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Each diagonal of an isosceles trapezoid is perpendicular to a leg. If the diagonal is 20 and the legs are each 15 find the length of each base.

Submitted by Anonymous on Wed, 03/14/2001 - 5:00 AM

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: Each diagonal of an isosceles trapezoid is perpendicular to a leg. If
: the diagonal is 20 and the legs are each 15 find the length of
: each base.(1) Rule 1 for geometry problems: DRAW IT!!!If you haven’t drawn it yet, stop here and try. I’ll post my answer after this, but if you look ahead at the answer, you’ve just wasted your time and mine.Hint: if the diagonal is perpendicular to a leg, where do you have to have a right angle? Where’s the diagonal in relation to the sides of the trapezoid? So what kind of angle does the trapezoid have to have there?Once it’s drawn and *labelled* with *all* info from the question, do you see any familiar patterns? Can you solve it yourself now?NOW check my answer

Submitted by Anonymous on Wed, 03/14/2001 - 5:00 AM

Permalink

: Each diagonal of an isosceles trapezoid is perpendicular to a leg. If
: the diagonal is 20 and the legs are each 15 find the length of
: each base.Here’s part 1 again — see below for beginning of answer.(1) Rule 1 for geometry problems: DRAW IT!!!If you haven’t drawn it yet, stop here and try. I’ll post my answer after this, but if you look ahead at the answer, you’ve just wasted your time and mine.Hint: if the diagonal is perpendicular to a leg, where do you have to have a right angle? Where’s the diagonal in relation to the sides of the trapezoid? So what kind of angle does the trapezoid have to have there?Once it’s drawn and *labelled* with *all* info from the question, do you see any familiar patterns? Can you solve it yourself now?NOW check my answer**************************************** Answer— first partThe diagonal has to be inside the trapezoid. In order to get those right angles inside, you need a nice wide obtuse angle.I’ll describe this vertically, then see if I can illustrate with x’s. If your computer is set up with a monospace font, like a typewriter’s that spaces evenly, the sktch will come out. (1) Start about 1/3 of the way down a clean sheet of paper (NEVER try to squish diagrams tiny — what’s worse for you and the environment, spending one cent on a sheet of new paper, or failing geometry and repeating the whole course?) Draw a line across the paper one or two inches long, to be the top “base” of the trapezoid.(2) From the ends of this line, draw lines sloping outwards towards the sides of the paper to be the “legs” of the trapezoid, about twice as long as the top. (3) Draw in those diagonals meeting the legs at right angles (4) when the diagonals and the opposite legs meet, you’ve found the bottom corners; draw the bottom base. (5) Label the lengths of known sides and the right angles.Now look at it and see if you see a very well-known pattern.(1) xxxxxxxxxxxxxx(2) xxxxxxxxxxxxxxx xx xx xx xx xx xx x(3) xxxxxxxxxxxxxxxxand x x x x (4) and (5) x 90ox x 90oxx x x xx xx xx x x x15 x x x x 15x x 20 20 x xx x x xx x x xx x x xx x x xx x x xx x x xx x xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxNow do you see the familiar pattern? What if you tried to cut out just part of the figure, a single shape? What’s the shape hidden in the picture?Try to see it and work on it, and the rest of the answer will be in the next post.

Submitted by Anonymous on Wed, 03/14/2001 - 5:00 AM

Permalink

: Each diagonal of an isosceles trapezoid is perpendicular to a leg. If
: the diagonal is 20 and the legs are each 15 find the length of
: each base.Here are part 1 AND 2 again — see below for more of the answer.(1) Rule 1 for geometry problems: DRAW IT!!!If you haven’t drawn it yet, stop here and try. I’ll post my answer after this, but if you look ahead at the answer, you’ve just wasted your time and mine.Hint: if the diagonal is perpendicular to a leg, where do you have to have a right angle? Where’s the diagonal in relation to the sides of the trapezoid? So what kind of angle does the trapezoid have to have there?Once it’s drawn and *labelled* with *all* info from the question, do you see any familiar patterns? Can you solve it yourself now?NOW check my answer**************************************** Answer— first partThe diagonal has to be inside the trapezoid. In order to get those right angles inside, you need a nice wide obtuse angle.I’ll describe this vertically, then see if I can illustrate with x’s. If your computer is set up with a monospace font, like a typewriter’s that spaces evenly, the sktch will come out. (1) Start about 1/3 of the way down a clean sheet of paper (NEVER try to squish diagrams tiny — what’s worse for you and the environment, spending one cent on a sheet of new paper, or failing geometry and repeating the whole course?) Draw a line across the paper one or two inches long, to be the top “base” of the trapezoid.(2) From the ends of this line, draw lines sloping outwards towards the sides of the paper to be the “legs” of the trapezoid, about twice as long as the top. (3) Draw in those diagonals meeting the legs at right angles (4) when the diagonals and the opposite legs meet, you’ve found the bottom corners; draw the bottom base. (5) Label the lengths of known sides and the right angles.Now look at it and see if you see a very well-known pattern.(1) xxxxxxxxxxxxxx(2) xxxxxxxxxxxxxxx xx xx xx xx xx xx x(3) xxxxxxxxxxxxxxxxand x x x x (4) and (5) x 90ox x 90oxx x x xx xx xx x x x15 x x x x 15x x 20 20 x xx x x xx x x xx x x xx x x xx x x xx x x xx x xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxNow do you see the familiar pattern? What if you tried to cut out just part of the figure, a single shape? What’s the shape hidden in the picture?****************************************************To draw this correctly, you had to remember that “perpendicular” means the same as right angle means the same as 90 degrees; and “isosceles” means two sides equal. Also if the two sides are equal and the angles of the diagonals are equal, the thing is symmetric (same left and right) — it would take a while and be a nuisance to prove this, but you can see it has to be true - so the two diagonals have to be equal, both 20 units long.You should see two RIGHT-ANGLED TRIANGLES. They have to be identical (congruent) because of the symmetries mentioned above.What do right-angled triangles make you think of?More in next post (work it out yourself — don’t depend on me!)

Submitted by Anonymous on Wed, 03/14/2001 - 5:00 AM

Permalink

: Each diagonal of an isosceles trapezoid is perpendicular to a leg. If
: the diagonal is 20 and the legs are each 15 find the length of
: each base.Here are part 1 AND 2 AND 3 again — see below for more of the answer.(1) Rule 1 for geometry problems: DRAW IT!!!If you haven’t drawn it yet, stop here and try. I’ll post my answer after this, but if you look ahead at the answer, you’ve just wasted your time and mine.Hint: if the diagonal is perpendicular to a leg, where do you have to have a right angle? Where’s the diagonal in relation to the sides of the trapezoid? So what kind of angle does the trapezoid have to have there?Once it’s drawn and *labelled* with *all* info from the question, do you see any familiar patterns? Can you solve it yourself now?NOW check my answer**************************************** Answer— first partThe diagonal has to be inside the trapezoid. In order to get those right angles inside, you need a nice wide obtuse angle.I’ll describe this vertically, then see if I can illustrate with x’s. If your computer is set up with a monospace font, like a typewriter’s that spaces evenly, the sktch will come out. (1) Start about 1/3 of the way down a clean sheet of paper (NEVER try to squish diagrams tiny — what’s worse for you and the environment, spending one cent on a sheet of new paper, or failing geometry and repeating the whole course?) Draw a line across the paper one or two inches long, to be the top “base” of the trapezoid.(2) From the ends of this line, draw lines sloping outwards towards the sides of the paper to be the “legs” of the trapezoid, about twice as long as the top. (3) Draw in those diagonals meeting the legs at right angles (4) when the diagonals and the opposite legs meet, you’ve found the bottom corners; draw the bottom base. (5) Label the lengths of known sides and the right angles.Now look at it and see if you see a very well-known pattern.(1) xxxxxxxxxxxxxx(2) xxxxxxxxxxxxxxx xx xx xx xx xx xx x(3) xxxxxxxxxxxxxxxxand x x x x (4) and (5) x 90ox x 90oxx x x xx xx xx x x x15 x x x x 15x x 20 20 x xx x x xx x x xx x x xx x x xx x x xx x x xx x xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxNow do you see the familiar pattern? What if you tried to cut out just part of the figure, a single shape? What’s the shape hidden in the picture?****************************************************To draw this correctly, you had to remember that “perpendicular” means the same as right angle means the same as 90 degrees; and “isosceles” means two sides equal. Also if the two sides are equal and the angles of the diagonals are equal, the thing is symmetric (same left and right) — it would take a while and be a nuisance to prove this, but you can see it has to be true - so the two diagonals have to be equal, both 20 units long.You should see two RIGHT-ANGLED TRIANGLES. They have to be identical (congruent) because of the symmetries mentioned above.What do right-angled triangles make you think of?***********************************************************They had better make you think of our old friend PYTHAGORAS. Now can you get part of this before I tell you?***Before you use the theorem of Pythagoras, you had better identify who is who. What is the hypotenuse? (In case you were mis-taught, it is NOT the longest side. Lots of things have longest sides that don’t have hypotenuses. The hypotenuse is the side of a right triangle that his opposite the right angle.)OK, the hypotenuse is the base, which we don’t know yet, so this is the easy form of Pythagoras.You can do 15 squared + 20 squared = base squared That’s all correct and above-board, and I’ll do it in a minute.But first, do you see another way to do this?Yippee! This is our old 3-4-5 right triangle in disguise.15 = 3x520 = 4x5 so our missing hypotenuse must be 5x5 or 25 units long.Going back to basic way to prove we do get the same answer and it’s correct15x15 + 20x20 = base squared225 + 400 = base squared625 = base squaredso base = square root of 625. Fact: 25x25 = 625 so base = 25 units, same as above.***First part of answer: Long base, on bottom of trapezoid in my sketch., is 25 units long.***You know what? I’m still working out the second base. I can see some ways to do it, but they are long and ugly. Will complete the answer when I find a reasonably neat way to get it. Back in a bit.

Submitted by Anonymous on Wed, 03/14/2001 - 5:00 AM

Permalink

: Here are part 1 AND 2 AND 3 again — see below for more of the
: answer.DARN — just looked at this and the reader wiped out all my carefully done spacing trying to make a sketch. If yours wipes it out also, sorry, ignore attempt at sketch in x’sI’ll try again with dash marks instead of spaces and see how it works.****************************************** ****************************************** A second thought: I sincerely hope this is not is not a test or a take-home exam question. If it is, what you are doing is CHEATING. Stop here and don’t read any further until you’ve handed the test in honestly. ******************************************* *******************************************: (1) Rule 1 for geometry problems: DRAW IT!!!: If you haven’t drawn it yet, stop here and try. I’ll post my answer
: after this, but if you look ahead at the answer, you’ve just
: wasted your time and mine.: Hint: if the diagonal is perpendicular to a leg, where do you have to
: have a right angle? Where’s the diagonal in relation to the sides
: of the trapezoid? So what kind of angle does the trapezoid have to
: have there?: Once it’s drawn and *labelled* with *all* info from the question, do
: you see any familiar patterns? Can you solve it yourself now?: NOW check my answer: **************************************** Answer— first part: The diagonal has to be inside the trapezoid. In order to get those
: right angles inside, you need a nice wide obtuse angle.: I’ll describe this vertically, then see if I can illustrate with x’s.
: If your computer is set up with a monospace font, like a
: typewriter’s that spaces evenly, the sktch will come out. (1)
: Start about 1/3 of the way down a clean sheet of paper (NEVER try
: to squish diagrams tiny — what’s worse for you and the
: environment, spending one cent on a sheet of new paper, or failing
: geometry and repeating the whole course?) Draw a line across the
: paper one or two inches long, to be the top “base” of
: the trapezoid.(2) From the ends of this line, draw lines sloping
: outwards towards the sides of the paper to be the “legs”
: of the trapezoid, about twice as long as the top. (3) Draw in
: those diagonals meeting the legs at right angles (4) when the
: diagonals and the opposite legs meet, you’ve found the bottom
: corners; draw the bottom base. (5) Label the lengths of known
: sides and the right angles.: Now look at it and see if you see a very well-known pattern.: (1)–––––xxxxxxxxxxxxxx: (2)–––––-xxxxxxxxxxxxxx
: ––––––x––––—x
: –––––—x–––––-x
: –––––-x––––––x
: –––––x––––––—x
: ––––—x–––––––-x
: ––––-x––––––––x
: ––––x––––––––—x:-(3)–––––––xxxxxxxxxxxxxxxx
:-and––––––—x—x–––-x—x
:(4) and (5)––––x-90o-x––x-90o—x
:–––––––-x––—x—x–––x
:–––––––x–––—x–––—x
:––––––—x–––-x–x–––-x
:–––––-15 x–––x––-x ––—x 15
:––––––x––—x 20–—20 x––—x
:–––––—x––-x–––––x––-x
:–––––-x––x––––––-x––x
:–––––x–—x–––––––—x–—x
:––––—x–-x–––––––––x–-x
:––––-x–x––––––––––-x–x
:––––x—x–––––––––––—x—x
:–––—x-x–––––––––––––x-x
:–––-xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx: Now do you see the familiar pattern? What if you tried to cut out
: just part of the figure, a single shape? What’s the shape hidden
: in the picture?: ****************************************************: To draw this correctly, you had to remember that
: “perpendicular” means the same as right angle means the
: same as 90 degrees; and “isosceles” means two sides
: equal. Also if the two sides are equal and the angles of the
: diagonals are equal, the thing is symmetric (same left and right)
: — it would take a while and be a nuisance to prove this, but you
: can see it has to be true - so the two diagonals have to be equal,
: both 20 units long.: You should see two RIGHT-ANGLED TRIANGLES. They have to be identical
: (congruent) because of the symmetries mentioned above.: What do right-angled triangles make you think of?: ***********************************************************: They had better make you think of our old friend PYTHAGORAS. Now can
: you get part of this before I tell you?: ***Before you use the theorem of Pythagoras, you had better identify
: who is who. What is the hypotenuse? (In case you were mis-taught,
: it is NOT the longest side. Lots of things have longest sides that
: don’t have hypotenuses. The hypotenuse is the side of a right
: triangle that his opposite the right angle.): OK, the hypotenuse is the base, which we don’t know yet, so this is
: the easy form of Pythagoras.: You can do 15 squared + 20 squared = base squared That’s all correct
: and above-board, and I’ll do it in a minute.: But first, do you see another way to do this?: Yippee! This is our old 3-4-5 right triangle in disguise.: 15 = 3x5: 20 = 4x5 so our missing hypotenuse must be 5x5 or 25 units long.: Going back to basic way to prove we do get the same answer and it’s
: correct: 15x15 + 20x20 = base squared: 225 + 400 = base squared: 625 = base squared: so base = square root of 625. Fact: 25x25 = 625 so base = 25 units,
: same as above.: ***First part of answer: Long base, on bottom of trapezoid in my
: sketch., is 25 units long.***: You know what? I’m still working out the second base. I can see some
: ways to do it, but they are long and ugly. Will complete the
: answer when I find a reasonably neat way to get it. Back in a bit.****************************************************************

Submitted by Anonymous on Wed, 03/14/2001 - 5:00 AM

Permalink

: Here are parts 1 AND 2 AND 3 AND 4 again — see below for *rest* of the answer.DARN — just looked at this and the reader wiped out all my carefully done spacing trying to make a sketch. If yours wipes it out also, sorry, ignore attempt at sketch in x’s in previous postsI’ll try again with dash marks instead of spaces and see how it works.****************************************** ****************************************** A second thought: I sincerely hope this is not is not a test or a take-home exam question. If it is, what you are doing is CHEATING. Stop here and don’t read any further until you’ve handed the test in honestly. ******************************************* *******************************************: (1) Rule 1 for geometry problems: DRAW IT!!!: If you haven’t drawn it yet, stop here and try. I’ll post my answer
: after this, but if you look ahead at the answer, you’ve just
: wasted your time and mine.: Hint: if the diagonal is perpendicular to a leg, where do you have to
: have a right angle? Where’s the diagonal in relation to the sides
: of the trapezoid? So what kind of angle does the trapezoid have to
: have there?: Once it’s drawn and *labelled* with *all* info from the question, do
: you see any familiar patterns? Can you solve it yourself now?: NOW check my answer: **************************************** Answer— first part: The diagonal has to be inside the trapezoid. In order to get those
: right angles inside, you need a nice wide obtuse angle.: I’ll describe this verbally, then see if I can illustrate with x’s.
: If your computer is set up with a monospace font, like a
: typewriter’s that spaces evenly, the sketch will come out. (1)
: Start about 1/3 of the way down a clean sheet of paper (NEVER try
: to squish diagrams tiny — what’s worse for you and the
: environment, spending one cent on a sheet of new paper, or failing
: geometry and repeating the whole course?) Draw a line across the
: paper one or two inches long, to be the top “base” of
: the trapezoid.(2) From the ends of this line, draw lines sloping
: outwards towards the sides of the paper to be the “legs”
: of the trapezoid, about twice as long as the top. (3) Draw in
: those diagonals meeting the legs at right angles (4) when the
: diagonals and the opposite legs meet, you’ve found the bottom
: corners; draw the bottom base. (5) Label the lengths of known
: sides and the right angles.: Now look at it and see if you see a very well-known pattern.: (1)–––––xxxxxxxxxxxxxx: (2)–––––-xxxxxxxxxxxxxx
: ––––––x––––—x
: –––––—x–––––-x
: –––––-x––––––x
: –––––x––––––—x
: ––––—x–––––––-x
: ––––-x––––––––x
: ––––x––––––––—x:-(3)–––––––xxxxxxxxxxxxxxxx
:-and––––––—x—x–––-x—x
:(4) and (5)––––x-90o-x––x-90o—x
:–––––––-x––—x—x–––x
:–––––––x–––—x–––—x
:––––––—x–––-x–x–––-x
:–––––-15 x–––x––-x ––—x 15
:––––––x––—x 20–—20 x––—x
:–––––—x––-x–––––x––-x
:–––––-x––x––––––-x––x
:–––––x–—x–––––––—x–—x
:––––—x–-x–––––––––x–-x
:––––-x–x––––––––––-x–x
:––––x—x–––––––––––—x—x
:–––—x-x–––––––––––––x-x
:–––-xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx: Now do you see the familiar pattern? What if you tried to cut out
: just part of the figure, a single shape? What’s the shape hidden
: in the picture?: ****************************************************: To draw this correctly, you had to remember that
: “perpendicular” means the same as right angle means the
: same as 90 degrees; and “isosceles” means two sides
: equal. Also if the two sides are equal and the angles of the
: diagonals are equal, the thing is symmetric (same left and right)
: — it would take a while and be a nuisance to prove this, but you
: can see it has to be true - so the two diagonals have to be equal,
: both 20 units long.: You should see two RIGHT-ANGLED TRIANGLES. They have to be identical
: (congruent) because of the symmetries mentioned above.: What do right-angled triangles make you think of?: ***********************************************************: They had better make you think of our old friend PYTHAGORAS. Now can
: you get part of this before I tell you?: ***Before you use the theorem of Pythagoras, you had better identify
: who is who. What is the hypotenuse? (In case you were mis-taught,
: it is NOT the longest side. Lots of things have longest sides that
: don’t have hypotenuses. The hypotenuse is the side of a right
: triangle that his opposite the right angle.): OK, the hypotenuse is the base, which we don’t know yet, so this is
: the easy form of Pythagoras.: You can do 15 squared + 20 squared = base squared That’s all correct
: and above-board, and I’ll do it in a minute.: But first, do you see another way to do this?: Yippee! This is our old 3-4-5 right triangle in disguise.: 15 = 3x5: 20 = 4x5 so our missing hypotenuse must be 5x5 or 25 units long.: Going back to basic way to prove we do get the same answer and it’s
: correct: 15x15 + 20x20 = base squared: 225 + 400 = base squared: 625 = base squared: so base = square root of 625. Fact: 25x25 = 625 so base = 25 units,
: same as above.: ***First part of answer: Long base, on bottom of trapezoid in my
: sketch., is 25 units long.*******************************************************************Well, I got an answer for the second base, but I had to go out of the simpler geometry to do it. None of the concepts I’m about to use are difficult or beyond first year geometry, but they are combined in some weird and wonderful ways. I would not expect 99% of high school students to be able to do the rest of this.AGAIN, IF THIS IS A TEST OR A TAKE-HOME FOR GRADES, YOU ARE CHEATING. STOP HERE AND DO NOT LOOK AT THIS UNTIL YOU’VE HANDED IN YOUR *OWN* WORK.Go back to the diagram. Sorry, I can’t re-draw it, and the copy function doesn’t seem to work when I’m doing web posts, so you’ll just have to redraw it yourself on paper.I look at the diagram, and I’ve used up all the stuff I can see. So I try adding some lines to make new figures that may give me more information.(1A) Extend the two sides (legs) of the trapezoid towards the top; they meet and form a triangle out of the diagram. (2A) Because this thing is symmetric, I know that the two lines I’ve drawn have to be equal. Call the length we’ve added a. (3A) Now I look and see I have some *new* right triangles on the top of the figure. So I use Pythagoras again. (4A) I see we have (using the typing form ^2 to mean “squared)(a + 15)^2 = a^2 + 20^2 Note that the hypotenuse of this new triangle is (a + 15) and I must square the *whole thing* (a common error is to miss this) (5A) When I multiply this out, I get a^2 + 30a + 225 = a^2 + 400 Be careful about multiplying binomials! Another common error point. (6A) Oh look! I have a squared on both sides and so it can just be subtracted and this is a whole lot simpler, no quadratic. 30a + 225 = 400 Now subtract 225 and 30a = 175 or a = 175/30. Simplifying by dividing out 5/5, we get ** a = 35/6 **(DO NOT USE DECIMALS _— LONGER, HARDER, AND INTRODUCE ROUNDING ERRORS) (7A) Well, what good did all this do us? Good question. We look back at our diagram, and see what other common patterns we can find. Hey! There are some similar triangles. Since the original triangle is isosceles, and the angles of the diagonals are equal, it’s left-right symmetric, and so the top short base is parallel to the bottom long base. So the new little triangle we just added on top is similar to the big triangle made by adding the little one on top of the original trapezoid. Sorry I really can’t draw this — draw it yourself and look. (8A) We remember the rule of similarity that proportions of corresponding sides are equal. Call the top short base of the trapezoid s. Then s/a = long base over side of big triangle s/a = 25 / (15 + a) But we know what a is, so we can find s! (Phew!)s divided by 35/6 = 25 divided by sum(15 + 35/6) OK, true but ugly. Let’s make this a lot less of a chore by multiplying each of those fractions by 6/6 to get rid of the fraction-over-fraction thing. 6s divided by 6 times 35/6 = 6 times 25 divided by 6 times sum(15 + 35/6)(Remember to multiply *both* terms in that parenthesis by 6.)(6s)/35 = 150/(90 + 35)(6s)/35 = 150/125 Simplifying right side by dividing out 25/25 (6s)/35 = 6/5 Dividing both sides by 6 s/35 = 1/5 Multiplying both sides by 35 (DON’T “cross-multiply”. It’s not a very meaningful operation, and makes you take three messy steps instead of one simple one) s = 35 x (1/5) s = 7GOT IT!!! ***Long base = 25 units, above, short base = 7 units.***(There may be a way to shorten this down, but I’m not seeing it off-hand)************************************************* Got any more fun puzzles for me?? victoria [email protected]

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