Question 119304
Find the area of the right triangle whose base is 40cm and whose hypotenuse is twice the base or 80cm.
We will need to find the length of the third leg, or the height, of the triangle before we can find the area.
Using the Pythagorean theorem {{{c^2 = a^2+b^2}}} where the hypotenuse, c = 80, the base, a = 40, and b is h, the height.
{{{80^2 = 40^2+h^2}}} Simplify.
{{{6400 = 1600+h^2}}} Subtract 1600 from both sides.
{{{4800 = h^2}}} Take the square root of both sides.
{{{h = 40sqrt(3)}}}
Now we can calculate the area of the triangle which is given by:{{{A = (1/2)bh}}} where b is the length of the base or 40cm and h is the height, or {{{h = 40sqrt(3)}}}.
Making the apprpriate substitutions, we get:
{{{A = (1/2)(40)(40sqrt(3))}}} Simplifying this...
{{{A = 800sqrt(3)}}}
To the nearest tenth, this would be:
{{{A = 800(1.7321)}}}
{{{A = 1385.7}}}sq.cm