If base of a triangle is $6.5$ and its height is $6.5sqrt3$ , what is its area?

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Answer 1 · 2017-11-15T04:37:41

Area is $(42.25sqrt3)/2$ or $36.5885$ square units

Area of any triangle is $1/2xx"Base"xx"Height"$

Here $"Base"=6.5$

$"Height"=6.5sqrt3$

Hence $"Area"=1/2xx6.5xx6.5xxsqrt3=(6.5^2xxsqrt3)/2$

$=(42.25sqrt3)/2$

$=(42.25xx1.732)/2=36.5885$ square units

Answer 2 · 2017-11-15T04:38:27

It is $21.125sqrt3$ .

The formula for the area of a triangle is:

$A=B*H/2$

where B is the base and H is the Height.

Therefore, in your problem:

$A=6.5*6.5sqrt3/2$ = $21.125sqrt3$

If base of a triangle is 6.56.5 and its height is 6.5sqrt36.5sqrt3, what is its area?