If ABC~DEF and the perimeter of ABC is 4 times greater than the perimeter of DEF, what is the relationship between the areas of the
triangles?

The area of triangle ABC is 16 times larger than the area of triangle DEF

Step-by-step explanation:

Step 1: Re-state the input

Given two triangles ABC and DEF are similar with ratio k.

Step 2: Derive the information

Suppose that the triangle ABC is k times larger than triangle DEF.

=> It means that all sides and heights of triangle ABC are k times larger than corresponding sides and heights of triangle DEF.

For example:

AB = k x DE

AC = k x DF

BC = k x EF

=> AB + AC + BC = k x (DE + DF + EF)

=> Perimeter of triangle ABC is k times larger than perimeter of triangle DEF

Step 3: Solve the problem step-by-step

Suppose that AH is the height from vertex A to base BC of triangle ABC and DH' is the height from the vertex D to base EF of triangle DEF.

As the above assumption states:

AH = k x DH'

BC = k x EF

Multiply both sides of the two above equations:

=> AH x BC = DH x EF x k x k

Simplify k components:

=> AH x BC = DH x EF x k^2

Multiply (1/2) to both sides:

=> (1/2) x AH x BC = (1/2) x DH x EF x k^2

Apply the formula to calculate the area of a triangle:

=> Area of triangle ABC = Area of triangle DEF x k^2

In other words:

The area of triangle ABC is (k^2) times larger than area of triangle DEK.

Here, the perimeter of triangle ABC is 4 times larger than the perimeter of triangle DEF => k = 4

Step 4: Make the conclusion

=>The area of triangle ABC is (4^2 = 16) times larger than area of triangle DEK.

Hope this helps!

:)

12m

step-by-step explanation:

square root means have of a number so have of 3 is 1.5 then multiply it by 8 and you get your answer!