Answer:
The area of triangle ABC is 16 times larger than the area of triangle DEF
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!
:)