If ABCDE and LMNOP are similar polygons, then the ratio of AB to LM must be equal to the ratio of CD to NO. (Assume and are corresponding sides, as are and .)
True
To prove that the ratio of AB to LM is equal to the ratio of CD to NO, we can use the concept of similar polygons.
First of all, let's define what it means for two polygons to be similar. Two polygons are said to be similar if their corresponding angles are congruent and the lengths of their corresponding sides are proportional. In other words, if two polygons have the same shape, but possibly different sizes, they are considered similar.
In this case, we are given that ABCDE and LMNOP are similar polygons. This means that the corresponding angles of ABCDE and LMNOP are congruent. Additionally, it implies that the lengths of the corresponding sides of ABCDE and LMNOP are proportional.
Let's compare the ratios of AB and LM and CD and NO:
Ratio of AB to LM:
AB/LM
Ratio of CD to NO:
CD/NO
Since ABCDE and LMNOP are similar polygons, the corresponding sides AB and LM are proportional, and the corresponding sides CD and NO are also proportional. Therefore, we can write:
AB/LM = CD/NO
This shows that the ratio of AB to LM is equal to the ratio of CD to NO.
To summarize, if ABCDE and LMNOP are similar polygons, the ratio of AB to LM is indeed equal to the ratio of CD to NO. This can be proven by using the properties of similar polygons, specifically the fact that corresponding sides are proportional.
Yes, that statement is correct. If ABCDE and LMNOP are similar polygons, it means that their corresponding angles are equal and their corresponding sides are proportional.
In particular, if AB and LM are corresponding sides, and CD and NO are also corresponding sides, then the ratio of AB to LM must be equal to the ratio of CD to NO. This can be represented mathematically as:
AB/LM = CD/NO
So, if ABCDE and LMNOP are similar polygons, the ratio of AB to LM is indeed equal to the ratio of CD to NO.