Perimeter, Area, and Volume Practice Questions

The perimeter of a rectangle is the total distance around it:

\(P = 2l + 2w = 2(12) + 2(7) = 24 + 14\)\(\:= 38 \text{ cm}\)

Choice B adds the length and width only once (\(12 + 7 = 19\)) without accounting for both pairs of sides. Choice C multiplies the two dimensions (giving the area, not the perimeter).

The area of a triangle is:

\(A = \dfrac{1}{2}bh = \dfrac{1}{2}(10)(6) = 30 \text{ in}^2\)

Choice B forgets to multiply by \(\frac{1}{2}\), which would give the area of a rectangle, not a triangle. A triangle is always half the area of a rectangle with the same base and height.

The area of a parallelogram is:

\(A = bh = 9 \times 5 = 45 \text{ cm}^2\)

The formula is the same as a rectangle, where the height is the perpendicular distance between the two parallel bases, not the length of the slanted side.

Choice C applies the triangle formula (\(\frac{1}{2}bh\)) instead. Choice B adds all four sides, which would give the perimeter (not the area).

The volume of a rectangular prism is:

\(V = lwh = 4 \times 5 \times 3 = 60 \text{ cm}^3\)

Volume is measured in cubic units (cm³), while area uses square units (cm²) and perimeter uses linear units (cm). Choice D gives the surface area, not the volume.

A rectangular prism has three pairs of identical faces. The surface area is:

\(SA = 2lw + 2lh + 2wh\)

Substitute \(l = 6\), \(w = 4\), \(h = 3\):

\(SA = 2(6)(4) + 2(6)(3) + 2(4)(3)\)\(\:= 48 + 36 + 24 = 108 \text{ m}^2\)

Choice A gives the volume (\(6 \times 4 \times 3 = 72\)), not the surface area. Choice C computes only one of each face pair (\(24 + 18 + 12 = 54\)) and forgets to multiply by 2.

The volume of a cylinder is the area of its circular base times its height:

\(V = \pi r^2 h = \pi(3)^2(10) = \pi(9)(10)\)\(\:= 90\pi \text{ cm}^3\)

Choice B uses \(\pi r h\) (forgetting to square the radius). Choice C uses \(2\pi r h\) (the lateral surface area formula, not volume).

The total surface area of a cylinder includes two circular bases and the curved lateral surface:

\(SA = 2\pi r^2 + 2\pi rh\)

Substitute \(r = 5\) and \(h = 8\):

\(SA = 2\pi(5)^2 + 2\pi(5)(8)\)\(\:= 50\pi + 80\pi = 130\pi \text{ cm}^2\)

Choice B gives only the lateral (side) surface area without the two bases. Choice C gives only the two bases without the lateral surface.

The volume of a sphere is:

\(V = \dfrac{4}{3}\pi r^3\)

Substitute \(r = 6\):

\(V = \dfrac{4}{3}\pi(6)^3 = \dfrac{4}{3}\pi(216) = \dfrac{864\pi}{3}\)\(\:= 288\pi \text{ cm}^3\)

Choice B uses the surface area formula (\(4\pi r^2 = 144\pi\)) instead of volume. Choice C forgets the \(\frac{1}{3}\) factor.

The volume of a cone is one-third the volume of a cylinder with the same base and height:

\(V = \dfrac{1}{3}\pi r^2 h\)

Substitute \(r = 4\) and \(h = 9\):

\(V = \dfrac{1}{3}\pi(4)^2(9) = \dfrac{1}{3}\pi(144) = 48\pi \text{ cm}^3\)

Choice B gives \(\pi r^2 h = 144\pi\), which is the volume of a cylinder (forgetting the \(\frac{1}{3}\) factor). A cone always holds exactly one-third the volume of the corresponding cylinder.

The volume of any prism is the area of the base times the length (or depth) of the prism:

\(V = B \times l\)

First, find the area of the triangular base:

\(B = \dfrac{1}{2}(6)(4) = 12 \text{ cm}^2\)

Multiply by the length of the prism:

\(V = 12 \times 10 = 120 \text{ cm}^3\)

Choice B forgets to halve the base area (\(6 \times 4 \times 10 = 240\)), treating the base as a rectangle instead of a triangle.

Break the composite shape into two simpler parts:

Rectangle:

\(A_{\text{rect}} = 10 \times 6 = 60 \text{ cm}^2\)

Semicircle: The diameter is 6 cm, so the radius is 3 cm.

\(A_{\text{semi}} = \dfrac{1}{2}\pi r^2 = \dfrac{1}{2}(3.14)(3)^2 = \dfrac{1}{2}(28.26)\)\(\:= 14.13 \text{ cm}^2\)

Total area:

\(A = 60 + 14.13 = 74.13 \text{ cm}^2\)

Choice B adds a full circle instead of a semicircle. For composite shapes, always identify each component separately, find its area, and add them together.

When a circle is inscribed in a square, the diameter of the circle equals the side length of the square. With \(r = 5\), the diameter is 10, so the square has side length 10.

The shaded region is the area between the square and the circle, so subtract the circle from the square:

\(A_{\text{square}} = 10^2 = 100 \text{ cm}^2\)

\(A_{\text{circle}} = \pi(5)^2 = 3.14(25) = 78.50 \text{ cm}^2\)

\(A_{\text{shaded}} = 100 – 78.50 = 21.50 \text{ cm}^2\)

For shaded region problems, the strategy is almost always: larger area minus smaller area.

For the perimeter of a composite shape, trace along the outside edge only. The semicircle replaces one short side of the rectangle, so that side is no longer part of the perimeter.

The perimeter consists of: two long sides (12 each), one short side (8), and the curved semicircle edge.

The curved part of a semicircle is half the circumference:

\(\dfrac{1}{2} \times \pi d = \dfrac{1}{2}(3.14)(8) = 12.56 \text{ cm}\)

Add all the outer edges:

\(P = 12 + 12 + 8 + 12.56 = 44.56 \text{ cm}\)

Choice B includes both short sides plus the semicircle, double-counting the side that the semicircle replaced.

The volume of a pyramid is one-third the volume of a prism with the same base and height:

\(V = \dfrac{1}{3}Bh\)

The base is a square with side length 10, so \(B = 10^2 = 100 \text{ m}^2\). Substitute:

\(V = \dfrac{1}{3}(100)(12) = \dfrac{1{,}200}{3} = 400 \text{ m}^3\)

Choice C forgets the \(\frac{1}{3}\) factor and gives \(Bh = 1{,}200\). Just like a cone is \(\frac{1}{3}\) of a cylinder, a pyramid is \(\frac{1}{3}\) of a prism.

First, find the volume of the cylinder:

\(V = \pi r^2 h = 3.14(2)^2(5) = 3.14(4)(5)\)\(\:= 62.8 \text{ m}^3\)

Convert cubic meters to liters using the given conversion factor:

\(62.8 \text{ m}^3 \times 1{,}000 = 62{,}800 \text{ liters}\)

This is a two-step problem: first calculate the volume in the given units, then convert to the requested units. Choice D uses \(2 \times 5 \times 1{,}000 \times 2 = 20{,}000\), which does not correctly apply the cylinder volume formula.