Students of Grade 3 can get a strong foundation on mathematics concepts by referring to the Eureka Math Book. It was developed by highly professional mathematics educators and the solutions prepared by them are in a concise manner for easy grasping. To achieve high scores in Grade 3, students need to solve all questions and exercises included in Eureka’s Math Grade 3 Book.

## Engage NY Eureka Math 3rd Grade Module 4 Lesson 14 Answer Key

Eureka Math Answer Key for Grade 3 aids teachers to differentiate instruction, building, and reinforcing foundational mathematics skills that alter from the classroom to real life.
With the help of the Eureka Primary School Grade 3 Answer Key, You can think deeply regarding what you are learning, and you will really learn math easily just like that. So teachers and students can find this Eureka Answer Key for Grade 3 more helpful in raising students’ scores and supporting teachers to educate the students.

### Eureka Math Grade 3 Module 4 Lesson 14 Pattern Sheet Answer Key

Multiply

8 x 1 = 8, 8 x 2 = 16, 8 x 3 = 24, 8 x 4 = 32, 8 x 5 = 40, 8 x 6 = 48, 8 x 7 = 56, 8 x 8 = 64, 8 x 9 = 72, 8 x 10 = 80, 8 x 5 = 40, 8 x 6 = 48, 8 x 5 = 40, 8 x 7 = 56, 8 x 5 = 40, 8 x 8 = 64, 8 x 5 = 40, 8 x 9 = 72, 8 x 5 = 40, 8 x 10 = 80, 8 x 6 = 48, 8 x 5 = 40, 8 x 6 = 48, 8 x 7 = 56, 8 x 6 = 48, 8 x 7 = 56, 8 x 6 = 48, 8 x 7 = 56, 8 x 8 = 64, 8 x 7 = 56, 8 x 9 = 72, 8 x 7 = 56, 8 x 8 = 64, 8 x 6 = 48, 8 x 8 = 64, 8 x 7 = 56, 8 x 8 = 64, 8 x 9 = 72, 8 x 9 = 72, 8 x 6 = 48, 8 x 9 = 72, 8 x 7 = 56, 8 x 9 = 72, 8 x 8 = 64, 8 x 9 = 72, 8 x 8 = 64, 8 x 6 = 48, 8 x 9 = 72, 8 x 7 = 56, 8 x 9 = 72, 8 x 6 = 48, 8 x 8 = 64, 8 x 9 = 72, 8 x 7 = 56, 8 x 6 = 48, 8 x 8 = 64.

Explanation:
In the above-given question,
given that,
multiply with 8.
8 x 1 = 8, 8 x 2 = 16, 8 x 3 = 24, 8 x 4 = 32, 8 x 5 = 40, 8 x 6 = 48, 8 x 7 = 56, 8 x 8 = 64, 8 x 9 = 72, 8 x 10 = 80, 8 x 5 = 40, 8 x 6 = 48, 8 x 5 = 40, 8 x 7 = 56, 8 x 5 = 40, 8 x 8 = 64, 8 x 5 = 40, 8 x 9 = 72, 8 x 5 = 40, 8 x 10 = 80, 8 x 6 = 48, 8 x 5 = 40, 8 x 6 = 48, 8 x 7 = 56, 8 x 6 = 48, 8 x 7 = 56, 8 x 6 = 48, 8 x 7 = 56, 8 x 8 = 64, 8 x 7 = 56, 8 x 9 = 72, 8 x 7 = 56, 8 x 8 = 64, 8 x 6 = 48, 8 x 8 = 64, 8 x 7 = 56, 8 x 8 = 64, 8 x 9 = 72, 8 x 9 = 72, 8 x 6 = 48, 8 x 9 = 72, 8 x 7 = 56, 8 x 9 = 72, 8 x 8 = 64, 8 x 9 = 72, 8 x 8 = 64, 8 x 6 = 48, 8 x 9 = 72, 8 x 7 = 56, 8 x 9 = 72, 8 x 6 = 48, 8 x 8 = 64, 8 x 9 = 72, 8 x 7 = 56, 8 x 6 = 48, 8 x 8 = 64.

### Eureka Math Grade 3 Module 4 Lesson 14 Problem Set Answer Key

Question 1.
Find the area of each of the following figures. All figures are made up of rectangles.
a.

The area of figure 1 = 10 sq cm.
the area of figure 2 = 9 sq cm.

Explanation:
In the above-given question,
given that,
area of the figure 1 = 10 sq cm.
area = l x b.
where l = length and b = breadth.
area = 5 x 2.
area = 10.
area of the figure 2 = 9 sq cm.
area = l x b.
where l = length and b = breadth.
area = 3 x 3.
area = 9.

b.

The area of the shaded figure = 78 sq cm.

Explanation:
In the above-given question,
given that,
area of longer rectangle = 12 sq cm.
area = l x b.
area = 3 x 4.
area = 12 sq cm.
area of smaller rectangle = 2 sq cm.
area = l x b.
area = 1 x 2.
area = 2 sq cm.
area of shaded rectangle = area of longer rectangle – area of smaller rectangle.
area of shaded rectangle = 12 – 2.
area = 10 sq cm

Question 2.
The figure below shows a small rectangle in a big rectangle. Find the area of the shaded part of the figure.

The area of the shaded figure = 24 sq cm.

Explanation:
In the above-given question,
given that,
the area of the longer rectangle = 30 sq cm.
area = l x b.
area = 6 x 5.
area = 30 sq cm.
area of smaller rectangle = 6 sq cm.
area = l x b.
area = 3 x 2.
area = 6 sq cm.
area of shaded rectangle = area of longer rectangle – area of smaller rectangle.
area of shaded rectangle = 30 – 6.
area of shaded rectangle = 24.

Question 3.
A paper rectangle has a length of 6 inches and a width of 8 inches. A square with a side length of 3 inches was cut out of it. What is the area of the remaining paper?

The area of the remaining paper = 25 sq in.

Explanation:
In the above-given question,
given that,
A paper rectangle has a length of 6 inches and a width of 8 inches.
A square with a side length of 3 inches was cut out of it.
area = 6 x 8 = 48.
the area of remaining paper = 25.
8 – 3 = 5.
6 – 1 = 5.
area = 5 x 5.
area = 25.

Question 4.
Tila and Evan both have paper rectangles measuring 6 cm by 9 cm. Tila cuts a 3 cm by 4 cm rectangle out of hers, and Evan cuts a 2 cm by 6 cm rectangle out of his. Tila says she has more paper left over. Evan says they have the same amount. Who is correct? Show your work below.

Tila is correct.

Explanation:
In the above-given question,
given that,
Tila and Evan both have paper rectangles measuring 6 cm by 9 cm.
Tila cuts a 3 cm by 4 cm rectangle out of hers,
Evan cuts 2 cm by 6 cm rectangle out of his.
3 x 5 = 15.
4 x 3 = 12.
so Tila is correct.

### Eureka Math Grade 3 Module 4 Lesson 14 Exit Ticket Answer Key

Mary draws an 8 cm by 6 cm rectangle on her grid paper. She shades a square with a side length of 4 cm inside her rectangle. What area of the rectangle is left unshaded?

The area of the rectangle is left unshaded = 24 sq cm.

Explanation:
In the above-given question,
given that,
area of longer rectangle = 48 sq cm.
area = l x b.
area = 8 x 6.
area = 48 sq cm.
area of smaller rectangle = 24 sq cm.
area = l x b.
area = 6 x 4.
area = 24 sq cm.
area of shaded rectangle = area of longer rectangle – area of smaller rectangle.
area of shaded rectangle = 48 – 24.
area of shaded rectangle = 24 sq cm.

### Eureka Math Grade 3 Module 4 Lesson 14 Homework Answer Key

Question 1.
Find the area of each of the following figures. All figures are made up of rectangles.
a.

The area of figure 1 = 48 sq cm.
the area of figure 2 = 27 sq cm.

Explanation:
In the above-given question,
given that,
area of the figure 1 = 48 sq cm.
area = l x b.
where l = length and b = breadth.
area = 6 x 8.
area = 48.
area of the figure 2 = 27 sq cm.
area = l x b.
where l = length and b = breadth.
area = 9 x 3.
area = 27.

b.

The area of figure 1 = 64 sq in.
the area of figure 2 = 6 sq in.

Explanation:
In the above-given question,
given that,
area of the figure 1 = 64 sq in.
area = l x b.
where l = length and b = breadth.
area = 8 x 8.
area = 64.
area of the figure 2 = 6 sq in.
area = l x b.
where l = length and b = breadth.
area = 2 x 3.
area = 6.

Question 2.
The figure below shows a small rectangle cut out of a big rectangle.

a. Label the side lengths of the unshaded region.

The side lengths of the unshaded region is = 5 ft and 3 ft.

Explanation:
In the above-given question,
given that,
the area of the larger rectangle = 10 x 7.
area = 70 sq ft.
10 – 3 + 2 = 5.
7 – 2 + 2 = 4.
so the side lengths of the unshaded region = 5 ft and 3 ft.

b. Find the area of the shaded region.

The area of the shaded figure = 55 sq ft.

Explanation:
In the above-given question,
given that,
area of longer rectangle = 70 sq ft.
area = l x b.
area = 10 x 7.
area = 70 sq cm.
area of smaller rectangle = 15 sq ft.
area = l x b.
area = 5 x 3.
area = 15 sq ft.
area of shaded rectangle = area of longer rectangle – area of smaller rectangle.
area of shaded rectangle = 70 – 15.
area of shaded rectangle = 55.