Video Transcript
Segment 𝐶𝐷 has mirror symmetry in
the line 𝐴𝐹. Given that 𝐸𝐷 is equal to five
and 𝐵𝐶 is equal to 5.1, calculate the perimeters of 𝐴𝐶𝐹𝐷 and triangle
𝐵𝐶𝐷.
As the line 𝐴𝐹 is a line of
symmetry, we know that the lengths of several sides will be equal. The length 𝐴𝐶 will be equal to
the length 𝐴𝐷. Both of these have length 6.7. The lengths 𝐵𝐶 and 𝐵𝐷 are also
equal in length. We are told that 𝐵𝐶 is equal to
5.1. Therefore, 𝐵𝐷 is also equal to
5.1. The lengths 𝐸𝐷 and 𝐸𝐶 are also
equal in length. 𝐸𝐷 is equal to five. Therefore, 𝐸𝐶 is also equal to
five. Finally, the lengths 𝐷𝐹 and 𝐶𝐹
are also equal. We’re told in the diagram that 𝐶𝐹
is equal to 8.2. Therefore, 𝐷𝐹 is also equal to
8.2.
The first perimeter we’re asked to
calculate is that of 𝐴𝐶𝐹𝐷. This is the outside of the
shape. We can calculate this by adding the
four lengths 𝐴𝐶, 𝐶𝐹, 𝐹𝐷 and 𝐷𝐴. 𝐴𝐶 and 𝐴𝐷 or 𝐷𝐴 are both
equal to 6.7. 𝐶𝐹 and 𝐹𝐷 or 𝐷𝐹 are both
equal to 8.2. We need to add 6.7, 8.2, 8.2, and
6.7. 6.7 plus 8.2 is equal to 14.9. This means that 8.2 plus 6.7 is
also equal to 14.9. 14.9 plus 14.9 or two times 14.9 is
equal to 29.8. The perimeter of 𝐴𝐶𝐹𝐷 is
29.8.
We also need to calculate the
perimeter of the triangle 𝐵𝐶𝐷. This is equal to the three lengths
𝐵𝐶, 𝐶𝐷 and 𝐷𝐵. This can be split into four lengths
that we know, 𝐵𝐶, 𝐶𝐸, 𝐸𝐷 and 𝐷𝐵. 𝐵𝐶 and 𝐵𝐷 are both equal to
5.1. 𝐶𝐸 and 𝐸𝐷 are both equal to
five. We need to add 5.1, five, five, and
5.1. 5.1 plus five is equal to 10.1. We get the same answer when we add
them the other way round. 10.1 plus 10.1 is equal to
20.2. The perimeter of triangle 𝐵𝐶𝐷 is
20.2. Therefore, our two correct answers
are 29.8 and 20.2.
