5.1 Perimeter, Area, and Volume of Geometric Figures

Fundamental Concepts of Perimeter, Area, and Volume

Geometry on the SHSAT requires a firm command of formulas and the ability to apply them to multi-step word problems. Rather than simple calculation questions, you will often need to find the area of an irregular shape, calculate the volume of a partially filled container, or determine how changes in linear dimensions affect area or volume.

1. Two-Dimensional Figures: Perimeter and Area

Perimeter is the total distance around the outside of a two-dimensional shape, measured in linear units (e.g., inches, centimeters, feet). Area is the measure of the space inside the boundary of a flat object, expressed in square units (e.g., square inches, square feet).

The table below lists the essential formulas for two-dimensional figures tested on the SHSAT:

The Perpendicular Height Rule for Triangles

A common mistake on the SHSAT is using the side length of a non-right triangle as the height. The height (h) must always be the perpendicular distance from the base to the opposite vertex. In a right triangle, the height is simply one of the legs. In an obtuse or acute triangle, you must look for the height marked by a perpendicular line segment forming a 90° angle with the base.

Circles, Arc Lengths, and Sector Areas

For circles, the SHSAT frequently tests parts of a circle, known as sectors and arcs. A sector is a fraction of the circle's area, and an arc is a fraction of its circumference. These fractions are determined by the central angle θ (in degrees) of the sector.

• Arc Length Formula: L = (θ / 360) * 2πr
• Sector Area Formula: A_sector = (θ / 360) * πr²

For example, if a circular pizza has a radius of 6 inches and a slice is cut with a central angle of 60°:

• The length of the crust (arc length) is: (60 / 360) * 2π(6) = 1/6 * 12π = 2π inches.
• The area of the slice (sector area) is: (60 / 360) * π(6²) = 1/6 * 36π = 6π square inches.

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2. Three-Dimensional Figures: Volume and Surface Area

Volume measures the amount of three-dimensional space a solid occupies, measured in cubic units (e.g., cubic inches, cubic feet). Surface Area is the total area of all the outer faces of a three-dimensional object, measured in square units.

Prisms

A prism is a three-dimensional solid with two parallel, congruent bases. The cross-section of a prism parallel to the bases is the same throughout its height. The general formula for the volume of any prism is:

$$V = B * h$$

where B is the area of the base, and h is the height of the prism perpendicular to the bases.

• Rectangular Prism:
  • Volume: V = lwh (since the base is a rectangle with area lw)
  • Surface Area: SA = 2lw + 2lh + 2wh (representing the sum of the areas of the 6 rectangular faces)
• Triangular Prism:
  • Volume: V = (0.5 * b * h_tri) * h_prism (where b and h_tri are the base and height of the triangular base)
  • Surface Area: SA = 2 * (base area) + (perimeter of base * h_prism)

Cylinders

A cylinder is a solid with two parallel circular bases. Like a prism, its volume is the area of the base times the height.

• Volume: V = πr²h
• Lateral Surface Area (LSA): LSA = 2πrh (the area of the curved side face)
• Total Surface Area (SA): SA = 2πr² + 2πrh (the area of the two circular bases plus the lateral area)

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3. Worked SHSAT Applications and Strategies

Let's examine two standard SHSAT multi-step problems.

Application 1: Shaded Regions and Border Areas

Problem: A rectangular garden measuring 12 feet by 8 feet is surrounded by a concrete walkway that is 2 feet wide on all sides. What is the area of the walkway?

Strategy: Draw a diagram to visualize this. The inner rectangle is the garden. The outer rectangle includes the garden and the walkway.

• Inner Dimensions: l = 12 ft, w = 8 ft.
• Outer Dimensions: The walkway adds 2 feet to each side. Therefore, the outer length is 12 + 2 + 2 = 16 feet, and the outer width is 8 + 2 + 2 = 12 feet.
• Outer Area: A_outer = 16 * 12 = 192 sq ft.
• Inner Area: A_inner = 12 * 8 = 96 sq ft.
• Area of the Walkway: A_walkway = A_outer - A_inner = 192 - 96 = 96 square feet.

Trap Warning: A common mistake is adding the walkway width only once (e.g., 12 + 2 = 14 and 8 + 2 = 10). Remember, a border surrounds a shape on all sides, adding to both the left and right, and the top and bottom.

Application 2: Cylinder Volume and Height Changes

Problem: A cylindrical tank with a radius of 4 inches is filled with water to a height of 10 inches. If all of this water is poured into another cylindrical tank with a radius of 8 inches, what will the height of the water be in the second tank?

Strategy: Since the water is transferred, the volume of the water (V) remains constant.

• Volume of water in the first tank: V = π * (r_1)² * h_1 = π * 4² * 10 = 160π cubic inches.
• Set this equal to the volume formula for the second tank: V = π * (r_2)² * h_2
• 160π = π * 8² * h_2 => 160π = 64π * h_2
• Divide both sides by π: 160 = 64 * h_2
• Solve for h_2: h_2 = 160 / 64 = 2.5 inches.

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4. Dimensional Changes and Scale Factors in 3D

A common conceptual question on the SHSAT asks how scaling the sides of a shape affects its area or volume.

• If you multiply all the dimensions of a 2D shape by a scale factor k, its perimeter is multiplied by k, and its area is multiplied by k². For example, doubling the radius of a circle multiplies its area by 2² = 4.
• If you multiply all the dimensions of a 3D solid by a scale factor k, its surface area is multiplied by k², and its volume is multiplied by k³. For example, if you double all the dimensions of a rectangular box, its new volume is 2³ = 8 times the original volume.

Understanding these proportionalities allows you to solve scaling questions in seconds without calculating the actual dimensions, which is a major advantage on the timed exam.