## 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.  a. Write the ordered pairs of each of the vertices next to the vertex points.
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.
See figure.

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

d. Find the area of each triangle.
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.
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.  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.
See figure.

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

h. What do figures R and P have in common?
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. A = $$\frac{1}{2}$$ × 7 units × 4 units = 14 sq.units

Exercise 2.
Find the area of quadrilateral ABCD two different ways.  $$\frac{1}{2}$$ × 2 × 5 + 2 × 5 + $$\frac{1}{2}$$ × 1 × 5 = 5 + 10 + 2.5 = 17.5
The area is 17.5 sq.units. $$\frac{1}{2}$$ × (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. 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 $$\overline{B C}$$. 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. 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. Area = 13.5 sq. units

Question 2. Area = 4.5π sq. units ≈14.13 sq. units

Question 3. Area = 48 sq. units

Question 4. Area = (2π + 16) sq. units ≈ 22.28 sq. units

Question 5. Area = 68 sq. units

Question 6. 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  Question 8.
A parallelogram with an area of 50 sq. units  Question 9.
A triangle with an area of 25 sq. units  Find the unknown value labeled as x on each figure.
Question 10.
The rectangle has an area of 80 sq. units. x = 8

Question 11.
The trapezoid has an area of 115 sq. units. x = 10

Question 12.
Find the area of triangle ABC. 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. 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? 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. Question 1.
Find the area of rectangle ABCD.
Area = 4 units × 2 units = 8 sq. units

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

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