Eureka Math Grade 7 Module 3 Lesson 19 Answer Key

Engage NY Eureka Math 7th Grade Module 3 Lesson 19 Answer Key

Eureka Math Grade 7 Module 3 Lesson 19 Example Answer Key

Example: Area of a Parallelogram
The coordinate plane below contains figure P, parallelogram ABCD.
Engage NY Math 7th Grade Module 3 Lesson 19 Example Answer Key 1
Answer:
Engage NY Math 7th Grade Module 3 Lesson 19 Example Answer Key 2
a. Write the ordered pairs of each of the vertices next to the vertex points.
Answer:
See figure.

b. Draw a rectangle surrounding figure P that has vertex points of A and C. Label the two triangles in the figure as S and T.
Answer:
See figure.

c. Find the area of the rectangle.
Answer:
Base = 8 units
Height = 6 units
Area = 8 units × 6 units = 48 sq. units

d. Find the area of each triangle.
Answer:
Figure S
Base = 3 units
Height = 6 units
Area = \(\frac{1}{2}\) × 3 units × 6 units
= 9 sq. units
Figure T
Base = 3 units
Height = 6 units
Area = \(\frac{1}{2}\) × 3 units × 6 units
= 9 sq. units

e. Use these areas to find the area of parallelogram ABCD.
Answer:
Area P = Area of rectangle – Area S – Area T
= 48 sq. units – 9 sq. units – 9 sq. units = 30 sq. units

The coordinate plane below contains figure R, a rectangle with the same base as the parallelogram above.
Engage NY Math 7th Grade Module 3 Lesson 19 Example Answer Key 3
Answer:
Engage NY Math 7th Grade Module 3 Lesson 19 Example Answer Key 4
f. Draw triangles S and T and connect to figure R so that you create a rectangle that is the same size as the rectangle you created on the first coordinate plane.
Answer:
See figure.

g. Find the area of rectangle R.
Answer:
Base = 5 units
Height = 6 units
Area = 30 sq. units

h. What do figures R and P have in common?
Answer:
They have the same area. They share the same base and have the same height.

Eureka Math Grade 7 Module 3 Lesson 19 Exercise Answer Key

Exercise 1.
Find the area of triangle ABC.
Engage NY Math Grade 7 Module 3 Lesson 19 Exercise Answer Key 1
Answer:
A = \(\frac{1}{2}\) × 7 units × 4 units = 14 sq.units

Exercise 2.
Find the area of quadrilateral ABCD two different ways.
Engage NY Math Grade 7 Module 3 Lesson 19 Exercise Answer Key 2
Answer:
Engage NY Math Grade 7 Module 3 Lesson 19 Exercise Answer Key 3
\(\frac{1}{2}\) × 2 × 5 + 2 × 5 + \(\frac{1}{2}\) × 1 × 5 = 5 + 10 + 2.5 = 17.5
The area is 17.5 sq.units.

Engage NY Math Grade 7 Module 3 Lesson 19 Exercise Answer Key 4
\(\frac{1}{2}\) × (5 + 2) × 5 = 17.5
The area is 17.5 sq.units.

Exercise 3.
The area of quadrilateral ABCD is 12 sq. units. Find x.
Engage NY Math Grade 7 Module 3 Lesson 19 Exercise Answer Key 5
Answer:
Area = base×height
12 sq.units = 2x
6 units = x

Exercise 4.
The area of triangle ABC is 14 sq. units. Find the length of side \(\overline{B C}\).
Engage NY Math Grade 7 Module 3 Lesson 19 Exercise Answer Key 6
Answer:
Area = \(\frac{1}{2}\) × base×height
14 sq.units = \(\frac{1}{2}\) × BC × (7 units)
BC = 4 units

Exercise 5.
Find the area of triangle ABC.
Engage NY Math Grade 7 Module 3 Lesson 19 Exercise Answer Key 7
Answer:
Area of rectangle ARST = 11 units × 10 units = 110 sq.units
Area of triangle ARB = \(\frac{1}{2}\) × 7 units×10 units = 35 sq.units
Area of triangle BSC = \(\frac{1}{2}\) × 4 units×5 units = 10 sq.units
Area of triangle ATC = \(\frac{1}{2}\) × 11 units×5 units = 27.5 sq.units
Area of triangle ABC = Area of ARST – Area of ARB – Area of BSC – Area of ATC = 37.5 sq.units

Eureka Math Grade 7 Module 3 Lesson 19 Problem Set Answer Key

Find the area of each figure.
Question 1.
Eureka Math 7th Grade Module 3 Lesson 19 Problem Set Answer Key 1
Answer:
Area = 13.5 sq. units

Question 2.
Eureka Math 7th Grade Module 3 Lesson 19 Problem Set Answer Key 2
Answer:
Area = 4.5π sq. units ≈14.13 sq. units

Question 3.
Eureka Math 7th Grade Module 3 Lesson 19 Problem Set Answer Key 3
Answer:
Area = 48 sq. units

Question 4.
Eureka Math 7th Grade Module 3 Lesson 19 Problem Set Answer Key 4
Answer:
Area = (2π + 16) sq. units ≈ 22.28 sq. units

Question 5.
Eureka Math 7th Grade Module 3 Lesson 19 Problem Set Answer Key 5
Answer:
Area = 68 sq. units

Question 6.
Eureka Math 7th Grade Module 3 Lesson 19 Problem Set Answer Key 6
Answer:
Area = 46 sq. units

For Problems 7–9, draw a figure in the coordinate plane that matches each description.
Question 7.
A rectangle with an area of 18 sq. units
Eureka Math 7th Grade Module 3 Lesson 19 Problem Set Answer Key 7
Answer:
Eureka Math 7th Grade Module 3 Lesson 19 Problem Set Answer Key 8

Question 8.
A parallelogram with an area of 50 sq. units
Eureka Math 7th Grade Module 3 Lesson 19 Problem Set Answer Key 9
Answer:
Eureka Math 7th Grade Module 3 Lesson 19 Problem Set Answer Key 10

Question 9.
A triangle with an area of 25 sq. units
Eureka Math 7th Grade Module 3 Lesson 19 Problem Set Answer Key 11
Answer:
Eureka Math 7th Grade Module 3 Lesson 19 Problem Set Answer Key 12

Find the unknown value labeled as x on each figure.
Question 10.
The rectangle has an area of 80 sq. units.
Eureka Math 7th Grade Module 3 Lesson 19 Problem Set Answer Key 13
Answer:
x = 8

Question 11.
The trapezoid has an area of 115 sq. units.
Eureka Math 7th Grade Module 3 Lesson 19 Problem Set Answer Key 14
Answer:
x = 10

Question 12.
Find the area of triangle ABC.
Eureka Math 7th Grade Module 3 Lesson 19 Problem Set Answer Key 15
Answer:
Area = 6.5 sq. units

Question 13.
Find the area of the quadrilateral using two different methods. Describe the methods used, and explain why they result in the same area.
Eureka Math 7th Grade Module 3 Lesson 19 Problem Set Answer Key 16
Answer:
Area = 15 sq. units
One method is by drawing a rectangle around the figure. The area of the parallelogram is equal to the area of the rectangle minus the area of the two triangles. A second method is to use the area formula for a parallelogram (Area = base × height).

Question 14.
Find the area of the quadrilateral using two different methods. What are the advantages or disadvantages of each method?
Eureka Math 7th Grade Module 3 Lesson 19 Problem Set Answer Key 17
Answer:
Area = 60 sq. units
One method is to use the area formula for a trapezoid, A = \(\frac{1}{2}\) (base 1 + base 2)×height. The second method is to split the figure into a rectangle and a triangle. The second method requires more calculations. The first method requires first recognizing the figure as a trapezoid and recalling the formula for the area of a trapezoid.

Eureka Math Grade 7 Module 3 Lesson 19 Exit Ticket Answer Key

The figure ABCD is a rectangle. AB = 2 units, AD = 4 units, and AE = FC = 1 unit.
Eureka Math Grade 7 Module 3 Lesson 19 Exit Ticket Answer Key 1
Question 1.
Find the area of rectangle ABCD.
Answer:
Area = 4 units × 2 units = 8 sq. units

Question 2.
Find the area of triangle ABE.
Answer:
Area = \(\frac{1}{2}\) × 1 unit × 2 units = 1 sq. unit

Question 3.
Find the area of triangle DCF.
Answer:
Area = \(\frac{1}{2}\) × 1 unit × 2 units = 1 sq. unit

Question 4.
Find the area of the parallelogram BEDF two different ways.
Answer:
Area = Area of ABCD – Area of ABE – Area of DCF
= (8 – 1 – 1) sq. units = 6 sq. units

Area = base × height
= 3 units × 2 units = 6 sq. units

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