Area and Volume Pattern - Geometric Algebra

This pattern asks you to set up equations from geometric formulas where one or more dimensions are given as algebraic expressions or variables. It blends geometry with algebra.

 

Triangle Area with a Dimension Offset

When one dimension of a triangle is described in terms of another (like "the height is 5 feet longer than the base"), express both dimensions algebraically, then apply the area formula.

The base of a right triangle is 12 inches. The height is 5 inches shorter than the base. What is the area?

A) $12$
B) $42$
C) $84$
D) $102$

Height $= 12 - 5 = 7$. Area $= \dfrac{1}{2}(12)(7) = 42$. The answer is B. Option C forgets the $\dfrac{1}{2}$, and option D adds the dimensions ($12 + 7 = 19$) and multiplies incorrectly.

The height of a right triangle is 12 feet. The base is 5 feet longer than the height. What is the area?

A) $12$
B) $29$
C) $102$
D) $204$

Base $= 12 + 5 = 17$. Area $= \dfrac{1}{2}(12)(17) = 102$. The answer is C.

The height of a right triangle is 10 inches. The base is 6 inches shorter than the height. What is the area?

A) $4$
B) $14$
C) $20$
D) $40$

Base $= 10 - 6 = 4$. Area $= \dfrac{1}{2}(10)(4) = 20$. The answer is C. Option D forgets the $\dfrac{1}{2}$.

 

Scaling: Volume Ratio of Pyramids and Cones

A right square pyramid has volume $V = \dfrac{1}{3}s^2 h$, and a cone has volume $V = \dfrac{1}{3}\pi r^2 h$. If a second figure has volume $n$ times the first, test each answer choice by computing $S^2 \cdot H$ (or $R^2 \cdot H$) and checking whether it equals $n \cdot s^2 \cdot h$.

A right square pyramid has base side $s$ and height $h$. A second pyramid has volume 180 times as large. Which could be its dimensions?

A) $S = 5s$ and $H = 6h$
B) $S = 6s$ and $H = 5h$
C) $S = 12s$ and $H = 15h$
D) $S = 15s$ and $H = 12h$

We need $S^2 \cdot H = 180 \cdot s^2 h$. Check each option: - A) $(5s)^2(6h) = 150s^2h$ — not 180 - B) $(6s)^2(5h) = 180s^2h$ — yes!

The answer is B.

A cone has radius $r$ and height $h$. A second cone has volume 180 times the first. Which could be its dimensions?

A) $R = 5r$ and $H = 6h$
B) $R = 6r$ and $H = 5h$
C) $R = 10r$ and $H = 18h$
D) $R = 3r$ and $H = 10h$

We need $R^2 \cdot H = 180 \cdot r^2 h$ (the $\dfrac{1}{3}\pi$ cancels). Check B: $(6r)^2(5h) = 180r^2h$. The answer is B.

Tip: You don't need to check all four options. Once you find one that works, move on.

 

Scaling: Surface Area and Area Ratios

When dimensions scale by $k$, area and surface area scale by $k^2$.

Square P has side $4k$ and square Q has side $148k$ (where $k > 0$). The area of Q is how many times the area of P?

A) $37$
B) $74$
C) $1{,}369$
D) $148$

Ratio of sides $= \dfrac{148k}{4k} = 37$. Area ratio $= 37^2 = 1{,}369$. The answer is C.

Sphere P has radius $5k$ and sphere Q has radius $115k$. The surface area of Q is how many times that of P?

A) $23$
B) $115$
C) $529$
D) $46$

Radius ratio $= \dfrac{115k}{5k} = 23$. Surface area ratio $= 23^2 = 529$. The answer is C.

Circle P has a radius 84 times the radius of circle Q. The area of P is $m$ times the area of Q. What is $m$?

$m = 84^2 = 7{,}056$.

Cube X has edge lengths 21 times those of cube Y. The volume of X is $p$ times the volume of Y. What is $p$?

$p = 21^3 = 9{,}261$.

 

Two Cylinders Joined Along a Base

When two identical cylinders (each with surface area $S$) are glued together along one circular base, two circular faces are hidden. The combined surface area is $2S - 2\pi r^2$. If the problem states this equals $\dfrac{n}{m}S$, set up the equation and solve for $r$.

Two identical cylinders each have height 15 mm and surface area $S$ mm$^2$. Joined along one base, the combined surface area is $\dfrac{12}{7}S$. What is the radius?

A) $3$
B) $5$
C) $6$
D) $10$

Each cylinder: $S = 2\pi r^2 + 2\pi r(15) = 2\pi r^2 + 30\pi r$.
Combined: $2S - 2\pi r^2 = \dfrac{12}{7}S$.
$2S - \dfrac{12}{7}S = 2\pi r^2$
$\dfrac{2}{7}S = 2\pi r^2$
$S = 7\pi r^2$

But $S = 2\pi r^2 + 30\pi r$, so $7\pi r^2 = 2\pi r^2 + 30\pi r$. Divide by $\pi$: $7r^2 = 2r^2 + 30r$, giving $5r^2 = 30r$, so $r = 6$. The answer is C.

Two identical cylinders each have height 10 cm and surface area $S$. Joined along one base, the result has surface area $\dfrac{5}{3}S$. What is the radius?

A) $4$
B) $5$
C) $10$
D) $20$

$2S - 2\pi r^2 = \dfrac{5}{3}S$, so $\dfrac{1}{3}S = 2\pi r^2$, giving $S = 6\pi r^2$.
But $S = 2\pi r^2 + 20\pi r$, so $6\pi r^2 = 2\pi r^2 + 20\pi r$. Then $4r^2 = 20r$ and $r = 5$. The answer is B.

Strategy for joined-solid problems: (1) Write $S$ in terms of $r$ and $h$. (2) Express the combined surface area as $2S - 2(\text{hidden face area})$. (3) Set it equal to the given fraction of $S$. (4) Solve the resulting equation for $r$.

 

What to Do on Test Day

• Volume ratio problems: If the volume is $n$ times as large, check each answer choice by computing $S^2 \cdot H$ (for pyramids/cones) and seeing which equals $n \cdot s^2 \cdot h$. You don't need to check all four — stop when one works.
• Area/surface area ratio: Linear dimensions $\times k$ → area/surface area $\times k^2$. Compute $k$ first: $k = \dfrac{\text{new dimension}}{\text{old dimension}}$.
• Dimension offsets: "5 feet shorter" means subtract; "5 feet longer" means add. Get the translation right before plugging into the formula.
• Joined cylinders: The combined surface area is $2S - 2\pi r^2$ (two circular faces are hidden). Set this equal to the given fraction of $S$ and solve.
• Key shortcuts:
• $\dfrac{1}{3}\pi$ cancels when comparing two cones (or two pyramids) — just compare $r^2 h$ (or $s^2 h$)
• For cubes, volume ratio $= k^3$ and surface area ratio $= k^2$
• When $k$ has a variable like $k$ in the side length, it cancels in the ratio