Parallelogram

Mathematics · FE Reference Handbook section

Mathematics

6 formulas

10 exam-style examples

~57 min

All Mathematics lectures

Learning objectives

What you must be able to do before leaving this section.

This chapter section covers Parallelogram within Mathematics. Read it the way you would read a textbook chapter: the theory first so the relations mean something, then every equation with its use and its trap, then 10 fully worked examples with the arithmetic shown line by line, and finally a self-check you should be able to answer without notes.

Explain, in your own words, what parallelogram describes physically and when it applies.
State every one of the 6 relations the handbook lists here and name each symbol with its unit.
Select the correct relation from the wording of an exam stem within 20 seconds.
Carry a complete solution from givens to a "most nearly" answer with the correct unit.
Recognise the distractors generated by the unit trap: radians vs degrees — set the calculator before the first trig entry.

Lecture

Why this section exists. Parallelogram is the part of Mathematics that lets you connect an algebraic or calculus expression that must be evaluated exactly to a number you can defend. Before any equation is useful you must be able to picture the physical situation it describes; the schematic below is that picture.

How the theory is built. The handbook prints results, not derivations. Each relation in this section comes from one governing principle applied to the idealised system: state the principle, impose the stated assumptions, and the printed equation follows. Knowing which assumption each relation rests on is what lets you reject a wrong answer choice in seconds.

How it is examined. Items from this page are written as short symbolic manipulations with one numeric evaluation at the end. Roughly two thirds are direct substitution, one third require one intermediate quantity from a neighbouring relation, and a small number are conceptual — testing whether you know the assumption, not the arithmetic.

The habit that earns the points. Unit discipline. radians vs degrees — set the calculator before the first trig entry. Every relation below is dimensionally consistent only when that rule is honoured, and the distractor set is deliberately built from candidates who ignored it. Write the unit next to every number you substitute, every time.

How to study this page. Read the theory, then cover the formula cards and try to reproduce each relation from its description. Then work the examples with the solution hidden, revealing one line at a time. Finish with the self-check questions; if you cannot answer one, return to the matching formula card.

Three engineers in hard hats and safety vests reviewing drawings on a truck tailgate. — Photo 1. Where this shows up in practice: parallelogram.
Capstone Studio instructional photograph

Theory, developed

Read this before the equations — it is what makes them memorable.

The physical situation

Every item from this section describes an algebraic or calculus expression that must be evaluated exactly. Sketch it before you compute — a labelled sketch with the givens on it converts a wordy stem into a solvable problem and exposes the quantity the examiner left out on purpose.

The governing principle

The 6 relations on this page are consequences of one principle applied to that idealised system. Identify which quantity is conserved, balanced, or defined, and the correct equation follows without memorisation.

Assumptions and limits of validity

Each printed relation carries silent assumptions — linearity, steady state, uniformity, small deformation, or standard conditions, depending on the subject. Conceptual exam items are written by violating exactly one of these, so read the sentence above the equation as carefully as the equation itself.

Solution procedure you should automate

1) Read the last sentence of the stem to identify the requested quantity. 2) Locate the relation on this page whose left-hand side is that quantity. 3) Tabulate the givens with units and mark the missing symbol. 4) If a symbol is missing, find the one relation that produces it. 5) Rearrange symbolically, substitute once, evaluate, and round only at the end.

Notation used in this section

P	Quantity produced by "P = 2^a + bh" — read its definition and unit from the handbook line directly above the equation.
d1	Quantity produced by "d1 = a2 + b2 - 2ab^cos zh" — read its definition and unit from the handbook line directly above the equation.
d2	Quantity produced by "d2 = a2 + b2 + 2ab^cos zh" — read its definition and unit from the handbook line directly above the equation.
d 12 + d 22	Quantity produced by "d 12 + d 22 = 2 _ a2 + b2i" — read its definition and unit from the handbook line directly above the equation.
A	Quantity produced by "A = ah = ab^sin zh" — read its definition and unit from the handbook line directly above the equation.
If a	Quantity produced by "If a = b, the parallelogram is a rhombus." — read its definition and unit from the handbook line directly above the equation.

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1

Circular sector and circular segment areas — Parallelogram

A circular sedimentation basin of radius 3.0 ft is partitioned by a central angle of 40°. Compute the arc length, the area of the circular sector, and the area of the circular segment cut off by the chord.

Given

R = 3.0 ft
θ = 40° = 0.6981 rad

Find

Arc length s, sector area, and segment area

Start with the thinking

All mensuration formulas for sectors and segments require the central angle in radians.
The segment is the sector minus the triangle formed by the two radii and the chord.

Step-by-step solution

Formula
$s = R θ$
Substituting
$s = 3.0 \times 0.6981 = 2.09 f t$
Formula
$A_{s} ec t or = ½ R^{2} θ$
Substituting
$A_{s} ec t or = ½ (3.0)^{2} (0.6981) = 3.14 f t^{2}$
Formula
$A_{s} e g m e n t = ½ R^{2} (θ - sin θ)$
Substituting
$A_{s} e g m e n t = ½ (3.0)^{2} (0.6981 - 0.6428) = 0.25 f t^{2}$

Answer: s = 2.09 ft, sector = 3.14 ft², segment = 0.25 ft²

Why the other options are there

180.0 ft² (degrees used directly)
-1.36 ft² (wrong triangle area)

Reference: FE Reference Handbook — Mathematics → Parallelogram

Example 2

Circular sector and circular segment areas — Parallelogram (2)

A circular sedimentation basin of radius 10.5 ft is partitioned by a central angle of 117°. Compute the arc length, the area of the circular sector, and the area of the circular segment cut off by the chord.

Given

R = 10.5 ft
θ = 117° = 2.0420 rad

Find

Arc length s, sector area, and segment area

Start with the thinking

All mensuration formulas for sectors and segments require the central angle in radians.
The segment is the sector minus the triangle formed by the two radii and the chord.

Step-by-step solution

Formula
$s = R θ$
Substituting
$s = 10.5 \times 2.0420 = 21.44 f t$
Formula
$A_{s} ec t or = ½ R^{2} θ$
Substituting
$A_{s} ec t or = ½ (10.5)^{2} (2.0420) = 112.6 f t^{2}$
Formula
$A_{s} e g m e n t = ½ R^{2} (θ - sin θ)$
Substituting
$A_{s} e g m e n t = ½ (10.5)^{2} (2.0420 - 0.8910) = 63.45 f t^{2}$

Answer: s = 21.44 ft, sector = 112.6 ft², segment = 63.45 ft²

Why the other options are there

6,450 ft² (degrees used directly)
57.44 ft² (wrong triangle area)

Reference: FE Reference Handbook — Mathematics → Parallelogram

Example 3

Circular sector and circular segment areas — Parallelogram (3)

A circular sedimentation basin of radius 9.0 ft is partitioned by a central angle of 160°. Compute the arc length, the area of the circular sector, and the area of the circular segment cut off by the chord.

Given

R = 9.0 ft
θ = 160° = 2.7925 rad

Find

Arc length s, sector area, and segment area

Start with the thinking

All mensuration formulas for sectors and segments require the central angle in radians.
The segment is the sector minus the triangle formed by the two radii and the chord.

Step-by-step solution

Formula
$s = R θ$
Substituting
$s = 9.0 \times 2.7925 = 25.13 f t$
Formula
$A_{s} ec t or = ½ R^{2} θ$
Substituting
$A_{s} ec t or = ½ (9.0)^{2} (2.7925) = 113.1 f t^{2}$
Formula
$A_{s} e g m e n t = ½ R^{2} (θ - sin θ)$
Substituting
$A_{s} e g m e n t = ½ (9.0)^{2} (2.7925 - 0.3420) = 99.25 f t^{2}$

Answer: s = 25.13 ft, sector = 113.1 ft², segment = 99.25 ft²

Why the other options are there

6,480 ft² (degrees used directly)
72.60 ft² (wrong triangle area)

Reference: FE Reference Handbook — Mathematics → Parallelogram

Example 4

Circular sector and circular segment areas — Parallelogram (4)

A circular sedimentation basin of radius 14.0 ft is partitioned by a central angle of 86°. Compute the arc length, the area of the circular sector, and the area of the circular segment cut off by the chord.

Given

R = 14.0 ft
θ = 86° = 1.5010 rad

Find

Arc length s, sector area, and segment area

Start with the thinking

All mensuration formulas for sectors and segments require the central angle in radians.
The segment is the sector minus the triangle formed by the two radii and the chord.

Step-by-step solution

Formula
$s = R θ$
Substituting
$s = 14.0 \times 1.5010 = 21.01 f t$
Formula
$A_{s} ec t or = ½ R^{2} θ$
Substituting
$A_{s} ec t or = ½ (14.0)^{2} (1.5010) = 147.1 f t^{2}$
Formula
$A_{s} e g m e n t = ½ R^{2} (θ - sin θ)$
Substituting
$A_{s} e g m e n t = ½ (14.0)^{2} (1.5010 - 0.9976) = 49.34 f t^{2}$

Answer: s = 21.01 ft, sector = 147.1 ft², segment = 49.34 ft²

Why the other options are there

8,428 ft² (degrees used directly)
49.10 ft² (wrong triangle area)

Reference: FE Reference Handbook — Mathematics → Parallelogram

Example 5

Circular sector and circular segment areas — Parallelogram (5)

A circular sedimentation basin of radius 8.5 ft is partitioned by a central angle of 95°. Compute the arc length, the area of the circular sector, and the area of the circular segment cut off by the chord.

Given

R = 8.5 ft
θ = 95° = 1.6581 rad

Find

Arc length s, sector area, and segment area

Start with the thinking

All mensuration formulas for sectors and segments require the central angle in radians.
The segment is the sector minus the triangle formed by the two radii and the chord.

Step-by-step solution

Formula
$s = R θ$
Substituting
$s = 8.5 \times 1.6581 = 14.09 f t$
Formula
$A_{s} ec t or = ½ R^{2} θ$
Substituting
$A_{s} ec t or = ½ (8.5)^{2} (1.6581) = 59.90 f t^{2}$
Formula
$A_{s} e g m e n t = ½ R^{2} (θ - sin θ)$
Substituting
$A_{s} e g m e n t = ½ (8.5)^{2} (1.6581 - 0.9962) = 23.91 f t^{2}$

Answer: s = 14.09 ft, sector = 59.90 ft², segment = 23.91 ft²

Why the other options are there

3,432 ft² (degrees used directly)
23.77 ft² (wrong triangle area)

Reference: FE Reference Handbook — Mathematics → Parallelogram

Example 6

Circular sector and circular segment areas — Parallelogram (6)

A circular sedimentation basin of radius 12.0 ft is partitioned by a central angle of 112°. Compute the arc length, the area of the circular sector, and the area of the circular segment cut off by the chord.

Given

R = 12.0 ft
θ = 112° = 1.9548 rad

Find

Arc length s, sector area, and segment area

Start with the thinking

All mensuration formulas for sectors and segments require the central angle in radians.
The segment is the sector minus the triangle formed by the two radii and the chord.

Step-by-step solution

Formula
$s = R θ$
Substituting
$s = 12.0 \times 1.9548 = 23.46 f t$
Formula
$A_{s} ec t or = ½ R^{2} θ$
Substituting
$A_{s} ec t or = ½ (12.0)^{2} (1.9548) = 140.7 f t^{2}$
Formula
$A_{s} e g m e n t = ½ R^{2} (θ - sin θ)$
Substituting
$A_{s} e g m e n t = ½ (12.0)^{2} (1.9548 - 0.9272) = 73.99 f t^{2}$

Answer: s = 23.46 ft, sector = 140.7 ft², segment = 73.99 ft²

Why the other options are there

8,064 ft² (degrees used directly)
68.74 ft² (wrong triangle area)

Reference: FE Reference Handbook — Mathematics → Parallelogram

Example 7

Circular sector and circular segment areas — Parallelogram (7)

A circular sedimentation basin of radius 7.5 ft is partitioned by a central angle of 140°. Compute the arc length, the area of the circular sector, and the area of the circular segment cut off by the chord.

Given

R = 7.5 ft
θ = 140° = 2.4435 rad

Find

Arc length s, sector area, and segment area

Start with the thinking

All mensuration formulas for sectors and segments require the central angle in radians.
The segment is the sector minus the triangle formed by the two radii and the chord.

Step-by-step solution

Formula
$s = R θ$
Substituting
$s = 7.5 \times 2.4435 = 18.33 f t$
Formula
$A_{s} ec t or = ½ R^{2} θ$
Substituting
$A_{s} ec t or = ½ (7.5)^{2} (2.4435) = 68.72 f t^{2}$
Formula
$A_{s} e g m e n t = ½ R^{2} (θ - sin θ)$
Substituting
$A_{s} e g m e n t = ½ (7.5)^{2} (2.4435 - 0.6428) = 50.64 f t^{2}$

Answer: s = 18.33 ft, sector = 68.72 ft², segment = 50.64 ft²

Why the other options are there

3,938 ft² (degrees used directly)
40.60 ft² (wrong triangle area)

Reference: FE Reference Handbook — Mathematics → Parallelogram

Example 8

Circular sector and circular segment areas — Parallelogram (8)

A circular sedimentation basin of radius 13.5 ft is partitioned by a central angle of 73°. Compute the arc length, the area of the circular sector, and the area of the circular segment cut off by the chord.

Given

R = 13.5 ft
θ = 73° = 1.2741 rad

Find

Arc length s, sector area, and segment area

Start with the thinking

All mensuration formulas for sectors and segments require the central angle in radians.
The segment is the sector minus the triangle formed by the two radii and the chord.

Step-by-step solution

Formula
$s = R θ$
Substituting
$s = 13.5 \times 1.2741 = 17.20 f t$
Formula
$A_{s} ec t or = ½ R^{2} θ$
Substituting
$A_{s} ec t or = ½ (13.5)^{2} (1.2741) = 116.1 f t^{2}$
Formula
$A_{s} e g m e n t = ½ R^{2} (θ - sin θ)$
Substituting
$A_{s} e g m e n t = ½ (13.5)^{2} (1.2741 - 0.9563) = 28.96 f t^{2}$

Answer: s = 17.20 ft, sector = 116.1 ft², segment = 28.96 ft²

Why the other options are there

6,652 ft² (degrees used directly)
24.98 ft² (wrong triangle area)

Reference: FE Reference Handbook — Mathematics → Parallelogram

Example 9

Circular sector and circular segment areas — Parallelogram (9)

A circular sedimentation basin of radius 13.0 ft is partitioned by a central angle of 55°. Compute the arc length, the area of the circular sector, and the area of the circular segment cut off by the chord.

Given

R = 13.0 ft
θ = 55° = 0.9599 rad

Find

Arc length s, sector area, and segment area

Start with the thinking

All mensuration formulas for sectors and segments require the central angle in radians.
The segment is the sector minus the triangle formed by the two radii and the chord.

Step-by-step solution

Formula
$s = R θ$
Substituting
$s = 13.0 \times 0.9599 = 12.48 f t$
Formula
$A_{s} ec t or = ½ R^{2} θ$
Substituting
$A_{s} ec t or = ½ (13.0)^{2} (0.9599) = 81.11 f t^{2}$
Formula
$A_{s} e g m e n t = ½ R^{2} (θ - sin θ)$
Substituting
$A_{s} e g m e n t = ½ (13.0)^{2} (0.9599 - 0.8192) = 11.90 f t^{2}$

Answer: s = 12.48 ft, sector = 81.11 ft², segment = 11.90 ft²

Why the other options are there

4,648 ft² (degrees used directly)
-3.39 ft² (wrong triangle area)

Reference: FE Reference Handbook — Mathematics → Parallelogram

Example 10

Circular sector and circular segment areas — Parallelogram (10)

A circular sedimentation basin of radius 4.0 ft is partitioned by a central angle of 153°. Compute the arc length, the area of the circular sector, and the area of the circular segment cut off by the chord.

Given

R = 4.0 ft
θ = 153° = 2.6704 rad

Find

Arc length s, sector area, and segment area

Start with the thinking

All mensuration formulas for sectors and segments require the central angle in radians.
The segment is the sector minus the triangle formed by the two radii and the chord.

Step-by-step solution

Formula
$s = R θ$
Substituting
$s = 4.0 \times 2.6704 = 10.68 f t$
Formula
$A_{s} ec t or = ½ R^{2} θ$
Substituting
$A_{s} ec t or = ½ (4.0)^{2} (2.6704) = 21.36 f t^{2}$
Formula
$A_{s} e g m e n t = ½ R^{2} (θ - sin θ)$
Substituting
$A_{s} e g m e n t = ½ (4.0)^{2} (2.6704 - 0.4540) = 17.73 f t^{2}$

Answer: s = 10.68 ft, sector = 21.36 ft², segment = 17.73 ft²

Why the other options are there

1,224 ft² (degrees used directly)
13.36 ft² (wrong triangle area)

Reference: FE Reference Handbook — Mathematics → Parallelogram

Self-check

Answer these without notes before moving on.

Without looking, state the relation on this page whose left-hand side is the quantity most often requested, and name every symbol in it.
Which assumption, if violated, makes the main relation of this section invalid?
Given an algebraic or calculus expression that must be evaluated exactly, what is the first quantity you would compute, and why that one first?
Which unit conversion in this subject most often produces a wrong answer choice, and what is its numerical factor?
Rework Example 1 above from the givens alone, without reading the solution lines.

Chapter summary

Parallelogram contains 6 relations; you must be able to find this page in under 15 seconds.
Exam style: short symbolic manipulations with one numeric evaluation at the end.
Unit rule: radians vs degrees — set the calculator before the first trig entry.
Work the 10 examples until the solution path, not the answer, is automatic.

Common traps in this section

radians vs degrees — set the calculator before the first trig entry
Answering the intermediate quantity instead of the quantity requested.
Rounding intermediate values before the final step.
Using a relation from an adjacent handbook section that shares a symbol.
Skipping the sketch — most lost points on this page start with a misread geometry.

Parallelogram

Learning objectives

Lecture

Theory, developed

Notation used in this section

Core formulas for this FE topic

P=2a+bh

d1 = a2 + b2 - 2ab^cos zhNeeds technical review

d2 = a2 + b2 + 2ab^cos zhNeeds technical review

d 12 + d 22 = 2 _ a2 + b2iNeeds technical review

A = ah = ab^sin zhNeeds technical review

Ifa=b,theparallelogramisarhombus.

Worked exam-style examples

Self-check

Chapter summary

Common traps in this section