Geometry Applications: Area and Volume

[Music]

[Music]

Title: Geometry Applications: Area and Volume

Title: Geometry Applications: Area and Volume

Title: Geometry Applications: Volume and Density

Title: Geometry Applications: Volume and Density

Title: Geometry Applications: Volume and Density

ON APRIL 10, 1912 THE TITANIC WAS LAUNCHED.

AT THE TIME IT WAS THE WORLD'S LARGEST

PASSENGER STEAMSHIP.

IT WAS SCHEDULED TO SAIL FROM SOUTHAMPTON, ENGLAND

TO NEW YORK CITY, MAKING A FEW STOPS ALONG THE WAY.

BUT FOUR DAYS INTO THE VOYAGE

IN THE MIDDLE OF THE ATLANTIC OCEAN

THE TITANIC HIT AN ICEBERG

THAT PUNCTURED THE HULL OF THE SHIP

CAUSING IT TO TAKE ON WATER.

IN LESS THAN THREE HOURS

THE SHIP WAS SO WATERLOGGED THAT IT SANK.

NEARLY 2,000 PEOPLE DIED.

AS WE APPROACH THE 100TH ANNIVERSARY OF THIS

TRAGIC SHIPWRECK, ONE THING HASN'T CHANGED:

WHY THE TITANIC SANK.

BUT BEFORE ADDRESSING THAT, IT'S IMPORTANT

TO UNDERSTAND WHY A SHIP OF THE TITANIC'S SIZE

COULD EVEN FLOAT IN THE FIRST PLACE.

AFTER ALL, THE SHIP WAS MADE UP OF 46,000 TONS

OF METAL, WOOD AND OTHER MATERIALS.

THIS ILLUSTRATION SHOWS THE COMPARATIVE SIZE

OF THE TITANIC.

CLEARLY THIS WAS A MASSIVE SHIP.

BUT HOWEVER MUCH MASS A SHIP MIGHT HAVE,

SO LONG AS ITS DENSITY IS LESS THAN

THE DENSITY OF WATER, IT WILL FLOAT.

DENSITY IS THE RATIO OF THE MASS OF AN OBJECT

TO THE VOLUME OF THE OBJECT,

WHICH IS THE SCIENTIFIC DEFINITION OF DENSITY.

LET'S LOOK AT THE GEOMETRY BEHIND DENSITY.

HERE IS A CUBE.

IT HAS MASS M AND VOLUME V.

AS THE MASS INCREASES SO DOES THE VOLUME.

THE INCREASE IN THE MASS AND VOLUME

HAPPENS AT A CONSTANT RATE

AND CAN BE SUMMARIZED WITH THIS EQUATION: M=cV.

IN THE EQUATION, c IS A CONSTANT.

AND THIS EQUATION IS AN EXAMPLE OF A

DIRECT VARIATION.

THE GRAPH OF A DIRECT VARIATION

IS A LINEAR FUNCTION THAT CROSSES THE ORIGIN.

THE CONSTANT c IS CALLED THE CONSTANT OF VARIATION.

WHEN YOU TAKE THE RATIO OF M OVER V

THEN THE CONSTANT c IS ALSO THE SLOPE

OF THE LINEAR FUNCTION.

THIS CONSTANT OF VARIATION, THE SLOPE,

IS WHAT SCIENTISTS REFER TO AS THE DENSITY.

FOR SIMPLICITY, IN THE METRIC SYSTEM

THE DENSITY OF WATER IS EQUAL TO 1.

ANY MATERIAL THAT HAS A DENSITY GREATER THAN 1,

OR A LINEAR GRAPH WITH A SLOPE GREATER THAN 1

IS DENSER THAN WATER AND WILL SINK.

ANY MATERIAL THAT HAS A DENSITY LESS THAN 1

OR A LINEAR GRAPH WITH A SLOPE LESS THAN 1

IS LESS DENSE THAN WATER AND WILL FLOAT.

THE EQUATION FOR DENSITY IS A FUNCTION.

THE VARIABLE IS V AND FOR ANY GIVEN SUBSTANCE

M IS A CONSTANT.

THE FUNCTION IS WRITTEN THIS WAY AND IS READ THIS WAY:

D OF V EQUALS M OVER V.

RETURNING TO THE EXAMPLE OF THE CUBE SHAPED OBJECT

OF MASS M, WE KNOW THAT THE VOLUME OF A CUBE

IS S CUBED WHERE S IS THE LENGTH OF

ANY OF THE EDGES OF THE CUBE.

SO FOR A CUBE WE GET THE FUNCTION

D EQUALS M OVER S CUBED.

THIS IS AN EXAMPLE OF A RATIONAL FUNCTION

WHOSE GRAPH HAS THIS SHAPE.

LET'S EXPLORE THIS GRAPH ON THE NSPIRE AND SEE

WHAT WE CAN LEARN ABOUT DENSITY AND THE TITANIC.

TURN ON THE TI-NSPIRE.

CREATE A NEW DOCUMENT.

YOU MAY NEED TO SAVE A PREVIOUS DOCUMENT.

CREATE A GRAPH WINDOW.

YOU'LL BE GRAPHING y=m OVER x SQUARED

WHERE M CAN TAKE ON DIFFERENT VALUES.

TO DO THAT CREATE A SLIDER.

PRESS MENU AND UNDER ACTIONS SELECT INSERT SLIDER.

YOU'LL SEE THE SLIDER APPEAR ON THE UPPER LEFT CORNER OF

THE SCREEN WITH THE VARIABLE NAME ALREADY HIGHLIGHTED.

CHANGE THE VARIABLE TO m AND PRESS ENTER.

NEXT PRESS TAB TWICE TO GO TO THE FUNCTION ENTRY LINE.

PRESS CONTROL AND THE DIVISION SYMBOL TO CREATE

A PLACEHOLDER FOR THE RATIONAL EXPRESSION.

INPUT THE LETTER m IN THE NUMERATOR.

PRESS THE DOWN ARROW TO GO TO THE DENOMINATOR

AND INPUT X CUBED, THEN PRESS ENTER.

BY DEFAULT THE VALUE OF m IS 5

AND IT RANGES IN VALUE FROM 0 TO 10.

MOVE THE POINTER ABOVE THE SLIDER.

PRESS AND HOLD THE CLICK KEY

UNTIL THE OPEN HAND BECOMES A CLOSED HAND.

USE THE LEFT AND RIGHT ARROW KEYS

TO CHANGE THE VALUE OF m.

YOU'LL SEE THAT AS THE VALUE OF m INCREASES

THE GRAPH SHIFTS TO THE RIGHT.

PRESS CONTROL AND G TO BRING UP THE

FUNCTION ENTRY LINE FOR f2.

INPUT 1 AND PRESS ENTER.

THIS VALUE REPRESENTS THE DENSITY OF WATER.

ANY VALUES OF y ABOVE THE HORIZONTAL LINE

REPRESENT A CONFIGURATION TOO DENSE TO FLOAT.

NOW COMPARE THE GRAPH

AS THE VALUE OF THE VOLUME INCREASES.

CREATE A NEW SLIDER.

PRESS MENU AND UNDER ACTIONS SELECT INSERT SLIDER.

THIS SLIDER WILL APPEAR ABOVE THE FIRST SLIDER

WITH THE TEXT FLD HIGHLIGHTED.

INPUT THE LETTER C AND PRESS ENTER.

PRESS ESCAPE AND MOVE THE POINTER

ABOVE THE SECOND SLIDER.

PRESS AND HOLD THE CLICK KEY

TO SELECT THE SECOND SLIDER.

MOVE IT TO THE OTHER SIDE OF THE SCREEN

AND PRESS ESCAPE.

PRESS CONTROL AND G

TO BRING UP THE FUNCTION ENTRY LINE FOR f3.

PRESS CONTROL AND THE DIVISION SYMBOL TO CREATE

A PLACEHOLDER FOR ANOTHER RATIONAL EXPRESSION.

INPUT M IN THE NUMERATOR AND PRESS THE DOWN ARROW.

INPUT THE EXPRESSION C TIMES X CUBED.

THIS REPRESENTS A SITUATION WHERE THE MASS OF THE OBJECT

IS THE SAME BUT THE VOLUME CHANGES.

PRESS ENTER.

FOR VALUES OF C GREATER THAN 1

THE DENSITY DECREASES COMPARED TO THE OTHER GRAPH.

SO SUPPOSE YOU HAVE A CONTAINER

MADE OF A HEAVY MATERIAL.

IN ORDER FOR IT TO FLOAT

THE VOLUME OF AIR INSIDE THE CONTAINER

MUST BE LARGE ENOUGH THAT THE RATIO OF

MASS TO VOLUME IS LESS THAN OR EQUAL TO 1.

THIS IS WHY A SHIP MADE OUT OF METAL

CAN STILL FLOAT ON WATER.

THE HULL OF THE SHIP IS BASICALLY A HOLLOWED OUT

VOLUME OF AIR THAT KEEPS THE DENSITY OF THE SHIP

LESS THAN OR EQUAL TO THE DENSITY OF WATER.

RETURNING TO THE TITANIC, THE HULL OF THAT SHIP

CAN BE APPROXIMATED BY A TRIANGULAR PRISM.

A TRIANGULAR PRISM HAS TWO TRIANGULAR FACES

AND THREE RECTANGULAR SIDES.

A NET FOR A TRIANGULAR PRISM LOOKS LIKE THIS.

THE VOLUME OF A RECTANGULAR PRISM

IS THE AREA OF THE TRIANGULAR BASE

TIMES THE LENGTH OF THE RECTANGULAR LENGTH.

THE FORMULA V EQUALS A TIMES L

CAN BE USED TO CALCULATE THE VOLUME.

BUT WE ALSO KNOW THE AREA OF A TRIANGLE

IS ONE HALF THE BASE TIMES THE HEIGHT.

SO THE VOLUME FORMULA CHANGES TO

V EQUALS ONE HALF B TIMES H TIMES L.

CONTRAST THIS TO A RECTANGULAR PRISM

WHICH HAS TWO RECTANGULAR FACES.

IF THE TWO PRISMS, ONE RECTANGULAR

AND ONE TRIANGULAR HAVE SIMILAR DIMENSIONS,

THE LENGTH AND THE HEIGHT,

THE RECTANGULAR PRISM WILL HAVE MORE VOLUME.

IF A SHIP LIKE THE TITANIC NEEDS TO MAXIMIZE ITS VOLUME

WHY DOESN'T IT HAVE A HULL IN THE SHAPE OF A

RECTANGULAR PRISM INSTEAD OF A TRIANGULAR PRISM?

THE ANSWER HAS TO DO WITH THE PHYSICS OF SAILING.

A TRIANGULAR PRISM SHAPED HULL WILL SLICE THROUGH THE

WATER AND MOVE MUCH FASTER THAN A RECTANGULAR HULL.

THE TITANIC WAS CALLED THIS FOR A REASON.

ITS DIMENSIONS WERE:

WHENEVER YOU PLACE AN OBJECT IN WATER

IT DISPLACES AN AMOUNT OF WATER

EQUAL TO THE WEIGHT OF THE OBJECT.

THE TITANIC'S DISPLACEMENT WAS 46,000 TONS.

THE HEIGHT LISTED HERE IS JUST FOR THE HEIGHT OF THE

TRIANGULAR PRISM SECTION THAT MAKES UP THE HULL.

LET'S USE THE TI-NSPIRE TO CALCULATE THE VOLUME

AND DENSITY OF THE TITANIC'S HULL.

CREATE A CALCULATOR WINDOW.

PRESS THE HOME KEY AND SELECT CALCULATOR.

TO CALCULATE THE VOLUME OF THE

TRIANGULAR PRISM THAT WE ARE USING

TO APPROXIMATE THE VOLUME OF THE HULL

INPUT THE EXPRESSION 0.5 X 270 X 28 X 30

AND RATHER THAN PRESSING ENTER

ASSIGN THIS VALUE TO A VARIABLE.

PRESS THE CONTROL BUTTON AND THE VAR KEY.

YOU WILL SEE AN ARROW POINTING TO THE RIGHT.

INPUT THE VARIABLE NAMED VOLUME AND PRESS ENTER.

YOU HAVE ASSIGNED THE CALCULATED VALUE

OF THE VOLUME TO VARIABLE.

THE VOLUME THAT YOU CALCULATED

IS IN METERS CUBED.

THE UNITS OF DENSITY

THAT DEFINE WATER WITH A DENSITY OF 1

IS GRAMS PER CENTIMETERS CUBED.

WE NEED TO CONVERT FROM METERS CUBED

TO CENTIMETERS CUBED.

SO MULTIPLY THE RESULT BY 100 CUBED.

INPUT THE EXPRESSION VOLUME TIMES 100 CUBED

AND ASSIGN IT TO THE SAME VARIABLE VOLUME.

PRESS ENTER.

ASSIGN THE VALUE 46,000 TO A NEW VARIABLE CALLED MASS.

NOW WE NEED TO CONVERT TONS TO GRAMS.

THE CONVERSION FACTOR IS 1.01605 TIMES 10 TO THE SIXTH.

INPUT THIS EXPRESSION INTO THE CALCULATOR WINDOW.

MASS TIMES 1.01605 TIMES 10 TO THE SIXTH

AND ASSIGN THIS VALUE BACK INTO THE MASS VARIABLE.

PRESS ENTER.

YOU ARE NOW READY TO CALCULATE THE DENSITY.

INPUT THE EXPRESSION MASS OVER VOLUME

AND ASSIGN IT TO A NEW VARIABLE: DENSITY.

PRESS ENTER.

YOU'LL SEE THAT THE DENSITY OF THE TITANIC

BEFORE IT HIT THE ICEBERG WAS ROUGHLY 0.4 WHICH,

BECAUSE THE VALUE IS LESS THAN 1,

WOULD HAVE MEANT THAT THE SHIP WAS FLOATING.

WHICH IT WAS.

BUT ONCE THE TITANIC HIT THE ICEBERG

WATER STARTED FLOODING THE HULL.

WHEN A SHIP STARTS TAKING ON WATER ITS DENSITY CHANGES.

BUT IT CHANGES QUICKLY BECAUSE BOTH THE MASS

AND THE VOLUME ARE CHANGING.

THE INCREASING AMOUNT OF WATER INCREASES THE

OVERALL MASS OF THE SHIP WHICH INCREASES THE DENSITY.

THE INCREASE IN THE AMOUNT OF WATER

COMES AT THE EXPENSE OF THE VOLUME OF AIR IN THE HULL.

THE VOLUME DECREASES

MAKING THE DENSITY INCREASE EVEN MORE.

THE COMBINATION OF BOTH VARIABLES CHANGING RAPIDLY

INCREASES THE DENSITY.

IN FACT FOR ANY VOLUME DECREASE X

THAT AMOUNT OF VOLUME IS CONVERTED INTO

A CORRESPONDING AMOUNT OF WATER.

FOR ANY CUBIC CENTIMETER OF WATER MULTIPLY BY .001

TO FIND THE CORRESPONDING MASS OF WATER.

FOR EXAMPLE SUPPOSE ONE-TENTH OF THE VOLUME

OF THE HULL IS FILLED WITH WATER.

THIS EXPRESSION IS THE EXPRESSION

FOR THE NEW DENSITY.

INPUT THIS EXPRESSION INTO THE CALCULATOR WINDOW

AND PRESS ENTER.

THE NEW DENSITY IS ABOUT 0.46.

SO THE DENSITY HAS GONE UP

BUT IT IS STILL MUCH LOWER THAN ONE.

SO THE SHIP WOULD NOT HAVE SUNK

TAKING IN THIS AMOUNT OF WATER.

AT WHAT PERCENTAGE OF THE VOLUME HAS THE SHIP

TAKEN ON TOO MUCH WATER AND WILL SINK?

WE CAN USE A GENERAL FORM OF THE EXPRESSION.

IN THIS EXPRESSION THE X OVER 100 TERM REPRESENTS THE

PERCENTAGE OF THE VOLUME CONVERTED TO WATER.

LET'S GRAPH THE FUNCTION BASED ON THIS EXPRESSION

AND SEE WHERE IT INTERSECTS THE GRAPH OF y=1.

PRESS THE HOME KEY AND SELECT A GRAPH WINDOW.

AT THE f1 ENTRY LINE, INPUT 1

AND PRESS THE DOWN ARROW.

AT THE f2 FUNCTION ENTRY LINE INPUT THIS EXPRESSION.

PRESS CONTROL AND THE DIVISION SYMBOL TO CREATE

A PLACEHOLDER FOR THE RATIONAL EXPRESSION.

AFTER YOU HAVE INPUT THE FUNCTION PRESS ENTER.

CHANGE THE WINDOW SETTINGS, PRESS MENU

AND UNDER WINDOW/ZOOM CHANGE THE WINDOW SETTINGS

TO XMIN EQUALS 0, XMAX EQUALS 100,

YMIN EQUALS 0 AND YMAX EQUALS 5.

PRESS OKAY.

NOW FIND WHERE THE TWO GRAPHS INTERSECT.

PRESS MENU AND UNDER POINTS AND LINES

SELECT INTERSECTION POINT.

MOVE THE POINTER ABOVE EACH GRAPH

AND PRESS ENTER EACH TIME.

YOU'LL SEE THE INTERSECTION POINT AS SLIGHTLY ABOVE 41%

SO THE TITANIC HAD TO TAKE ON A PERCENTAGE OF THE

HULL'S VOLUME GREATER THAN 41% BEFORE IT WOULD SINK.

NOW THE TITANIC DID HAVE SEPARATE

WATERTIGHT COMPARTMENTS SO THAT

IF THERE WERE A HOLE IN THE HULL OF THE SHIP

NOT ALL OF THE HULL WOULD BE FLOODED WITH WATER.

CONTAINING THE AMOUNT OF WATER WOULD HAVE

PREVENTED THE SHIP FROM SINKING.

SO WHY DID IT SINK?

AS IT TURNS OUT, OF THE 16 WATERTIGHT COMPARTMENTS

ON THE TITANIC SIX OF THEM WERE DAMAGED

AS THE ICEBERG SCRAPED THE SIDE OF THE HULL.

ENOUGH COMPARTMENTS WERE DAMAGED TO LET IN

ENOUGH WATER FOR THE SHIP TO REACH THE THRESHOLD

WHERE IT WOULD SINK.

HAD THE ICEBERG DAMAGED FEWER OF THE COMPARTMENTS

IT'S POSSIBLE THAT THE TITANIC WOULD HAVE

TAKEN ON WATER BUT NOT SUNK.

SO AS YOU CAN SEE SUBTLE CHANGES IN THE MEASURE

OF DENSITY CAN HAVE ENORMOUS CONSEQUENCES.

THE LOUVRE MUSEUM IN PARIS IS ONE OF

THE MOST IMPORTANT MUSEUMS IN THE WORLD.

IN IT YOU'LL FIND THE MONA LISA, THE VENUS DE MILO

AND MANY IMPORTANT WORKS OF ART FROM WORLD HISTORY.

IN 1984 THE MUSEUM WENT THROUGH AN EXTENSIVE

RENOVATION WHICH INCLUDED WHAT IS NOW

THE MOST NOTEWORTHY CHANGE TO THE MUSEUM:

THE GLASS PYRAMID.

IT IS A SQUARE PYRAMID,

MEANING THAT THE BASE OF THE PYRAMID IS A SQUARE.

A SQUARE PYRAMID HAS FOUR TRIANGULAR SIDES.

THE PYRAMID IS MADE UP OF A NUMBER OF GLASS PANELS

SO WHEN DESIGNING THE PYRAMID, THE BUILDERS

NEEDED TO KNOW THE SURFACE AREA OF THIS PYRAMID.

LET'S LOOK AT A NET FOR A SQUARE PYRAMID.

THE SURFACE AREA IS MADE UP OF THE AREA OF THE SQUARE

BASE AND THE AREAS OF THE FOUR TRIANGULAR SECTIONS.

SINCE THE TRIANGLES ARE CONGRUENT,

THEN THE SURFACE AREA OF THE TRIANGULAR SIDES

IS EQUAL TO FOUR TIMES THE AREA OF ONE OF THE TRIANGLES.

IF THE SQUARE SIDE HAS BASE b AND THE TRIANGULAR SIDES

HAVE HEIGHT h THEN THE SURFACE AREA OF

THE PYRAMID IS b SQUARED PLUS 4 TIMES 1/2 bh.

SIMPLIFYING WE GET: b SQUARED PLUS 2bh

FOR THE GLASS PYRAMID AT THE LOUVRE

THE BASE IS PART OF THE ENTRYWAY INTO THE MUSEUM.

SO ONLY THE SURFACE OF THE TRIANGULAR PORTIONS

ARE RELEVANT.

SO THE SURFACE AREA OF THIS PYRAMID IS 2bh.

NOW TURNING TO ONE OF THE TRIANGULAR SIDES

OF THE LOUVRE PYRAMID YOU'LL SEE THAT EACH SIDE

CONSISTS OF A TESSELLATION MADE UP OF RHOMBUSES.

ALONG THE SIDE OF THE TRIANGLE

THERE ARE 18 RHOMBUS SIDES.

ALONG THE BASE THERE ARE 18 RHOMBUS DIAGONALS.

WITH A RHOMBUS ALL FOUR SIDES ARE CONGRUENT.

THIS MEANS THAT A DIAGONAL DIVIDES A RHOMBUS

INTO TWO CONGRUENT ISOSCELES TRIANGLES.

FOCUS ON THE RHOMBUS AT THE APEX OF THE PYRAMID.

SPLIT THIS RHOMBUS INTO TWO CONGRUENT ISOSCELES

TRIANGLES BY DEFINING THIS HORIZONTAL SEGMENT.

NOTICE THAT THE LARGER TRIANGULAR FACE

OF THE PYRAMID AND THE TOP ANGLE OF THE SMALLER

ISOSCELES TRIANGLE SHARE THE SAME ANGLE.

SINCE THE LARGE AND SMALL TRIANGLES

ARE BOTH ISOSCELES TRIANGLES

THEN IT FOLLOWS THAT THE BASE TRIANGLES

ARE CONGRUENT TO EACH OTHER AS SHOWN HERE.

THIS MEANS THAT THE LARGER TRIANGLE

AND THE SMALLER ONE ARE SIMILAR TO EACH OTHER.

THIS IS IMPORTANT FOR FINDING

PROPORTIONAL SIDES OF THE TWO TRIANGLES.

ALSO MAKE A NOTE THAT SINCE THE SMALLER TRIANGLES

ARE SIMILAR TO THE LARGER TRIANGLE,

THE CORRESPONDING SIDES OF THE SMALLER TRIANGLE

ARE PARALLEL TO THE CORRESPONDING SIDES

OF THE LARGER TRIANGLE.

NOW EACH RHOMBUS SHAPED GLASS PANEL

IS A DIAMOND SHAPE

MEANING THAT THE RHOMBUS HAS FOUR 45 DEGREE ANGLES.

WE CAN USE THE PROPERTIES OF SIMILAR TRIANGLES

TO DETERMINE HOW MANY OF THESE GLASS PANELS

ARE NEEDED TO CREATE THE

TRIANGULAR SIDES OF THE PYRAMID.

LET'S USE THE TI-NSPIRE TO SOLVE THIS PROBLEM.

TURN ON THE TI-NSPIRE.

CREATE A NEW DOCUMENT.

YOU MAY NEED TO SAVE A PREVIOUS DOCUMENT.

CREATE A GEOMETRY WINDOW.

CREATE A LINE SEGMENT.

PRESS MENU AND UNDER POINTS AND LINES SELECT SEGMENT.

MOVE THE POINTER TO THE LOWER LEFT

PART OF THE SCREEN.

PRESS ENTER.

PRESS AND HOLD THE RIGHT ARROW KEY

TO CREATE THE SEGMENT.

WHEN THE SEGMENT COVERS MOST OF THE

LOWER PART OF THE SCREEN PRESS ENTER.

THIS LINE SEGMENT IS THE BASE OF ONE OF THE

TRIANGULAR SIDES OF THE PYRAMID.

IT IS THE BASE OF AN ISOSCELES TRIANGLE.

TO CONSTRUCT AN ISOSCELES TRIANGLE FIRST CONSTRUCT

THE PERPENDICULAR BISECTOR OF THE BASE.

PRESS MENU AND UNDER CONSTRUCTION

SELECT PERPENDICULAR BISECTOR.

MOVE THE POINTER ABOVE THE SEGMENT AND PRESS ENTER.

YOU'LL SEE THE PERPENDICULAR BISECTOR CONSTRUCTED.

EXTEND THE LENGTH OF THE PERPENDICULAR BISECTOR.

PRESS ESCAPE THEN MOVE THE POINTER

TO THE TOP END OF THE BISECTOR.

PRESS AND HOLD THE CLICK KEY TO SELECT IT.

PRESS AND HOLD THE UP ARROW TO INCREASE THE

LENGTH OF THE PERPENDICULAR BISECTOR.

NOW ADD TWO POINTS TO THE PERPENDICULAR BISECTOR

TO DEFINE THE TRIANGLE'S HEIGHT.

PRESS MENU AND UNDER POINTS AND LINES SELECT POINTS.

MOVE THE POINTER TO THE HORIZONTAL LINE

WHERE IT INTERSECTS THE BISECTOR.

YOU'LL SEE AN ONSCREEN LABEL THAT SAYS

"INTERSECTION POINT".

PRESS ENTER THEN PRESS AND HOLD THE UP ARROW

TO MOVE THE POINTER TO THE END OF THE LINE.

PRESS ENTER AGAIN.

NEXT CONSTRUCT THE TWO REMAINING

SIDES OF THE TRIANGLE.

PRESS MENU AND UNDER POINTS AND LINES SELECT SEGMENT.

MOVE THE POINTER TO THE TOP POINT

ON THE PERPENDICULAR BISECTOR.

PRESS ENTER.

THEN MOVE THE POINTER

TO ONE OF THE ENDPOINTS OF THE BASE.

PRESS ENTER AGAIN.

REPEAT THIS TO CONSTRUCT THE OTHER SIDE OF THE TRIANGLE.

WE KNOW THAT THE ANGLE AT THE TOP VERTEX

OF THE TRIANGLE IS 45 DEGREES.

THE PERPENDICULAR BISECTOR

RESULTS IN A 90 DEGREE ANGLE.

THIS MEANS THAT THE BASE ANGLES OF THE TRIANGLE

NEED TO BE 45 DEGREES.

SO MEASURE THE BASE ANGLES.

PRESS MENU AND UNDER MEASUREMENT SELECT ANGLE.

MOVE THE POINTER TO THE TOP POINT OF THE TRIANGLE

AND PRESS ENTER.

THEN MOVE THE POINTER TO ONE OF THE OTHER VERTICES

OF THE TRIANGLE AND PRESS ENTER AGAIN.

FINALLY MOVE THE POINTER ABOVE THE THIRD VERTEX

AND PRESS ENTER ONE MORE TIME.

YOU'LL SEE THE ANGLE MEASURE APPEAR.

REPEAT THIS PROCESS

FOR THE OTHER BASE ANGLE OF THE TRIANGLE.

TRY TO GET YOUR SCREEN TO LOOK LIKE THIS.

PRESS ESCAPE AND MOVE THE POINTER

TO THE TOP VERTEX OF THE TRIANGLE.

PRESS AND HOLD THE CLICK KEY TO SELECT THE POINT.

USE THE UP OR DOWN ARROW

TO CHANGE THE POSITION OF THE POINT.

NOTICE HOW THE BASE ANGLE MEASURES CHANGE.

YOU WANT TO MOVE THE POINT SO THAT

THE BASE ANGLES ARE 45 DEGREES.

YOU NEED THE MEASURE TO BE PRECISE

SO YOU MAY NEED TO MOVE ALL THE POINTS

ON THE TRIANGLE IN ORDER TO GET TO 45 DEGREES.

THIS TRIANGLE IS A MODEL OF ONE OF THE

TRIANGULAR SIDES OF THE LOUVRE PYRAMID.

MEASURE THE BASE AND HEIGHT OF THE PYRAMID.

THESE ARE THE MEASUREMENTS

WE WILL USE TO FIND THE CORRESPONDING AREA

OF THE RHOMBUS SHAPED PANELS.

PRESS MENU AND UNDER MEASUREMENT SELECT LENGTH.

MOVE THE POINTER ABOVE THE BASE AND PRESS ENTER

TO SEE THE LENGTH MEASUREMENT.

MOVE THE POINTER BELOW THE SEGMENT

AND PRESS ENTER AGAIN

TO PLACE THE MEASUREMENT ON SCREEN.

NEXT MOVE THE POINTER ABOVE ONE OF THE ENDPOINTS

OF THE PERPENDICULAR BISECTOR AND PRESS ENTER.

MOVE THE POINTER TO THE OTHER ENDPOINT

AND PRESS ENTER AGAIN.

YOU'LL SEE THE LENGTH MEASUREMENT.

MOVE THE POINTER TO THE SIDE OF THE VERTICAL SEGMENT

AND PRESS ENTER ONE MORE TIME.

ASSIGN THESE LENGTH MEASUREMENTS TO VARIABLES.

PRESS ESCAPE AND MOVE THE POINTER ABOVE

ONE OF THE MEASUREMENTS AT THE BASE OF THE TRIANGLE.

PRESS CONTROL AND MENU.

SELECT THE STORE OPTION.

CREATE A VARIABLE CALLED LARGEBASE AND PRESS ENTER.

NEXT, MOVE THE POINTER ABOVE

THE OTHER MEASUREMENT.

PRESS CONTROL AND MENU

AND SELECT THE STORE OPTION ONCE AGAIN.

CREATE A VARIABLE CALLED LARGEHEIGHT

AND PRESS ENTER.

NOW WE CAN CONSTRUCT ONE OF THE RHOMBUS SHAPED REGIONS.

PRESS MENU AND UNDER POINTS AND LINES SELECT POINT.

MOVE THE POINTER ALONG THE BASE OF THE TRIANGLE

AND ADD TWO POINTS NEAR EACH OF THE CORNER VERTICES

PRESSING ENTER EACH TIME TO ADD THE POINT.

MEASURE THE DISTANCES ALONG THESE SHORT SEGMENTS.

PRESS MENU AND UNDER MEASUREMENT SELECT LENGTH.

MOVE THE POINTER ABOVE ONE OF THE CORNER VERTICES

AND PRESS ENTER.

MOVE THE POINTER ABOVE ONE OF THE POINTS YOU JUST ADDED

AND PRESS ENTER AGAIN.

PLACE THE MEASUREMENT ONSCREEN.

REPEAT THIS FOR THE OTHER SHORT SEGMENT.

EACH OF THESE SHORT SEGMENTS WILL CORRESPOND

TO ONE OF THE RHOMBUS SHAPED SIDES.

AS WE SAW EARLIER THERE ARE 18 SUCH UNITS

ALONG EACH SIDE OF THE TRIANGLE.

PRESS ESCAPE AND MOVE THE POINTER ABOVE ONE OF THE

LENGTH MEASUREMENTS OF ONE OF THE SHORT SEGMENTS.

PRESS ENTER TO GO INTO EDIT MODE.

REPLACE THE MEASUREMENT WITH THE FORMULA

LARGEBASE DIVIDED BY 18 AND PRESS ENTER.

THE LENGTH CHANGES TO THE CORRESPONDING LENGTH

ON THE LOUVRE PYRAMID.

REPEAT THIS PROCESS FOR THE OTHER SHORT SEGMENT.

MOVE THE POINTER ABOVE THE MEASUREMENT

FOR THAT SEGMENT AND PRESS ENTER.

REPLACE THE MEASUREMENT WITH THE FORMULA

LARGEBASE DIVIDED BY 18 AND PRESS ENTER.

WE CAN NOW CREATE LINES

PARALLEL TO THE TRIANGULAR SIDES

THROUGH THE ENDPOINTS OF THE SHORT SEGMENTS.

PRESS MENU AND UNDER CONSTRUCTION

SELECT PARALLEL.

MOVE THE POINTER ABOVE ONE OF THE SIDES OF THE TRIANGLE

AND PRESS ENTER.

THEN MOVE THE POINTER ABOVE THE END POINT OF

THE SHORT SEGMENT ON THE BASE AND PRESS ENTER AGAIN.

YOU'LL SEE A LINE PARALLEL TO THE SIDE OF THE TRIANGLE.

REPEAT THIS PROCESS FOR THE OTHER SIDE OF THE TRIANGLE.

PRESS ESCAPE AND MOVE THE POINTER

TO THE END OF EACH PARALLEL LINE.

PRESS AND HOLD THE CLICK KEY TO SELECT THE ENDPOINT.

USE THE NAVIGATION ARROWS TO EXTEND THE PARALLEL LINES.

NOTICE THAT YOU NOW HAVE ONE OF THE OUTLINES

OF THE RHOMBUS SHAPED PANEL AT THE APEX OF THE PYRAMID.

CREATE INTERSECTION POINTS AMONG THE VARIOUS LINES.

PRESS MENU AND UNDER POINTS AND LINES

SELECT INTERSECTION POINTS.

CLICK ON PAIRS OF LINES TO CREATE INTERSECTION POINTS.

TRY TO GET YOUR SCREEN TO LOOK LIKE THIS.

NOW CONSTRUCT THE BASE AND HEIGHT

OF THE SMALLER ISOSCELES TRIANGLE

THAT IS SIMILAR TO THE LARGER TRIANGLE.

PRESS MENU AND UNDER POINTS AND LINES

SELECT SEGMENT.

CONSTRUCT THE TWO DIAGONALS OF THE RHOMBUS

AND CREATE AN INTERSECTION POINT FOR THESE DIAGONALS.

MEASURE THE LENGTHS OF THE BASE AND HEIGHT

OF THE SMALL TRIANGLE.

PRESS MENU AND UNDER MEASUREMENT SELECT LENGTH.

CLICK ON THE ENDPOINTS THAT THE DEFINE THE BASE

AND HEIGHT OF THIS TRIANGLE TO RECORD THE MEASUREMENTS.

ASSIGN EACH MEASUREMENT TO A NEW SET OF VARIABLES.

PRESS ESCAPE AND MOVE THE POINTER

ABOVE THE BASE MEASUREMENT.

PRESS CONTROL AND MENU AND ASSIGN THIS MEASUREMENT

TO THE VARIABLE SMALL BASE AND PRESS ENTER.

REPEAT THIS PROCESS FOR THE HEIGHT OF THE TRIANGLE

AND ASSIGN IT TO VARIABLE SMALLHEIGHT.

WE CAN NOW CALCULATE THE AREA OF EACH TRIANGLE.

CREATE A NEW CALCULATOR WINDOW.

PRESS THE HOME KEY

AND SELECT THE CALCULATOR OPTION.

START WITH THE AREA OF THE LARGER TRIANGLE.

USE THE VARIABLES YOU DEFINED EARLIER.

INPUT THE EXPRESSION

ONE HALF TIMES LARGE BASE TIMES LARGEHEIGHT.

ASSIGN IT TO VARIABLE AREA 1

BY PRESSING CONTROL AND THE VARIABLE KEY

AND INPUTTING THE VARIABLE NAME.

PRESS ENTER.

NOW INPUT THE EXPRESSION

ONE HALF TIMES SMALLBASE TIMES SMALLHEIGHT

AND ASSIGN IT TO VARIABLE AREA2.

PRESS ENTER.

INPUT THE EXPRESSION AREA1 DIVIDED BY AREA2

AND PRESS ENTER.

THERE ARE 324 TRIANGULAR PANELS.

TO FIND THE NUMBER OF RHOMBUS SHAPES

DIVIDE THIS VALUE BY TWO.

AND TO FIND THE TOTAL NUMBER OF RHOMBUS PANELS

FOR ALL FOUR SIDES,

MULTIPLY THAT RESULT BY FOUR.

YOUR RESULT SHOULD BE 648 PANELS.

NOTE HOWEVER THAT SOME OF THE RHOMBUS SHAPED PANELS

ARE SPLIT INTO TWO TRIANGLES

ALONG THE BASE OF THE PYRAMID.

IN ADDITION, THE MAIN ENTRANCE TO THE MUSEUM

THROUGH THE GLASS PYRAMID HAS SOME RHOMBUSES AND

TRIANGLES MISSING TO MAKE ROOM FOR AN ENTRYWAY.

SO THE FINAL COUNT IS 603 FULL RHOMBUS SHAPED PANELS

AND 70 TRIANGULAR SHAPED PANELS.

SO THE SIMPLICITY OF THE DESIGN OF THE LOUVRE PYRAMID

REVEALS AN INTRICATE MATHEMATICAL PATTERN

THAT RELIES ON SIMILARITY.

THE CITIGROUP BUILDING IN NEW YORK CITY

HAS A DISTINCTIVE OUTLINE.

FROM TOP TO BOTTOM THERE ARE SOME CLEVER INNOVATIONS.

AT THE TIME THE BUILDING WAS COMMISSIONED

THERE WAS A CHURCH WHERE THE TOWER WOULD STAND

AND AS PART OF THE ARRANGEMENT

THE TOWER NEEDED TO LEAVE ROOM FOR THE CHURCH.

THIS IS WHY THE BASE OF THE TOWER CONSISTS OF

FOUR COLUMNS THAT CLEAR A SPACE

ON THE GROUND FLOOR FOR THE CHURCH.

THE TOP OF THE TOWER HAS A SLANTED ROOF WHERE SOLAR

PANELS ARE INSTALLED AS AN ENERGY SAVING MEASURE.

THE SHAPE OF THE TOWER IS A COMBINATION

OF A SQUARE PRISM AND A TRIANGULAR PRISM.

A SQUARE PRISM IS A TYPE OF RECTANGULAR PRISM

WHERE THE TOP AND BOTTOM FACES ARE SQUARES.

THE TRIANGULAR PRISM IS THE ROOF SECTION OF THE TOWER.

WHILE NOT THE TALLEST SKYSCRAPER IN NEW YORK CITY,

THE CITIGROUP TOWER STILL HAS TO CONTEND WITH

A PROBLEM COMMON TO TALL BUILDINGS:

THE CONTINUAL LOSS OF HEAT.

KEEPING A BUILDING LIKE THIS COOL IN THE SUMMER

AND WARM IN THE WINTER CAN GET EXPENSIVE IF THE TOWER

CHANGES ITS INTERNAL TEMPERATURE TOO QUICKLY.

THE ABILITY FOR A BUILDING TO RETAIN HEAT

IS AFFECTED BY ITS SHAPE.

ONE OF THE BEST INDICATORS OF THE

POTENTIAL FOR HEAT LOSS IS TO FIND THE RATIO

OF THE BUILDING SURFACE AREA TO VOLUME.

TO GET A BETTER UNDERSTANDING

OF WHY THIS RATIO WAS IMPORTANT,

LET'S LOOK AT EXAMPLES FROM NATURE.

SNAKES ARE LONG AND THIN.

THIS SHAPE GIVES THEM A LOT OF SURFACE AREA

BUT NOT A LOT OF VOLUME.

SO THE RATIO OF SURFACE AREA TO VOLUME

IS RELATIVELY LARGE.

WITH MORE SURFACE AREA RELATIVE TO VOLUME

A SNAKE WILL LOSE HEAT QUICKLY.

THIS IS WHY SNAKES SPEND TIME IN THE SUN

ABSORBING HEAT TO MAKE UP FOR WHAT THEY LOST.

ON THE OTHER HAND, A POLAR BEAR HAS A

LARGER VOLUME COMPARED TO ITS SURFACE AREA.

THE RATIO OF SURFACE AREA TO VOLUME

FOR THE POLAR BEAR IS RELATIVELY SMALL WHICH

ALLOWS THE POLAR BEAR TO RETAIN HEAT LONGER.

THIS IS IMPORTANT SINCE THE POLAR BEAR

LIVES IN AN ENVIRONMENT WHERE

RETAINING HEAT IS VERY IMPORTANT.

SO LET'S TAKE A LOOK AT THE RATIO

OF SURFACE AREA TO VOLUME

FOR A TALL BUILDING LIKE THE CITIGROUP TOWER.

FOR SIMPLICITY LET'S SAY THAT THE SHAPE

IS THAT OF A SQUARE PRISM.

THE NET FOR A SQUARE PRISM SHOWS THAT THE SURFACE AREA

IS MADE UP OF TWO SQUARES AND FOUR RECTANGLES.

SUPPOSE WE LABEL THE SQUARE SIDES X.

RATHER THAN USING A DIFFERENT VARIABLE

FOR THE RECTANGULAR SIDE LETS DEFINE

THE LENGTH OF THE PRISM AS CX FOR SOME NUMBER C.

IN OTHER WORDS THE LENGTH OF THE SQUARE PRISM

IS SOME MULTIPLE OF THE SQUARE SIDE.

THIS MEANS THAT THE SURFACE AREA OF THE TOWER

BECOMES 2X SQUARED PLUS 4CX SQUARED.

THE VOLUME OF THE RECTANGULAR PRISM

IS THE AREA OF THE SQUARE BASE

TIMES THE LENGTH OR CX CUBED.

LET'S USE THE NSPIRE TO EXPLORE THE RATIO

OF SURFACE AREA TO VOLUME FOR A SQUARE PRISM.

TURN ON THE TI-NSPIRE.

CREATE A NEW DOCUMENT.

YOU MAY NEED TO SAVE A PREVIOUS DOCUMENT.

CREATE A GRAPH WINDOW.

BEFORE INPUTTING A GRAPH CREATE A SLIDER.

THIS WILL BE USED TO VARY THE VALUES OF THE CONSTANT C

FROM OUR SURFACE AREA TO VOLUME RATIO.

PRESS MENU AND UNDER ACTIONS SELECT INSERT SLIDER.

SINCE THE SLIDER LABEL IS HIGHLIGHTED

CHANGE IT TO C AND PRESS ENTER.

NEXT PRESS THE TAB KEY TWICE TO GET TO

THE f1 FUNCTION ENTRY LINE.

INPUT THIS EXPRESSION.

PRESS ENTER.

THIS IS THE GRAPH OF A RATIONAL FUNCTION.

AS X APPROACHES ZERO THE GRAPH APPROACHES INFINITY.

SINCE X IS THE VALUE FOR THE SIDE LENGTH

OF THE SQUARE THEN THE GRAPH IS UNDEFINED

FOR X EQUALS ZERO.

WE'RE ONLY INTERESTED IN THE BEHAVIOR

OF THIS GRAPH IN THE FIRST QUADRANT.

PRESS MENU AND UNDER WINDOW/ZOOM

CHANGE THE SETTINGS TO THESE:

XMIN=0. XMAX=50, YMIN=0, YMAX=5.

PRESS OKAY AFTER YOU HAVE MADE THESE CHANGES.

THE VALUE OF C IS THE MULTIPLE

OF THE SQUARE SIDE LENGTH

THAT MAKES UP THE LENGTH OF THE SQUARE PRISM.

BY DEFAULT C EQUALS 5.

SO THIS GRAPH IS THE RATIO OF SURFACE AREA TO VOLUME

FOR A BUILDING THAT IS FIVE TIMES TALLER

THAN IT IS WIDE.

FOR SMALL VALUES OF X

THE RATIO OF SURFACE AREA TO VOLUME IS HIGH

MEANING THAT THE BUILDING LOSES HEAT.

BUT AS X INCREASES IN VALUE...

IN OTHER WORDS AS A BUILDING GETS WIDER AND TALLER

THE RATIO GOES DOWN AND IT DOES SO RATHER DRAMATICALLY.

TALL BUILDINGS THAT MAINTAIN A 5 TO 1 RATIO

FOR THE SIDE LENGTH AND HEIGHT

CAN ACHIEVE A GOOD RATIO WHERE HEAT IS RETAINED.

BUT LET'S LOOK AT VALUES THAT ARE CLOSER

TO THOSE FOUND WITH SKYSCRAPERS.

THE CITIGROUP TOWER IS ROUGHLY SIX TIMES

THE HEIGHT OF THE SQUARE BASE.

THE SQUARE BASE IS ROUGHLY 46 METERS IN LENGTH.

FIRST MOVE THE POINTER ABOVE THE SLIDER.

PRESS AND HOLD THE CLICK KEY OVER THE SLIDING ARROW

UNTIL THE OPEN HAND TURNS TO A GRASPING HAND.

USE THE RIGHT ARROW KEY TO MOVE THE SLIDER

TO VALUE C EQUALS 6.

PRESS ESCAPE.

NEXT PRESS MENU AND UNDER TRACE SELECT GRAPH TRACE.

MOVE THE POINTER TO X EQUALS 46

OR SIMPLY INPUT 46 AND PRESS ENTER.

YOU'LL SEE THAT THE VALUE OF SURFACE AREA OVER VOLUME

FOR THE CITIGROUP BUILDING IS QUITE SMALL.

SO THE CITIGROUP BUILDING DOES HAVE

A GOOD SURFACE AREA TO VOLUME RATIO.

THE SURFACE AREA TO VOLUME RATIO FOR TALL BUILDINGS

IS SOMETHING AN ARCHITECT NEEDS TO BE AWARE OF.

KEEPING A SKYSCRAPER WARM IN WINTER AND COOL IN SUMMER

REQUIRES A GREAT DEAL OF ENERGY.

AS A RESULT, AN AREA OF CONCERN FOR ARCHITECTS

IS THE ENERGY EFFICIENCY OF TALL BUILDINGS.

WHILE TALL, SLEEK BUILDINGS HAVE VISUAL APPEAL,

BALANCING THE ARTISTIC NEEDS WITH THE ECONOMIC ONES

IS IMPORTANT.

NEWER BUILDINGS HAVE TAKEN THESE CONSIDERATIONS

INTO ACCOUNT AND HAVE RESULTED IN INCREASING

INTEREST IN SO-CALLED GREEN ARCHITECTURE.

MODIFICATIONS TO EXISTING BUILDINGS

LIKE THE CITIGROUP TOWER

ARE OFTEN TO MAKE THEM MORE ENERGY EFFICIENT.