Christmas Addition to 10 | Math Manipulatives | Worksheets
Sherry Clements
14.2k Followers
Grade Levels
K - 2nd
Resource Type
Standards
CCSSK.OA.A.1
CCSSK.OA.A.5
CCSS1.OA.C.6
Formats Included
- PDF
Pages
6 pages
Sherry Clements
14.2k Followers
What educators are saying
My students enjoyed using this resource to practice their math skills. It was engaging without being overwhelming and overstimulating.
A great way to get students excited about math and addition. The manipulatives included helped to practice fine motor skills too.
Description
This December addition resource for Christmas includes 6 pages with manipulatives. Each page contains 10 addition problems to solve (with sums to 10).
There are ten pictures/manipulatives at the bottom of each page for students to use to solve each addition problem. Students can color the manipulatives and use them to solve the problems.
The 6 pages include the following manipulatives: Santa, candy cane, gifts, gingerbread man, Christmas tree, and reindeer.
Related Products
⭐ Addition and Subtraction Word Problems Bundle
⭐ Christmas Boom Cards™ Santa Addition to 30
⭐ Christmas Boom Cards™ Santa Subtraction from 30
⭐ Christmas Addition to 10 | Reindeer
Thank you for stopping by!
Sherry Clements
Total Pages
6 pages
Answer Key
N/A
Teaching Duration
N/A
Last updated Nov 28th, 2021
Report this resource to TPT
Reported resources will be reviewed by our team. Report this resource to let us know if this resource violates TPT’s content guidelines.
Standards
to see state-specific standards (only available in the US).
CCSSK.OA.A.1
Represent addition and subtraction with objects, fingers, mental images, drawings, sounds (e.g., claps), acting out situations, verbal explanations, expressions, or equations.
CCSSK.OA.A.5
Fluently add and subtract within 5.
CCSS1.OA.C.6
Add and subtract within 20, demonstrating fluency for addition and subtraction within 10. Use strategies such as counting on; making ten (e.g., 8 + 6 = 8 + 2 + 4 = 10 + 4 = 14); decomposing a number leading to a ten (e.g., 13 - 4 = 13 - 3 - 1 = 10 - 1 = 9); using the relationship between addition and subtraction (e.g., knowing that 8 + 4 = 12, one knows 12 - 8 = 4); and creating equivalent but easier or known sums (e.g., adding 6 + 7 by creating the known equivalent 6 + 6 + 1 = 12 + 1 = 13).
Reviews
Questions & Answers
14.2k Followers
