What is the area of a triangle whose sides have length square root of 70, square root of 70 and square root of 28?
+9481 
From the Pythagorean theorem...
h2 + [ √28 / 2 ]2 = [ √70 ]2
h2 + 28 / 4 = 70
h2 + 7 = 70
h2 = 63
h = √63
And...
area of triangle = (1/2)(base)(height)
area of triangle = (1/2)(√28)(√63)
area of triangle = (1/2)(√1764)
area of triangle = (1/2)(42)
area of triangle = 21 sq. units
+633 I can tell this is an isosceles triangle, since 2 of the sides have the same length. This means that the base of the triangle is \(2\sqrt{7}\). Because of this, I know that the base in each of the sides (if you figutre out the length via the Pythagorean Theorem) is the square root of 14 I subtract from the number inside the square root to get my height, which is \(2\sqrt{14}\). By multiplying these togther and simplifying the root, I get \(28\sqrt{2}\).
+9481
From the Pythagorean theorem...
h2 + [ √28 / 2 ]2 = [ √70 ]2
h2 + 28 / 4 = 70
h2 + 7 = 70
h2 = 63
h = √63
And...
area of triangle = (1/2)(base)(height)
area of triangle = (1/2)(√28)(√63)
area of triangle = (1/2)(√1764)
area of triangle = (1/2)(42)
area of triangle = 21 sq. units
