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# Help plz

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203
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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

#2
+7266
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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
#1
+555
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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
+7266
+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