Question

Find area of a triangle with side lengths 15 15 and 8

Answer 1

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

Answer 2

Thank you!

Answer 3

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.