In a certain isosceles right triangle, the altitude to the hypotenuse has length \(4\sqrt2\). What is the area of the triangle?
drawing the triangle and setting the two equal sides as x and the height is 4sqrt2
we see that the height splits the 90 degree angle into two 45 degree angles and makes a 90 degree angle at the base
now we can use sin of o/h or sin 45 = 4sqrt2 / x which gives us x = 8
now we can use pythagorean to see that 8^2+ 8^2 = B^2 or the base is 11.3
now we apply the area formula bh/2 for area.
4sqrt2 * 11.3 = 64/2 = 32 so area is 32