Question

What must be the sides of an equilateral triangle so that it's area should be equal to the area of an isosceles triangle with base 12m and equal sides 10m

Answer 1

The sides must be around 10.53 m

Explanation:

You can calculate the area of isosceles dividing into 2 equal right triangles

Once you know that the area of the isosceles is 48 m²,solve for the side of the equilateral triangle

Formula for the Area: (A= area a = side)

A = (√3* a² ) / 4

48= (√3* a² ) / 4

Multiply both sides by 4

192 =√3* a²

Divide both sides by√3 ( √3 = 1.732050....)

110.851303 = a²

Square root on both sides

10 .5385...( 10.53 ) = a

Answer 2

The sides of the equilateral triangle are 11.7 m.

Explanation:

Let's find the area of the isosceles triangle:

`A_(i) = (bh)/(2)`

Where:

b: is the base = 12 m

h: is the height

We can find the height by using Pitagoras:

`x^(2) = (b^(2))/(2) + h^(2)`

Where:

x is the hypotenuse = side of the triangle = 10 m

`h = \sqrt{x^(2) - (b^(2))/(2)} = \sqrt{(10)^(2) - 6^(2)} = 8 m`

Then, the area is:

`A_(i) = (12*8)/(2) = 48 m^(2)`

Now, since the area of the isosceles triangle is equal to the area of the equilateral triangle:

`A_(e) = A_(i) = 48 m^(2)`

`A_(e) = (bh)/(2)`

The height of the equilateral triangle is given by:

`b^(2) = (b^(2))/(2) + h^(2)`

`h = \sqrt{b^(2) - (b^(2))/(2)} = (b)/(√(2))`

Hence, the sides are:

`A_(e) = (1)/(2)b(b)/(√(2)) = (b^(2))/(2√(2))`

`b = \sqrt{A*2√(2)} = \sqrt{48*2√(2)} = 11.7 m`

Therefore, the sides of the equilateral triangle are 11.7 m.

I hope it helps you!