Question

Question #d8f28

Answer 1

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`

Answer 2

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!!

Answer 3

`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