Mid-Unit Assessment

1.

What is the area of this parallelogram?

A.

30 cm²

B.

24 cm²

C.

22 cm²

D.

12 cm²

Answer:

24 cm²

Teaching Notes

Students who select Choice A may be multiplying the one side of the parallelogram by the other side, rather than multiplying one of the bases by the height. Students who fail to select Choice B may be confused about which measurements to use to calculate the area of a parallelogram. Students who select Choice C are calculating the perimeter of the parallelogram and may need to revisit the conceptual differences between area and perimeter. Students who select D are likely applying the formula for the area of a triangle.

2.

Select all the triangles that have an area of 45 square units.

A.

B.

C.

D.

E.

Answer: D, E

Teaching Notes

Students who select Triangle A are calculating the perimeter rather than the area. Students who select Triangle B are using the side with length 10 as the height. Students who select Triangle C are multiplying the base and the height but are not dividing that product by 2. Students who fail to select D may have a major misconception about the area of a triangle or may have neglected to divide by 2. Students who fail to select E may be using the incorrect base or height to calculate the area.

3.

Select all the parallelograms that have an area of 12 square units.

A.

B.

C.

D.

E.

Answer:

A,B,E

Teaching Notes

Students who fail to select Parallelogram A may think that a rectangle is not a parallelogram. Students who fail to select Parallelogram B may be confusing the side length with the height. Students who select Parallelogram C may be calculating the perimeter instead of the area. Students who select Parallelogram D are not applying the formula for finding the area of a parallelogram.

4.

On each triangle, draw a segment to represent the height that corresponds to the given base. Label each height with the word “height.”

Answer:

Teaching Notes

Identifying the height for a chosen base is important for calculating the area of a triangle.

5.

Draw two parallelograms, each with an area of 16 square units. The two parallelograms should not be identical copies of each other.

Answer:

Sample responses:

Minimal Tier 1 response:

Work is complete and correct.
The two parallelograms may have the same base and height as long as they are not congruent.
Sample: Two parallelograms (rectangles allowed) with base and height lengths that when multiplied equal 16.

Tier 2 response:

Work shows general conceptual understanding and mastery, with some errors.
Sample errors: Only one parallelogram is drawn. Only one parallelogram has the correct area. An equation such as “ $8 \boldcdot 2 =16$ ” is written, but the base or height of the parallelogram is slightly off.

Tier 3 response:

Significant errors in work demonstrate lack of conceptual understanding or mastery.
Sample errors: A length or a side that is not perpendicular to the base is used as the height. Shapes drawn are not parallelograms.

Teaching Notes

This question purposely admits the possibility that students might draw rectangles if they know that a rectangle is a particular kind of parallelogram. The question could be modified to instruct students to draw parallelograms that are not rectangles.

6.

Find the area of the figure. Explain or show your reasoning.

Answer:

38 square units. Possible strategies:

Decompose the figure into rectangles and triangles with known bases and heights. Find the sum of the areas of the components.
Enclose the figure in a 6-by-9 rectangle. From 54 square units, subtract the areas of the right triangles that are not part of the figure.

Minimal Tier 1 response:

Work is complete and correct.
A diagram is included.
Acceptable errors: No units are included.
Sample: A box around the shape has an area of 54. The triangles on the outside have areas of 12, 3, and 1, respectively. The area of the box minus combined areas of the triangles is 38.

Tier 2 response:

Work shows good conceptual understanding and mastery, with either minor errors or correct work with insufficient explanation or justification.
Sample errors: The area of a small number of the partitioned shapes are calculated incorrectly. Addition or subtraction to find the area of the polygon is done incorrectly. The area of one of the partitioned shapes cannot be correctly calculated because the base or height does not lie on a vertical or horizontal line. The partitioned shapes are not shown on the diagram.

Tier 3 response:

Significant errors in work demonstrate lack of conceptual understanding or mastery.
Sample errors: The partitioned shapes are not shapes for which students can find the area. Incorrect area formulas for triangles or rectangles are used. Calculations show many errors.

Teaching Notes

Watch for students who break the shape into parts in unusual ways. These students may not be recalling the methods developed in the lessons, and may create parts for which they cannot determine the area.

7.

The figure is a diagram of a sign. Lengths are given in inches.

A painter is going to paint the sign on both the front and the back. The painter has a small can of paint that can cover 800 square inches. Is that enough to paint both sides of the sign? Explain or show your reasoning.

Answer:

Yes, that is enough paint. Sample reasoning: The sign is composed of a rectangle that is 26 inches by 10 inches and a triangle with a base of 20 inches and a height of $36-26$ , which is 10 inches.

The area of the rectangle is 260 square inches because $26 \boldcdot 10 = 260$ .
The area of the triangle is 100 square inches because $\frac{1}{2} \boldcdot 20 \boldcdot 10 = 100$ .
The combined area is $260+100$ or 360 square inches.
The area of both faces of the sign (front and back) is twice that amount, $2 \boldcdot 360$ , which is 720 square inches. This is less than 800 square inches.

Minimal Tier 1 response:

Work is complete and correct, with complete explanation or justification.
Sample: The area of the sign is the area of a rectangle that is 36 inches by 20 inches minus the area of 2 extra rectangles and 2 extra right triangles.
- Area of the enclosing rectangle: $36 \boldcdot 20 = 720$ , or 720 square inches
- Area of the 2 extra rectangles: $2 \boldcdot (5 \boldcdot 26) = 260$ , or 260 square inches
- Area of the 2 right triangles: $2 \boldcdot (\frac{1}{2} \boldcdot 10 \boldcdot 10) = 100$ , or 100 square inches
- Area of each face of the sign: $720 - 260 - 100$ , which is 360 square inches. For both faces, double that to 720 square inches, which is less than 800 square inches.

Tier 2 response:

Work shows good conceptual understanding and mastery, with either minor errors or correct work with insufficient explanation or justification.
Sample errors: Units omitted; arithmetic errors in calculating area. Acceptable errors: Incorrect total area due to an error in computing a partial area.

Tier 3 response:

Work shows a developing but incomplete conceptual understanding, with significant errors.
Sample errors: Plausible but flawed strategy for partitioning; incorrect area formulas with correct partitioning strategy.

Tier 4 response:

Work includes major errors or omissions that demonstrate a lack of conceptual understanding and mastery.
Sample errors: No reasonable strategy for partitioning.

Teaching Notes

Strategies for this problem range from decomposing the sign into rectangles and triangles, to enclosing the sign and subtracting the areas of the extra regions. Watch for students who struggle to find the measurements of the tip of the arrow. Others may struggle with doubling the area in the second part.