# Question Video: Translating a Triangle Using Congruent Triangles Mathematics • 11th Grade

Fill in the blank: In the figure, the triangles 𝑋𝐿𝑀, 𝐿𝑌𝑁, 𝐿𝑁𝑀, and 𝑀𝑁𝑍 are congruent. Then, the image of triangle 𝑀𝑁𝑍 by translation of magnitude 𝑋𝑀 in the direction of the ray 𝑍𝑀 is ＿.

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### Video Transcript

Fill in the blank. In the figure, the triangles 𝑋𝐿𝑀, 𝐿𝑌𝑁, 𝐿𝑁𝑀, and 𝑀𝑁𝑍 are congruent. Then, the image of triangle 𝑀𝑁𝑍 by translation of magnitude 𝑋𝑀 in the direction of the ray 𝑍𝑀 is what.

In this question, we have been asked to find the image of the triangle 𝑀𝑁𝑍 after a translation of magnitude 𝑋𝑀 in the direction of the ray 𝑍𝑀. Let’s start by considering the four congruent triangles we have been given in the question. These are triangle 𝑋𝐿𝑀, triangle 𝐿𝑌𝑁, triangle 𝐿𝑁𝑀, and triangle 𝑀𝑁𝑍.

Using the congruent triangles, we can label the congruent angles in the figure. We see that the three angles at 𝑀 are three internal angles of any one of the triangles. Since the angles in a triangle sum to 180 degrees, the three angles at 𝑀 must also sum to 180 degrees. Hence, 𝑍𝑀𝑋 is a straight line segment. Similarly, the angles at 𝑁 must sum to 180 degrees. So, 𝑍𝑁𝑌 is also a straight line segment.

We have been asked to translate the triangle 𝑀𝑁𝑍. We can do this by translating each of the points 𝑀, 𝑁, and 𝑍 to find the image of the triangle after the translation. Let’s start the translation by translating point 𝑀. We need to translate it with a magnitude of 𝑋𝑀 and in the direction of the ray 𝑍𝑀. Since 𝑍𝑀𝑋 is a straight line segment, this is the same as the direction of the ray 𝑍𝑋. Translating the point 𝑀 in the direction of the ray 𝑍𝑋 by a magnitude of 𝑋𝑀 will move it to the point 𝑋. Hence, the image of the point 𝑀 is the point 𝑋. Since the length of the line segment 𝑋𝑀 and 𝑀𝑍 are the same, we find that translating the point 𝑍 with magnitude 𝑋𝑀 or 𝑀𝑍 in the direction of the ray 𝑍𝑀 gives us that the image of point 𝑍 is the point 𝑀.

Now all we need to do is translate the point 𝑁. Since 𝑍𝑁𝑌 is a straight line segment and the angles 𝐿𝑁𝑌 and 𝑀𝑍𝑁 are equal, the direction of the ray 𝑍𝑀 is equal to the direction of the ray 𝑁𝐿. Also, the length of 𝑋𝑀 is equal to the length of 𝐿𝑁. Therefore, translating the point 𝑁 with a magnitude of 𝑋𝑀 in the direction of the ray 𝑍𝑀 will move it to the point 𝐿. Here, we reach our solution, which is that the image of triangle 𝑀𝑁𝑍 after a translation of magnitude 𝑋𝑀 in the direction of the ray 𝑍𝑀 is the triangle 𝑋𝐿𝑀.