51 basic shapes and formulas

Basic Geometrical Shapes and Formulas

If we connect the two ends of a rope that’s resting flat in a plane, we obtain a loop.

If we connect the two ends of a rope that’s resting flat in a plane, we obtain a loop.The loop forms a perimeter or border that encloses a flat area, or a plane-shape.

If we connect the two ends of a rope that’s resting flat in a plane, we obtain a loop.The loop forms a perimeter or border that encloses a flat area, or a plane-shape.The length of the border, i.e. the length of the rope, is also referred to as the perimeter of the area.

If we connect the two ends of a rope that’s resting flat in a plane, we obtain a loop.The loop forms a perimeter or border that encloses a flat area, or a plane-shape.The length of the border, i.e. the length of the rope, is also referred to as the perimeter of the area.All the areas above are enclosed by the same rope, so they have equal perimeters.

Following shapes are polygons:A plane-shape is a polygon if it is formed by straight lines.

Following shapes are polygons: These are not polygons:A plane-shape is a polygon if it is formed by straight lines.

Three sided polygons are triangles.

If the sides of a triangle are labeled as a, b, and c, then the perimeter isP = a + b + c.

Triangles with three equal sides are equilateral triangles. s

The perimeter of an equilateral triangle is P = 3s.s

Rectangles are 4-sided polygons where the sides are joint at a right angle as shown.

Rectangles are 4-sided polygons where the sides are joint at a right angle as shown. s

ss A square

Rectangle with four equal sides are squares.

Rectangles are 4-sided polygons where the sides are joint at a right angle as shown. s

ss A square

Rectangle with four equal sides are squares.The perimeter of a squares is P = s + s + s + s = 4s

Basic Geometrical Shapes and FormulasExample A. We have to fence in an area with two squares and an equilateral triangle as shown. How many feet of fences do we need?

The area consists of two squares and an equilateral triangle so all the sides, measured from corner to corner, are equal.

The area consists of two squares and an equilateral triangle so all the sides, measured from corner to corner, are equal. There are 9 sections where each is 20 ft hence we need 9 x 20 = 180 ft of fence.

If we know two adjacent sides of a rectangle, then we know all four sides because their opposites sides are identical.

If we know two adjacent sides of a rectangle, then we know all four sides because their opposites sides are identical.We will use the word “height” for the vertical side and “width” for the horizontal side.

width (w)

height (h)

If we know two adjacent sides of a rectangle, then we know all four sides because their opposites sides are identical.We will use the word “height” for the vertical side and “width” for the horizontal side. The perimeter of a rectangle is h + h + w + w or that

width (w)

height (h)

P = 2h + 2w

Example B. a. We want to rope off a 50-meter by 70-meter rectangular area and also rope off sections of area as shown. How many meters of rope do we need?

70 mWe have three heights where each requires 50 meters of rope,

and three widths where each requires 70 meters of rope. 50 m

and three widths where each requires 70 meters of rope.Hence it requires3(50) + 3(70) = 150 + 210 = 360 meters of rope.

b. What is the perimeter of the following step-shape if all the short segments are 2 feet?

2 ftThe perimeter of the step-shape is the same as the perimeter of the rectangle that boxes it in as shown.

The height of the rectangle is 6 ft and the width is 10 ft, so the perimeter P = 2(6) +2(10) = 32 ft.

AreaIf we connect the two ends of a rope that’s resting flat in a plane, the rope form a loopthat encloses an area.

AreaIf we connect the two ends of a rope that’s resting flat in a plane, the rope form a loopthat encloses an area.The word “area” also denotes the amount of surface enclosed.

AreaIf we connect the two ends of a rope that’s resting flat in a plane, the rope form a loopthat encloses an area.The word “area” also denotes the amount of surface enclosed. If each side of a square is 1 unit, we define the area of the square to be 1 unit x 1 unit = 1 unit2, i.e. 1 square-unit.

If each side of a square is 1 unit, we define the area of the square to be 1 unit x 1 unit = 1 unit2, i.e. 1 square-unit.Hence the areas of the following squares are:

1 square-inch

1 square-inch 1 square-meter

1 m2 1 mi2

1 square-inch 1 square-meter 1 square-mile

1 m2 1 mi2

1 square-inch 1 square-meter 1 square-mileWe find the area of rectangles by cutting them into squares.

3 miAreaA 2 mi x 3 mi rectangle may be cut into six 1 x 1 squares so it covers an area of 2 x 3 = 6 mi2 (square miles).

= 6 mi2 2 x 3

3 miAreaA 2 mi x 3 mi rectangle may be cut into six 1 x 1 squares so it covers an area of 2 x 3 = 6 mi2 (square miles).In general, given the rectangle with height = h (units) width* = w (units), h

= 6 mi2 2 x 3

A = h x w (unit2) then its area A = h x w (unit2).

= 6 mi2 2 x 3

The area of a square is s*s = s2.

= 6 mi2 2 x 3

The area of a square is s*s = s2.By cutting and pasting, we may find areas of other shapes.

= 6 mi2 2 x 3

The area of a square is s*s = s2.By cutting and pasting, we may find areas of other shapes.Example C. a. Find the area of R as shown. Assume the unit is meter.

= 6 mi2 2 x 3

There are two basic approaches. R

= 6 mi2 2 x 3

I. We may view R as a 12 x 12 square with a 4 x 8 corner removed.

= 6 mi2 2 x 3

12 12Hence the area of R is12 x 12 – 4 x 8

= 6 mi2 2 x 3

12 12Hence the area of R is12 x 12 – 4 x 8 = 144 – 32 = 112 m2.

AreaIl. We may dissect R into two rectangles labeled I and II.

4 4 I II

I IIArea of I is 12 x 8 = 96,

I IIArea of I is 12 x 8 = 96,area of II is 4 x 4 = 16.

I IIArea of I is 12 x 8 = 96,area of II is 4 x 4 = 16.The area of R is the sum of the two or 96 + 16 = 112 m2.

b. Find the area of the following shape R where all the short segments are 2 ft.

b. Find the area of the following shape R where all the short segments are 2 ft.Let’s cut R into three rectangles as shown.

III 2 ft

b. Find the area of the following shape R where all the short segments are 2 ft.Let’s cut R into three rectangles as shown. Area of I is 2 x 2 = 4,

III 2 ft

b. Find the area of the following shape R where all the short segments are 2 ft.Let’s cut R into three rectangles as shown. Area of I is 2 x 2 = 4,area of II is 2 x 6 = 12,

III 2 ft

b. Find the area of the following shape R where all the short segments are 2 ft.Let’s cut R into three rectangles as shown. Area of I is 2 x 2 = 4,area of II is 2 x 6 = 12,and area of III is 2 x 5 = 10.

III 2 ft

b. Find the area of the following shape R where all the short segments are 2 ft.Let’s cut R into three rectangles as shown. Area of I is 2 x 2 = 4,area of II is 2 x 6 = 12,and area of III is 2 x 5 = 10.Hence the total area is 4 + 12 + 10 = 16 ft2.

III 2 ft

b. Find the area of the following shape R where all the short segments are 2 ft.Let’s cut R into three rectangles as shown. Area of I is 2 x 2 = 4,area of II is 2 x 6 = 12,and area of III is 2 x 5 = 10.Hence the total area is 4 + 12 + 10 = 16 ft2.

III 2 ft

By cutting and pasting we obtain the following area formulas.

AreaA parallelogram is a shape enclosed by two sets of parallel lines.

AreaA parallelogram is a shape enclosed by two sets of parallel lines. By cutting and pasting, we may arrange a parallelogram into a rectangle.

Hence the area of the parallelogram is A = h x b whereh = height and b = base.

For example, the area of all the parallelograms shown here is 8 x 12 = 96 ft2, so they are the same size.

12 ft8 ft

12 ft8 ft 12 ft8 ft

12 ft8 ft

8 ft8 ft12 ft12 ft

12 ft8 ft

AreaA triangle is half of a parallelogram.

AreaA triangle is half of a parallelogram.Given a triangle, we may copy it and paste to itself to make a parallelogram,

AreaA triangle is half of a parallelogram.Given a triangle, we may copy it and paste to itself to make a parallelogram, and the area of the triangle is half of the area of the parallelogram formed.

Therefore the area of a triangle is h x b

2A = (h x b) ÷ 2 or A = where h = height and b = base.

For example, the area of all the triangles shown here is (8 x 12) ÷ 2 = 48 ft2, i.e. they are the same size. 12 ft

8 ft 8 ft

For example, the area of all the triangles shown here is (8 x 12) ÷ 2 = 48 ft2, i.e. they are the same size. 12 ft

8 ft8 ft

12 ft12 ft

AreaA trapezoid is a 4-sided figure with one set of opposite sides parallel.

Example D. Find the area of the following trapezoid R. Assume the unit is meter.

A trapezoid is a 4-sided figure with one set of opposite sides parallel.

By cutting R parallel to one side as shown, we split R into two areas, one parallelogram and one triangle.

5Example D. Find the area of the following trapezoid R. Assume the unit is meter.

A trapezoid is a 4-sided figure with one set of opposite sides parallel. R

The parallelogram has base = 8 and height = 5, hence its area is 8 x 5 = 40 m2.

The parallelogram has base = 8 and height = 5, hence its area is 8 x 5 = 40 m2.The triangle has base = 4 and height = 5, hence its area is (4 x 5) ÷ 2 = 10 m2.

Therefore the area of the trapezoid is 40 + 10 = 50 m2.

We may find the area of any trapezoid by cutting it into one parallelogram and one triangle.

Circumference and Area of CirclesA circle has a center x,

Circumference and Area of CirclesA circle has a center x, and the distance from any location on the circle to C is a fixed number r, r is called the radius of the circle.

xThe diameter d of the circle is the length any straight line that goes through the center x connecting two opposite points.

d (diameter)

The diameter d of the circle is the length any straight line that goes through the center x connecting two opposite points.

d (diameter)

The diameter d of the circle is the length any straight line that goes through the center x connecting two opposite points. Hence d = 2r.

d (diameter)

The diameter d of the circle is the length any straight line that goes through the center x connecting two opposite points. Hence d = 2r. The perimeter C of a circle is called the circumference and C = πd or C = 2πr where π ≈ 3.14…

d (diameter)

The diameter d of the circle is the length any straight line that goes through the center x connecting two opposite points. Hence d = 2r. The perimeter C of a circle is called the circumference and C = πd or C = 2πr where π ≈ 3.14…We may use 3 as an under–estimation for π.

d (diameter)

The diameter d of the circle is the length any straight line that goes through the center x connecting two opposite points. Hence d = 2r. The perimeter C of a circle is called the circumference and C = πd or C = 2πr where π ≈ 3.14…We may use 3 as an under–estimation for π.Example D. Is 25 feet of rope enough to mark off a circle of radius r = 9 ft on the ground?

d (diameter)

The diameter d of the circle is the length any straight line that goes through the center x connecting two opposite points. Hence d = 2r. The perimeter C of a circle is called the circumference and C = πd or C = 2πr where π ≈ 3.14…We may use 3 as an under–estimation for π.Example D. Is 25 feet of rope enough to mark off a circle of radius r = 9 ft on the ground?No, 25 ft is not enough since the circumference C is at least 3 x 9 = 27 ft.

Circumference and Area of Circles

The area A (enclosed by) of a circle isA = πr2 where π ≈ 3.14… A

The area A (enclosed by) of a circle isA = πr2 where π ≈ 3.14…Example E. a. Approximate the area ofthe circle with a 5–meter radius using 3 as the estimated value of π.

Estimating using 3 in stead of 3.14 …we have the area A to be at least 3 x 52

Estimating using 3 in stead of 3.14 …we have the area A to be at least 3 x 52 = 3 x 25 = 75 m2.

b. Approximate the area of the circle with a 5–meter radius using π = 3.14 as the estimated value of π.

The better approximate answer using π = 3.14 is 3.14 x 52 = 3.14 x 75 = 78.5 m2

b. Approximate the area of the circle with a 5–meter radius using π = 3.14 as the estimated value of π.

VolumeThe volume of a solid is the measurement of the amount of “room” or “space” the solid occupies.

Volume

The volume of a solid is the measurement of the amount of “room” or “space” the solid occupies.A cube is a square–box, i.e. a box whose edges are the same. s

sA cube

Volume

The volume of a solid is the measurement of the amount of “room” or “space” the solid occupies.

We define the volume of a cube whose sides are 1 unit to be 1 unit x 1 unit x 1 unit = 1 unit3, i.e. 1 cubic unit.

A cube is a square–box, i.e. a box whose edges are the same. s

sA cube

Volume

sA cube

1 in 1 in1 in3

1 cubic inch

Volume

sA cube

1 in 1 in1 in3

1 cubic inch

1 m 1 m1 m3

1 cubic meter

Volume

sA cube

1 in 1 in1 in3

1 cubic inch

1 m 1 m1 m3

1 cubic meter

1 mi 1 mi1 mi31 mi

1 cubic mile

Volume

sA cube

1 in 1 in1 in3

1 cubic inch

1 m 1 m1 m3

1 cubic meter

1 mi 1 mi1 mi31 mi

1 cubic mile

We can cut larger cubes into smaller cubes to calculate their volume.

VolumeA 2 x 2 x 2 cube has volume 2 x 2 x 2 = 23 = 8, a 3 x 3 x 3 cube has volume 33 = 27, a 4 x 4 x 4 cube has volume 43 = 64 (unit3).

Volume

w = width

A rectangular box is specified by three sides: the length, the width, and the height. We say that the dimension of the box is “l by w by h”. l = length

h = height

A 2 x 2 x 2 cube has volume 2 x 2 x 2 = 23 = 8, a 3 x 3 x 3 cube has volume 33 = 27, a 4 x 4 x 4 cube has volume 43 = 64 (unit3).

Volume

w = width

h = height

A 2 x 2 x 2 cube has volume 2 x 2 x 2 = 23 = 8,

Here is a “4 by 3 by 2” box.

a 3 x 3 x 3 cube has volume 33 = 27, a 4 x 4 x 4 cube has volume 43 = 64 (unit3).

Volume

w = width

h = height

Here is a “4 by 3 by 2” box.Assuming the unit is inch, then the box may be cut into 2 x 3 x 4 = 241–inch cubes so its volume is 24 in3. 4 3

Volume

w = width

h = height

Here is a “4 by 3 by 2” box.Assuming the unit is inch, then the box may be cut into 2 x 3 x 4 = 241–inch cubes so its volume is 24 in3.

We define the volume V of a box whose sides are l, w, and h to be V = l x w x h unit3. In particular the volume of a cube whose sides equal to s is V = s x s x s = s3 unit3.

VolumeExample F. a. How many cubic inches are there in a cubic foot?(There are 12 inches in 1 foot.)

VolumeExample F. a. How many cubic inches are there in a cubic foot?(There are 12 inches in 1 foot.) One cubic foot is 1 ft x 1 ft x 1ft

VolumeExample F. a. How many cubic inches are there in a cubic foot?(There are 12 inches in 1 foot.) One cubic foot is 1 ft x 1 ft x 1ft or 12 in x 12 in x 12 in = 1728 in3.

Example b. How many cubic feet are there in the followingsolid?

One cubic foot is 1 ft x 1 ft x 1ft or 12 in x 12 in x 12 in = 1728 in3.

We may view the solid is consisted oftwo solids I and II as shown.

Method 1.

IThe volume of I is 3 x 3 x 3 = 27,the volume of II is 10 x 3 x 6 = 180.

Method 1.

IThe volume of I is 3 x 3 x 3 = 27,the volume of II is 10 x 3 x 6 = 180. Hence the volume of the entire solid is180 + 27 = 207 ft3.

Method 1.

Volume

The solid may be viewed as a boxwith volume 3 x 10 x 9 = 270 with a top portion removed.

Method 2.

10 ft 3 ft

Volume

The solid may be viewed as a boxwith volume 3 x 10 x 9 = 270 with a top portion removed. The dimension of the removed portion is also a box with volume 3 x 3 x 7 = 63.

Method 2.

10 ft 3 ft

3 ft 7 ft

Volume

The solid may be viewed as a boxwith volume 3 x 10 x 9 = 270 with a top portion removed. The dimension of the removed portion is also a box with volume 3 x 3 x 7 = 63. Hence the volume of the given solid is270 – 63 = 207 ft3.

Method 2.

10 ft 3 ft

3 ft 7 ft

51 basic shapes and formulas

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