Prove algebraically that the ratio of perimeters of two triangles is 1:4 if the ratio of their corresponding sides is 1:4
Square the ratios, and what do you get?
sides a, b c
perimeter = a+b+c
perimeter 4 a + 4 b + 4 c =4(a+b+c)
not area, perimeter
Oh sorry, I mis-read. Thank you for pointing it out.
You are welcome :)
To prove algebraically that the ratio of perimeters of two triangles is 1:4, given that the ratio of their corresponding sides is 1:4, we can use the fact that the perimeter of a triangle is simply the sum of the lengths of its three sides.
Let's consider two triangles, Triangle ABC and Triangle DEF, with corresponding sides AB and DE having a ratio of 1:4. We can express the lengths of the other corresponding sides as BC = 4x and EF = 4y, where x and y are positive real numbers.
The perimeter P1 of Triangle ABC is given by:
P1 = AB + BC + AC = 1x + 4x + AC = 5x + AC
The perimeter P2 of Triangle DEF is given by:
P2 = DE + EF + DF = 1y + 4y + DF = 5y + DF
Since we want to prove that the ratio of their perimeters is 1:4, we need to show that P2 = 4P1.
Substituting the expressions for the perimeters of the triangles, we have:
5y + DF = 4(5x + AC)
Expand the right side:
5y + DF = 20x + 4AC
Rearrange the equation:
DF = 20x + 4AC - 5y
To prove that the ratio of the perimeters is 1:4, we need to show that the value of DF is four times the value of AC.
We know that the ratios of the sides are 1:4, so we can express AC in terms of x:
AC = 4x
Substituting this into the equation above, we get:
DF = 20x + 4(4x) - 5y
DF = 20x + 16x - 5y
DF = 36x - 5y
Now let's substitute DF = 36x - 5y into the perimeter equation for Triangle DEF:
P2 = 5y + (36x - 5y)
P2 = 36x
We can see that P2 = 36x and P1 = 5x + AC = 5x + 4x = 9x
Therefore, P2 = 4P1, proving that the ratio of perimeters of two triangles is 1:4 when the ratio of their corresponding sides is 1:4.