simplest way:
Heron's Formula
A = √(s(s-a)(s-b)(s-c)), where s = (1/2)the perimeter.
s = (1/2)(15 + 15 + 8) = 19
s-a = 19-15 = 4
s-b = 19-15 = 4
s-c = 19-8 - 11
area = √(19(4)(4)(11) = √3344
= appr 57.83
2nd way: works in this case because it is isosceles.
Sketch the triangle, let the angle between the two equal sides be 2θ
Draw a perpendicular from that angle to the base.
sinθ = 4/15
θ = 15.466..
2θ = 30.932..
area = (1/2)(15)(15)sin30.932..
= 57.83 , same as above
There are other ways, but these two work nicely. The 2nd way of course only worked so fine, because we had an isosceles triangle
Find area of a triangle with side lengths 15 15 and 8
3 answers
Thank you!
An easier way would be to find the perpendicular and use pythag. Thm
c^2 = a^2 + b^2
15^2 = 4^2 + b^2
b = 14.45
A = 1/2 bh
A= 1/2 8 x 14.45
you will get the same answer as above.
Depending on the level you are working at this might be what your teacher is looking for.
c^2 = a^2 + b^2
15^2 = 4^2 + b^2
b = 14.45
A = 1/2 bh
A= 1/2 8 x 14.45
you will get the same answer as above.
Depending on the level you are working at this might be what your teacher is looking for.
0 answers