Area and Volume Pattern - Multi Step Composite

This pattern requires multiple calculation steps — finding intermediate values, combining shapes, or converting between formulas before reaching the answer.

 

Combined Areas: Adding or Subtracting

When a figure is made of multiple shapes, find each area separately and add (or subtract for a cutout).

Rectangle P has area 60 ft$^2$. Square Q has side length 5 ft. What is the total area?

A) $35$
B) $55$
C) $65$
D) $85$

Square area $= 5^2 = 25$. Total $= 60 + 25 = 85$. The answer is D.

A circular garden has area 64 m$^2$. A square fountain with area 16 m$^2$ is built in the center. What area is covered in grass?

A) $50$
B) $56$
C) $48$
D) $80$

$64 - 16 = 48$ m$^2$. The answer is C.

 

Combining Sphere Volumes

Compute each sphere's volume individually, then add.

Sphere P has radius 3 cm. Sphere Q has volume $288\pi$ cm$^3$. What is the total volume?

A) $18\pi$
B) $300\pi$
C) $9\pi$
D) $324\pi$

Volume of P $= \dfrac{4}{3}\pi(3)^3 = 36\pi$. Total $= 36\pi + 288\pi = 324\pi$. The answer is D.

 

Cone Inside a Cylinder

When a cone is inscribed in a cylinder with the same base and height, the space between them equals the cylinder volume minus the cone volume.

A cone is inscribed in a cylinder. Both have base radius 8 cm and height 15 cm. What is the volume of the space inside the cylinder but outside the cone, to the nearest cubic centimeter?

A) $251$
B) $1{,}005$
C) $2{,}011$
D) $3{,}016$

Cylinder $= \pi(8)^2(15) = 960\pi$. Cone $= \dfrac{1}{3}\pi(8)^2(15) = 320\pi$. Difference $= 640\pi \approx 2{,}011$. The answer is C.

Shortcut: The cone is always $\dfrac{1}{3}$ of the cylinder, so the remaining space is $\dfrac{2}{3}$ of the cylinder.

 

Cone: Base Area Given, Find Volume

When the problem gives the base area instead of the radius, use $V = \dfrac{1}{3} \times (\text{base area}) \times h$ directly.

The base of a right circular cone has area $36\pi$ in$^2$ and height 7 inches. What is the volume?

A) $14\pi$
B) $252\pi$
C) $84\pi$
D) $336\pi$

$V = \dfrac{1}{3}(36\pi)(7) = \dfrac{252\pi}{3} = 84\pi$. The answer is C. Option B forgets the $\dfrac{1}{3}$.

 

Area of a Triangle from Side Lengths with Radicals

Some problems give side lengths involving $\sqrt{3}$ or other radicals. Identify the two legs of a right triangle and use $A = \dfrac{1}{2}(\text{leg}_1)(\text{leg}_2)$.

A right triangle has side lengths $4\sqrt{3}$, $5\sqrt{3}$, and $\sqrt{123}$. What is the area?

A) $60$
B) $9\sqrt{3} + \sqrt{123}$
C) $30$
D) $15\sqrt{41}$

Check: $(4\sqrt{3})^2 + (5\sqrt{3})^2 = 48 + 75 = 123 = (\sqrt{123})^2$. So the legs are $4\sqrt{3}$ and $5\sqrt{3}$.

$A = \dfrac{1}{2}(4\sqrt{3})(5\sqrt{3}) = \dfrac{1}{2}(20 \cdot 3) = 30$. The answer is C.

 

Three Points Define a Circle

When three points are on a circle, find the center and radius, then compute area or circumference.

If two points share the same $y$-coordinate (like $(2, 10)$ and $(12, 10)$), they form a horizontal chord. The center's $x$-coordinate is the midpoint of that chord. Use the third point to find the radius.

Three points $(2, 10)$, $(12, 10)$, and $(7, 15)$ lie on a circle. The circumference is $k\pi$. What is $k$?

A) $5$
B) $7$
C) $10$
D) $25$

The horizontal chord from $(2, 10)$ to $(12, 10)$ has midpoint $(7, 10)$. The center must be at $x = 7$ (on the perpendicular bisector). Let the center be $(7, c)$.

Distance to $(2, 10)$: $\sqrt{(7-2)^2 + (c-10)^2} = \sqrt{25 + (c-10)^2}$
Distance to $(7, 15)$: $|15 - c|$

Set equal: $25 + (c-10)^2 = (15-c)^2$. Expanding: $25 + c^2 - 20c + 100 = 225 - 30c + c^2$. Simplify: $10c = 100$, so $c = 10$. Wait — that makes the center $(7, 10)$, with radius $= 5$. Check: distance to $(7, 15) = 5$. Circumference $= 2\pi(5) = 10\pi$, so $k = 10$. The answer is C.

 

Line Segment as One Leg of a Right Triangle

Some problems give a line segment (via endpoint coordinates) as one leg and the area, then ask for the other leg.

A segment from $(2, 5)$ to $(8, 7)$ is one leg of a right triangle with area 30. What is the other leg?

A) $2\sqrt{10}$
B) $3\sqrt{2} \cdot \sqrt{10}$
C) $3\sqrt{10}$
D) $\dfrac{15}{\sqrt{2}}$

Leg length $= \sqrt{(8-2)^2 + (7-5)^2} = \sqrt{36 + 4} = \sqrt{40} = 2\sqrt{10}$.

$A = \dfrac{1}{2}(\text{leg}_1)(\text{leg}_2)$, so $30 = \dfrac{1}{2}(2\sqrt{10})(\text{leg}_2)$. Then $\text{leg}_2 = \dfrac{30}{\sqrt{10}} = 3\sqrt{10}$. The answer is C.

Tip: In multi-step problems, organize your work clearly. Find intermediate values one at a time and label them before combining.

 

What to Do on Test Day

• Break the problem into small steps. Multi-step problems are just several easy steps chained together. Write down each intermediate result.
• Combined areas: Add areas for joined shapes; subtract for cutouts. "Area of the region between" always means subtraction.
• Cone inside a cylinder: The cone is always $\dfrac{1}{3}$ of the cylinder (same base and height). The remaining space is $\dfrac{2}{3}$ of the cylinder.
• "Base area given": If the problem gives you the base area directly (e.g., $36\pi$), use $V = \dfrac{1}{3} \times \text{base area} \times h$ without finding the radius first.
• Distance formula for side lengths: When vertices are given as coordinates, use $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ to find side lengths before computing area.
• Simplify radicals as you go. Don't wait until the end — products of radicals are easier to simplify step by step (e.g., $(4\sqrt{3})(5\sqrt{3}) = 20 \cdot 3 = 60$).