Question
Find area of a triangle with side lengths 15 15 and 8
Answers
Reiny
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
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
Anonymous
Thank you!
John
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.