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

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

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.

To find the area of a triangle, you can use Heron's formula. Heron's formula states that the area of a triangle with side lengths a, b, and c is equal to the square root of s multiplied by (s - a) multiplied by (s - b) multiplied by (s - c), where s is half of the sum of the side lengths.

In this case, the side lengths of the triangle are 15, 15, and 8. To find the area, we first need to calculate s:

s = (15 + 15 + 8) / 2 = 19

Now, we can calculate the area using Heron's formula:

Area = sqrt(s * (s - 15) * (s - 15) * (s - 8))

Area = sqrt(19 * (19 - 15) * (19 - 15) * (19 - 8))

Area = sqrt(19 * 4 * 4 * 11)

Area = sqrt(304)

Hence, the area of the triangle is approximately 17.46 square units.