Question

Calculate the area of triangle ABC if the sides are 10, 14, and 15. I just dont get it??

Answer 1

simple
s=(10+14+15)/2=19.5
\[area=\sqrt{(19.5*9.5*5.5*4.5)}\]

Answer 2

=67.7 hope its correct

Answer 3

the area is \[\sqrt{s(s-a)(s-b)(s-c)} \] a,b,c are the sides

Answer 4

Area of triangle = base by height over 2

Answer 5

ok thank you so much :)

Answer 6

1) Let a=10, b=14 and c=15 and A, B and C are the angles directly opposite the sides in triangle ABC
2) Use cosine rule to work out an angle (A) \[a ^{2}=b ^{2}+c ^{2}-2bc \cos A\]
or \[\cos A=(b ^{2}+c ^{2}-a^{2})/(2bc)\]
\[\cos A=(14 ^{2}+15 ^{2}-10^{2})/(2\times14\times15)\]
\[\cos A=(196+225-100)/420\]
\[\cos A = 0.76428....\]
\[A=\cos^{-1} (0.764286...)=40.15^{o}\]
3) Area of triangle = \[(1/2)bc \sin A\]
\[0.5\times14\times15\times \sin 40.15^{o}\]
\[Area = 67.7\] square units