What is the area of a triangle whose sides have length square root of 70, square root of 70 and square root of 28?

Guest Nov 28, 2017

#2**+2 **

From the Pythagorean theorem...

h^{2} + [ √28 / 2 ]^{2} = [ √70 ]^{2}

h^{2} + 28 / 4 = 70

h^{2} + 7 = 70

h^{2} = 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

hectictar
Nov 28, 2017

#1**+1 **

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 \(28\sqrt{2}\).

helperid1839321
Nov 28, 2017

#2**+2 **

Best Answer

From the Pythagorean theorem...

h^{2} + [ √28 / 2 ]^{2} = [ √70 ]^{2}

h^{2} + 28 / 4 = 70

h^{2} + 7 = 70

h^{2} = 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

hectictar
Nov 28, 2017