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Unit 8: Geometry and Trigonometry

Topic 7

Area, Perimeter, and Volume

This topic covers the area and perimeter formulas for common two-dimensional shapes, plus the surface area and volume formulas for three-dimensional solids. The SAT provides most of these formulas on the reference sheet, but you should know them well enough that you do not need to look them up under time pressure.
Two-dimensional shapes:
The area of a rectangle is A=lwA = lw, and its perimeter is P=2l+2wP = 2l + 2w. A square with side ss has area s2s^2 and perimeter 4s4s.
The area of a triangle is:
A=12bhA = \frac{1}{2}bh
where bb is the base and hh is the height (the perpendicular distance from the base to the opposite vertex). Any side can serve as the base, as long as the height is measured perpendicular to it.
The area of a parallelogram is A=bhA = bh, where hh is the perpendicular height, not the slant side. The area of a trapezoid is:
A=12(b1+b2)hA = \frac{1}{2}(b_1 + b_2)h
where b1b_1 and b2b_2 are the two parallel sides (bases) and hh is the perpendicular height between them.
Three-dimensional solids:
A rectangular prism (box) with length ll, width ww, and height hh has volume V=lwhV = lwh and surface area SA=2(lw+lh+wh)SA = 2(lw + lh + wh).
A cylinder with radius rr and height hh has:
V=πr2h,SA=2πr2+2πrhV = \pi r^2 h, \qquad SA = 2\pi r^2 + 2\pi r h
A cone with radius rr and height hh has:
V=13πr2hV = \frac{1}{3}\pi r^2 h
The volume of a cone is exactly one-third the volume of a cylinder with the same base and height.
A sphere with radius rr has:
V=43πr3,SA=4πr2V = \frac{4}{3}\pi r^3, \qquad SA = 4\pi r^2
A pyramid with a rectangular base of area BB and height hh has:
V=13BhV = \frac{1}{3}Bh
Like a cone, a pyramid's volume is one-third of the prism with the same base and height.
On the SAT, volume problems often require you to solve for a dimension (radius, height, or side length) given the volume, rather than simply computing the volume from given dimensions. This means being comfortable rearranging these formulas.

Core Skills

  • Calculate area and perimeter of rectangles, triangles, parallelograms, and trapezoids.
  • Calculate the volume of prisms, cylinders, cones, spheres, and pyramids.
  • Solve for an unknown dimension given the area or volume.
  • Combine multiple shapes or solids (composite figures) to find total area or volume.
  • Apply area and volume formulas in word-problem contexts involving real-world measurements.

Example 1: Area of a Trapezoid

A trapezoid has parallel sides of length 8 and 14, and a height of 5. What is its area? Step 1: Apply the trapezoid area formula.
A=12(b1+b2)h=12(8+14)(5)=12(22)(5)=55A = \frac{1}{2}(b_1 + b_2)h = \frac{1}{2}(8 + 14)(5) = \frac{1}{2}(22)(5) = 55
The area is 55\boxed{55}.

Example 2: Volume of a Cylinder

A cylindrical water tank has a radius of 4 feet and a height of 10 feet. What is the volume of the tank? Express your answer in terms of π\pi. Step 1: Apply the cylinder volume formula.
V=πr2h=π(4)2(10)=160πV = \pi r^2 h = \pi(4)^2(10) = 160\pi
The volume is 160π\boxed{160\pi} cubic feet.

Example 3: Solving for a Dimension

A cone has a volume of 48π48\pi cubic centimeters and a radius of 4 cm. What is the height of the cone? Step 1: Substitute into the volume formula and solve for hh.
13πr2h=48π  ⟹  13π(16)h=48π\frac{1}{3}\pi r^2 h = 48\pi \implies \frac{1}{3}\pi(16)h = 48\pi
16πh3=48π  ⟹  16h=144  ⟹  h=9\frac{16\pi h}{3} = 48\pi \implies 16h = 144 \implies h = 9
The height is 9\boxed{9} cm.

Example 4: Volume of a Sphere

A basketball has a diameter of 9.4 inches. What is the volume of the basketball, to the nearest cubic inch? Step 1: The radius is 9.42=4.7\dfrac{9.4}{2} = 4.7 inches. Step 2: Apply the sphere volume formula.
V=43π(4.7)3=43π(103.823)≈43(326.01)≈434.68V = \frac{4}{3}\pi(4.7)^3 = \frac{4}{3}\pi(103.823) \approx \frac{4}{3}(326.01) \approx 434.68
To the nearest cubic inch, the volume is 435\boxed{435}.

Example 5: Composite Figure

A rectangular room is 12 feet long, 10 feet wide, and 8 feet tall. What is the total surface area of the four walls (not including the floor or ceiling)? Step 1: The four walls consist of two pairs of rectangles. Two walls are 12×812 \times 8 and two walls are 10×810 \times 8.
SAwalls=2(12)(8)+2(10)(8)=192+160=352SA_{\text{walls}} = 2(12)(8) + 2(10)(8) = 192 + 160 = 352
The total wall area is 352\boxed{352} square feet.

Example 6: Working Backward from Volume

A rectangular prism has a volume of 360 cubic inches. Its length is 12 inches and its width is 6 inches. What is the height? Step 1: Substitute and solve.
V=lwh  ⟹  360=(12)(6)h  ⟹  360=72h  ⟹  h=5V = lwh \implies 360 = (12)(6)h \implies 360 = 72h \implies h = 5
The height is 5\boxed{5} inches.

Key Takeaways

  • Cones and pyramids have volume equal to 13\dfrac{1}{3} of the corresponding prism or cylinder with the same base and height. Remembering this ``one-third'' relationship prevents formula mix-ups.
  • When a problem gives you the volume or area and asks for a dimension, substitute everything you know into the formula and solve the resulting equation. This is algebra, not geometry, and it is where most errors happen.
  • For composite figures, break the shape into simpler pieces, compute each area or volume separately, and then add (or subtract, if one shape is removed from another).
  • The SAT reference sheet provides these formulas, but knowing them cold saves valuable time. In particular, memorize the sphere formulas V=43πr3V = \dfrac{4}{3}\pi r^3 and SA=4πr2SA = 4\pi r^2, since looking them up costs the most time under pressure.