VTCAP Grade 7 Geometry Practice — Sports Academy Games

What the test measures

Geometry skills

Draw, construct, and describe geometrical figures and describe the relationships between them
Solve real-life and mathematical problems involving angle measure, area, surface area, and volume

Standards basis

Common Core State Standards for Mathematics (CCSS-M) — Vermont

How the VTCAP reports it

Vermont administered the Smarter Balanced summative through spring 2022, then replaced it with the Cognia-developed Vermont Comprehensive Assessment Program (VTCAP) beginning spring 2023 — this landing slug predates the switch. VTCAP mathematics is an adaptive, CCSS-M-aligned assessment in grades 3-9. It is NOT a Smarter Balanced administration: its reporting categories are groupings of the CCSS grade-level domains (e.g. 'The Number System, Expressions and Equations'), not SBAC claims, though two cross-grade Mathematical Practices indicators ('Problem Solving, Reasoning, and Argument'; 'Modeling, Patterns, and Structure') echo the old claim structure. Per the test specifications, the grades 3-8 test may assess any CCSS-M standard at the grade; the grade 9 test assesses a prioritized subset of the high school CCSS-M within the conceptual categories of Number and Quantity, Algebra, Functions, Geometry, and Statistics and Probability (the assessed focus areas are published in the grade 9 Performance Level Descriptors).

Blueprint & weighting

VTCAP Test Specifications (Spring 2026) operational test blueprint, target percent of points per grade — Grade 3: OA 24-35%, NBT 10-14%, NF 16-20%, MD 22-29%, G 6-10%, constructed-response Mathematical Practices 12%. Grade 4: OA 20-31%, NBT 16-20%, NF 20-31%, MD 12-20%, G 6-10%, practices 12%. Grade 5: OA 14-22%, NBT 14-25%, NF 22-29%, MD 20-27%, G 8-16%, practices 12%. Grade 6: RP 15-22%, NS 15-22%, EE 15-22%, G 11-19%, SP 11-19%, practices 11%. Grade 7: RP 15-22%, NS 11%, EE 15-30%, G 11-19%, SP 19-22%, practices 11%. Grade 8: F 18-29%, NS 7%, EE 20-31%, G 18-29%, SP 18-22%, practices 11%. Grade 9 (all machine-scored): Number and Quantity/Algebra 18-27%, Algebra/Functions 31-40%, Geometry 20-29%, Statistics and Probability 13-20%; the two practices indicators are dual-aligned targets of >18% of points each. Identical figures appear in the Spring 2025 specifications.

Official sources

VTCAP Test Specifications, Spring 2026 (Vermont AOE / Cognia)
VTCAP Test Specifications, Spring 2025 (Vermont AOE / Cognia)
VTCAP Performance Level Descriptors: Mathematics, Grades 3-9 (Vermont AOE)

Practice by skill

Topics mapped to this domain

Try it now

8 free questions · 0/0 correct

Practice playeasy

A cylinder has radius $2$ and height $3$ . Using $\pi = 3$ , what is its volume?

$V = \pi r^2 h = 3 \times 4 \times 3 = 36$ .

Practice playeasy

Similar triangles $RST$ and $UVW$ have perimeters $28$ and $42\textsf{.}$ Side $RS$ corresponds to side $UV\textsf{,}$ and $RS = 10\textsf{.}$ What is the length of $UV\textsf{?}$

The scale factor from $\triangle RST$ to $\triangle UVW$ is $\dfrac{42}{28} = \dfrac{3}{2}\textsf{.}$ So $UV = 10 \cdot \dfrac{3}{2} = 15\textsf{.}$

Practice playeasy

A triangle has side lengths $7$ and $3\textsf{.}$ Which inequality represents the possible lengths, $x\textsf{,}$ of the third side?

The triangle inequality requires $|7 - 3| < x < 7 + 3\textsf{.}$ Simplifying gives $4 < x < 10\textsf{.}$

Practice playeasy

Two lines cross. One angle formed is $75°$ . What is the vertical angle?

Vertical angles are equal, so the angle is $75°$ .

Practice playeasy

A circle has a diameter of $16$ inches. What is the area, in square inches, of the circle?

The radius is half the diameter: $r = \dfrac{16}{2} = 8$ . The area is $A = \pi r^{2} = \pi(8)^{2} = 64\pi\textsf{.}$

Practice playeasy

A circle has radius $2$ . Using $\pi = 3$ , what is its circumference?

$C = 2\pi r = 2 \times 3 \times 2 = 12$ .

Practice playhard

A rectangular prism has volume $72$ cubic centimeters and height $4$ centimeters. The length is twice the width. What is the width, in centimeters?

$V = \ell w h$ and $\ell = 2w$ , so $72 = (2w)(w)(4) = 8w^2$ . Then $w^2 = 9$ and $w = 3\text{ cm}\textsf{.}$

Practice playmedium

A museum display is shaped like a closed square pyramid with a $6$ -centimeter-by- $6$ -centimeter base. Each triangular face has slant height $5$ centimeters. What is the total exterior surface area, in square centimeters, of the display?

The base area is $6^2 = 36$ . The four congruent triangular faces each have area $\dfrac{1}{2}(6)(5) = 15$ , for a total of $4 \times 15 = 60$ . The surface area is $36 + 60 = 96$ square centimeters.