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Geometry: areas, volumes and triangles

Geometry gives the area, perimeter and volume formulas used every day in engineering (the section of a beam, the volume of a tank) and the laws to solve any triangle. This topic gathers them, explains how to solve for a measurement from the area or volume and how to handle the ambiguous case of triangles.

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Areas and perimeters

Figure	Area	Perimeter
Rectangle $b \times h$	$bh$	$2 (b + h)$
Triangle	$\frac{1}{2} bh$	$a + b + c$
Trapezoid	$\frac{( B + b )}{2} h$	sum of sides
Circle	$π r^{2}$	$2 π r$
Sector of angle $θ$	$\frac{θ}{36 0 ^{\circ}} π r^{2}$	$2 r + \frac{θ}{36 0 ^{\circ}} 2 π r$
Regular polygon of $n$ sides $a$	$\frac{1}{2} P a_{p}$ (apothem $a_{p}$ )	$na$

Volumes and surface areas of solids

Solid	Volume	Total surface area
Prism (base $A_{b}$ )	$A_{b} h$	$2 A_{b} + P_{b} h$
Cylinder	$π r^{2} h$	$2 π r (r + h)$
Pyramid	$\frac{1}{3} A_{b} h$	base + faces
Cone (slant height $g$ )	$\frac{1}{3} π r^{2} h$	$π r (r + g)$
Sphere	$\frac{4}{3} π r^{3}$	$4 π r^{2}$

Note the $\frac{1}{3}$ : a pyramid or a cone takes up one third of the prism or cylinder with the same base and height. Scaling every measurement by $k$ multiplies lengths by $k$ , areas by $k^{2}$ and volumes by $k^{3}$ .

Solving for a measurement: if the area or volume is known and a measurement is missing, set up the formula and solve. For example, a circle of area 50: $π r^{2} = 50$ → $r = 50/ π$ .

Solving triangles

A triangle is determined by three data, at least one of them a side. The angles add up to $18 0^{\circ}$ , and two laws relate sides and angles:

Notation: each side is named with the lowercase letter of the opposite angle.

Data	Use	Solutions
three sides (SSS)	law of cosines for each angle	one (if they satisfy the triangle inequality)
two sides and the included angle (SAS)	cosines for the third side	one
two angles and a side (ASA, AAS)	180° sum and law of sines	one
two sides and a non-included angle (SSA)	law of sines	zero, one or two (ambiguous case)

The area is $\frac{1}{2} ab sin C$ or, with the three sides, Heron's formula: $A = s (s - a) (s - b) (s - c)$ , with $s$ the semiperimeter.

The Geometry tool writes each formula, substitutes the data and gives the result; if given the area, perimeter or volume, it solves for the missing measurement and checks that it reproduces the data. In the ambiguous case it shows both triangles.

Common mistakes

Using the diameter where the radius goes (or vice versa).
Forgetting the $\frac{1}{3}$ of the cone and the pyramid.
Confusing the height $h$ of the cone with its slant height $g$ .
In the SSA case, keeping only one triangle: if $sin B < 1$ and the supplement of $B$ fits, there are two.
Having the calculator in radians when using angles in degrees.

Worked examples

Example 1Two sides and the included angle

A triangle has

a = 7

b = 9

m and

C = 4 0^{\circ}

. Find side

c

, the other angles and the area.

Solution

Law of cosines: $c^{2} = 7^{2} + 9^{2} - 2 \cdot 7 \cdot 9 cos 4 0^{\circ} = 130 - 96.52 = 33.48$ → $c = 5.78605$ m.
$cos A = \frac{b ^{2} + c ^{2} - a ^{2}}{2 b c}$ → $A = 51.045 8^{\circ}$ ; $B = 18 0^{\circ} - 4 0^{\circ} - 51.045 8^{\circ} = 88.954 2^{\circ}$ .
Area: $\frac{1}{2} ab sin C = \frac{1}{2} \cdot 7 \cdot 9 \cdot sin 4 0^{\circ} = 20.2478$ m².

c = 5.78605

A = 51.045 8^{\circ}

B = 88.954 2^{\circ}

, area

20.2478

m².

Solve it in Simuveliq

In Math → Geometry, Solve triangles: type Side a = 7, Side b = 9 and Angle C = 40. The solution (SAS) gives $c = 5.78605$ m, $A = 51.045 8^{\circ}$ , $B = 88.954 2^{\circ}$ and area 20.2478 m², plus the altitudes, medians and inscribed and circumscribed radii.

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Example 2The ambiguous case

Solve the triangle with

a = 8

b = 10

and

A = 4 0^{\circ}

Solution

Law of sines: $sin B = \frac{b sin A}{a} = \frac{10 sin 4 0 ^{\circ}}{8} = 0.8035$ .
Two angles have that sine: $B_{1} = 53.464 1^{\circ}$ and $B_{2} = 18 0^{\circ} - 53.464 1^{\circ} = 126.53 6^{\circ}$ . Both fit with $A = 4 0^{\circ}$ (they add up to less than 180°).
Triangle 1: $C = 86.535 9^{\circ}$ , $c = \frac{a sin C}{sin A} = 12.423$ . Triangle 2: $C = 13.464 1^{\circ}$ , $c = 2.89784$ .

Two triangles:

B = 53.464 1^{\circ}

C = 86.535 9^{\circ}

c = 12.423

; or

B = 126.53 6^{\circ}

C = 13.464 1^{\circ}

c = 2.89784

Solve it in Simuveliq

In Math → Geometry, Solve triangles: type Side a = 8, Side b = 10 and Angle A = 40. The tool says «SSA (ambiguous case) · two possible triangles» and shows both, with $c = 12.423$ and $c = 2.89784$ .

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Example 3Volume and area of a cone

A cone has radius 3 m and height 4 m. Find the slant height, the volume and the total surface area.

Solution

The height, the radius and the slant height form a right triangle: $g = 3^{2} + 4^{2} = 5$ m.
$V = \frac{1}{3} π \cdot 3^{2} \cdot 4 = 12 π = 37.6991$ m³.
Lateral area $π r g = 15 π = 47.1239$ m²; total $π r (r + g) = 24 π = 75.3982$ m².

g = 5

V = 37.6991

m³,

A = 75.3982

m².

Solve it in Simuveliq

In Math → Geometry, Solids: choose Cone and type Radius r = 3 and Height h = 4. It gives slant height 5 m, volume 37.6991 m³, lateral area 47.1239 m² and total surface area 75.3982 m²; the steps substitute each value in its formula.

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Example 4The radius from the area

A circle has an area of 50 m². What are its radius and circumference?

Solution

$π r^{2} = 50$ → $r = 50/ π = 3.98942$ m.
$C = 2 π r = 25.0663$ m.

r = 3.98942

m, circumference

25.0663

Solve it in Simuveliq

In Math → Geometry, Plane figures: choose Circle, leave the radius empty and type the area, 50. The tool solves for the radius, 3.98942 m (marked «solved for»), and gives the circumference 25.0663 m.

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Example 5A 1000 m³ tank

A 10 m tall cylindrical tank must hold 1000 m³. What radius does it need and how much sheet metal does it take (total area)?

Solution

$π r^{2} \cdot 10 = 1000$ → $r = 100/ π = 5.6419$ m.
Total area: $2 π r (r + h) = 2 π \cdot 5.6419 \cdot 15.6419 = 554.491$ m² (each base is 100 m²).

r = 5.6419

m; total area

554.491

m².

Solve it in Simuveliq

In Math → Geometry, Solids: choose Cylinder, leave the radius empty, type Height h = 10 and the volume, 1000. It gives the solved radius 5.6419 m, the area of one base 100 m² and the total area 554.491 m².

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Geometry: areas, volumes and triangles

Areas and perimeters

Volumes and surface areas of solids

Solving triangles

Common mistakes

Worked examples

Example 1Two sides and the included angle

Solution

Solve it in Simuveliq

Example 2The ambiguous case

Solution

Solve it in Simuveliq

Example 3Volume and area of a cone

Solution

Solve it in Simuveliq

Example 4The radius from the area

Solution

Solve it in Simuveliq

Example 5A 1000 m³ tank

Solution

Solve it in Simuveliq

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