18.5 x ( 18.5 - 15 ) x ( 18.5 - 12 ) x ( 18.5 - 10 ) ) = √ ( 18 x 3.5 x 6.5 x 8.5 )
= √ 3480.75 ≈ 59.0
This formula is known as Hero's Formula or Heron's Formula.
+130477 S = the semi-perimeter = (15 + 12+ 10) / 2 = 18.5
So we have
A = √[(18.5)*(18.5-15)*(18.5-12)*(18.5-10)]= √[18.5 * 3.5 * 6.5 * 8.5]= 59.8 m2
BTW......this "formula" is known as "Hero's Formula" or "Heron's Formula".....however, many mathematicians believe that it bears the distinct "pawprint" of Archimedes...the proof of it, geometrically, is somewhat involved......something at which Archimedes excelled !!!
