Question #d8f28

3 Answers
Oct 7, 2017

Area, A = #16 root\ 3#

Explanation:

We can apply Heron's formula to find area of the given triangle.

Area of triangle A = # root \ ((s(s-a)(s-b)(s-c))#

where, #a,b and c# are sides of the triangle and #s# is the semiperimeter of the triangle and is given as:

#s= (a+b+c)/ 2#

In equilateral triangle #a=b=c#

Here #a# is given as 8.

So semiperimeter #s = (a+a+a)/2 =(3a)/2# = # (3\timescancel8^4)/cancel2^1 = 12#

Area, A = # root \ ((s(s-a)(s-b)(s-c))#

A =# root \ ((3a)/2((3a)/2-a)((3a)/2-a)((3a)/2-a))#

A = # root\ ((3a)/2((3a-2a)/2)((3a-2a)/2)((3a-2a)/2))#

A = # root \ ((3a)/2((a)/2)((a)/2)((a)/2))#

A= # root \ ((3a^4)/2^4# =# root \ ((3a^4)/16#

A = #root \ 3 \times a^2/4#

Here #a# =8

A = #root \ 3 \times 8^2/4#

A = #root \ 3 \times 64/4#

A = #root \ 3 \times 16# = #16 root\ 3#

or
A = # root \ ((s(s-a)(s-b)(s-c))#

A = # root \ ((12(12-8)(12-8)(12-8))#

A = # root \ ((12(4)(4)(4))#

A = # root \ ((3\times(4)(4)(4)(4))#

A = # root \ ((3\times 256))#

A = #16 root\ 3#

Oct 7, 2017

Let's make it easy.

Explanation:

Let, each side of length #8#units be #"a"#.
So, #a=8#.

Now, Area of a triangle, #A=1/2xxhxxb#....(1)
where, #h# is height & #b# is base.
Here, #b=a=8#....(2)

Now, by Hypotenuse Principle #rarr#
#h=sqrt(a^2-(b/2)^2)#.
#:.h=sqrt(a^2-(a/2)^2)#
...From (2).
#:.h=sqrt((3a^2)/4)#.

So, #h=(asqrt(3))/2#.

Therefore, from (1) #rarr#
#A=1/2.a.(sqrt(3)a)/2#.

#:.A=(sqrt(3)a^2)/4#.

Now, substituting the respective values #rarr#
#A=(sqrt(3).8^2)/4#.
#:.A=16sqrt(3)#.
(Answer).

Hope it Helps!!

Oct 7, 2017

#A= 27.71# square units

Explanation:

The simplest and most direct method of finding the area of an equilateral triangle is by using the trig area rule.

All the sides are #60°# and all the sides are of length #8# units.

Therefore the requirements of two sides and the included angle are met.

#Area = 1/2 a*b*sinC#

#A= 1/2 *8*8*sin60#

#A=1/2*8*8*sqrt3/2#

#A=16sqrt3#

#A= 27.71# square units