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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

Best Answer 

 #2
avatar+5589 
+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
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2+0 Answers

 #1
avatar+505 
+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
avatar+5589 
+2
Best Answer

 

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

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