Question

the diagram shows the lengths of corresponding sides of similar triangles abc and abc. which expression gives the perimeter of abc?
a. 4(4 + b + c)
b. 2(a + b + c)
c. 4(a + b + c)
d. 8(a + b + c)

Explanation:

Step1: Find the scale factor

The corresponding sides of the similar triangles: in triangle \(ABC\), side \(a = 4\), and in triangle \(A'B'C'\), the corresponding side is \(8\). The scale factor from \(ABC\) to \(A'B'C'\) is \(\frac{8}{4}=2\).

Step2: Perimeter of \(ABC\)

The perimeter of triangle \(ABC\) is \(P_{ABC}=a + b + c=4 + b + c\) (since \(a = 4\)).

Step3: Perimeter of \(A'B'C'\)

For similar triangles, the ratio of perimeters is equal to the scale factor. Let the perimeter of \(A'B'C'\) be \(P_{A'B'C'}\). Since the scale factor is \(2\), \(P_{A'B'C'}=2\times P_{ABC}\). But wait, no—wait, actually, we need to check the direction. Wait, the side \(a = 4\) in \(ABC\) corresponds to \(8\) in \(A'B'C'\), so \(A'B'C'\) is larger by a scale factor of \(2\). Wait, no, the question is about the perimeter of \(A'B'C'\)? Wait, no, the question says "Which expression gives the perimeter of \(A'B'C'\)?" Wait, let's re - read. The diagram shows similar triangles \(ABC\) and \(A'B'C'\). Side \(a = 4\) (in \(ABC\)) and the corresponding side in \(A'B'C'\) is \(8\). So the scale factor \(k\) from \(ABC\) to \(A'B'C'\) is \(\frac{8}{4}=2\). The perimeter of \(ABC\) is \(P_{ABC}=a + b + c=4 + b + c\). Then the perimeter of \(A'B'C'\) is \(k\times P_{ABC}=2\times(4 + b + c)\)? Wait, no, wait, \(a = 4\), so \(P_{ABC}=4 + b + c\), and the scale factor is \(2\), so \(P_{A'B'C'}=2\times(4 + b + c)\)? Wait, no, wait, the options are in terms of \(a + b + c\). Wait, \(a = 4\), so \(4 + b + c=a + b + c\) (since \(a = 4\)). Wait, no, \(a = 4\), so \(P_{ABC}=a + b + c\), and the scale factor is \(2\), so \(P_{A'B'C'}=2\times(a + b + c)\)? Wait, no, wait the side \(a = 4\) (length of \(BC\)) and \(B'C'=8\), so the scale factor is \(2\). So all sides of \(A'B'C'\) are twice the sides of \(ABC\). So perimeter of \(ABC\) is \(a + b + c\), perimeter of \(A'B'C'\) is \(2(a + b + c)\)? Wait, no, wait \(a = 4\), so \(a + b + c=4 + b + c\), and the perimeter of \(A'B'C'\) is \(2\times(4 + b + c)=2(a + b + c)\) (since \(a = 4\)). Wait, but let's check the options. Option B is \(2(a + b + c)\), option C is \(4(a + b + c)\), option D is \(8(a + b + c)\), option A is \(4(4 + b + c)\). Wait, maybe I made a mistake. Wait, the perimeter of \(ABC\) is \(4 + b + c\), and the scale factor is \(2\), so perimeter of \(A'B'C'\) is \(2\times(4 + b + c)\). But \(4 + b + c=a + b + c\) (because \(a = 4\)), so \(2\times(a + b + c)\) is the same as \(2(4 + b + c)\). Wait, but let's re - calculate the scale factor. The length of \(BC\) is \(a = 4\), length of \(B'C'\) is \(8\). So the scale factor \(k=\frac{8}{4}=2\). The perimeter of a triangle is the sum of its sides. So if triangle \(ABC\) has sides \(a = 4\), \(b\), \(c\), its perimeter \(P_{ABC}=4 + b + c\). Then the perimeter of triangle \(A'B'C'\) (similar triangle) is \(k\times P_{ABC}=2\times(4 + b + c)\). But \(4 + b + c=a + b + c\) (since \(a = 4\)), so \(P_{A'B'C'}=2(a + b + c)\). Wait, but let's check the options again. Wait, maybe I misread the question. Wait, the question is "Which expression gives the perimeter of \(A'B'C'\)?" Let's check the options:

Option A: \(4(4 + b + c)\)

Option B: \(2(a + b + c)\)

Option C: \(4(a + b + c)\)

Option D: \(8(a + b + c)\)

Since \(a = 4\), \(4 + b + c=a + b + c\). The scale factor is \(2\), so perimeter of \(A'B'C'\) is \(2\times\) perimeter of \(ABC\). Perimeter of \(ABC\) is \(a + b + c\), so perimeter of \(A'B'C'\) is \(2(a + b + c)\). Wait, but let's confirm. If \(ABC\) has sides \(4\), \(b\), \(c\), perimeter \(4 + b + c\). \(A'B'C'\) has sides \(8\), \(2b\), \(2c\), so perimeter \(8 + 2b + 2c=2(4 + b + c)=2(a + b + c)\) (since \(a = 4\)). So the correct expression is \(2(a + b + c)\).

Answer:

B. \(2(a + b + c)\)