116k views
4 votes
In the figure, △ABC is congruent to △ADC. If the square ABCD is dilated by a factor of 2 to form A'B'C'D', what is the ratio of the area of A'B'C'D' to the area of ABCD?

a) 2:1
b) 3:1
c) 4:1
d) 5:1

In the figure, △ABC is congruent to △ADC. If the square ABCD is dilated by a factor-example-1
User Jfmatt
by
8.9k points

2 Answers

4 votes

Answer:

4 : 1

Explanation:

User Jblejder
by
7.9k points
3 votes
The answer to this question would be: c) 4:1

In this question, the square is dilated by a factor of 2 to form A'B'C'D'. That mean the new square sides will be twice as much as the old square. The formula for area is side^2. Then, the area of the new square should be

a'=2a

old square area= a*a= a^2
new square area= a'*
a'= 2a*2a= 4a^2

new square area:old square area = 4a^2:a^2= 4:1
User Can Aksoy
by
8.4k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories