+0

# Help plz

+1
105
2

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
+6588
+2

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

hectictar  Nov 28, 2017
Sort:

#1
+541
+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
+6588
+2

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

hectictar  Nov 28, 2017

### 25 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details