How can I deduce the hypotenuse from the information given?












4












$begingroup$


I'm going into Machine Learning and am currently brushing up on some Calculus on Coursera. Everything was going smoothly until I got to this word problem:




A ladder rests against a wall. The top of the ladder touches the wall at height $12$ meters. The length of the ladder is $4$ meters longer than the distance from the base of the ladder to the wall. Find the length of the ladder.




I am confused as to how to deduce the hypotenuse from the information given above. And have sat here trying different things with no success. What am I missing?






calculus algebra-precalculus trigonometry







share|cite|improve this question











$endgroup$












  • $begingroup$
    I added the "algebra-precalculus" tag to your post. Cheers!
    $endgroup$
    – Robert Lewis
    Dec 10 '18 at 2:47










  • $begingroup$
    Draw a picture with the ladder. Label the sides of the triangle formed by the ladder, the wall, and the floor. If you call the base x, the ladder is x+4. Then use the Pythagorean theorem.
    $endgroup$
    – Joel Pereira
    Dec 10 '18 at 2:47
















4












$begingroup$


I'm going into Machine Learning and am currently brushing up on some Calculus on Coursera. Everything was going smoothly until I got to this word problem:




A ladder rests against a wall. The top of the ladder touches the wall at height $12$ meters. The length of the ladder is $4$ meters longer than the distance from the base of the ladder to the wall. Find the length of the ladder.




I am confused as to how to deduce the hypotenuse from the information given above. And have sat here trying different things with no success. What am I missing?






calculus algebra-precalculus trigonometry







share|cite|improve this question











$endgroup$












  • $begingroup$
    I added the "algebra-precalculus" tag to your post. Cheers!
    $endgroup$
    – Robert Lewis
    Dec 10 '18 at 2:47










  • $begingroup$
    Draw a picture with the ladder. Label the sides of the triangle formed by the ladder, the wall, and the floor. If you call the base x, the ladder is x+4. Then use the Pythagorean theorem.
    $endgroup$
    – Joel Pereira
    Dec 10 '18 at 2:47














4












4








4


2



$begingroup$


I'm going into Machine Learning and am currently brushing up on some Calculus on Coursera. Everything was going smoothly until I got to this word problem:




A ladder rests against a wall. The top of the ladder touches the wall at height $12$ meters. The length of the ladder is $4$ meters longer than the distance from the base of the ladder to the wall. Find the length of the ladder.




I am confused as to how to deduce the hypotenuse from the information given above. And have sat here trying different things with no success. What am I missing?






calculus algebra-precalculus trigonometry







share|cite|improve this question











$endgroup$




I'm going into Machine Learning and am currently brushing up on some Calculus on Coursera. Everything was going smoothly until I got to this word problem:




A ladder rests against a wall. The top of the ladder touches the wall at height $12$ meters. The length of the ladder is $4$ meters longer than the distance from the base of the ladder to the wall. Find the length of the ladder.




I am confused as to how to deduce the hypotenuse from the information given above. And have sat here trying different things with no success. What am I missing?






calculus algebra-precalculus trigonometry




calculus algebra-precalculus trigonometry






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 12 '18 at 4:12









Key Flex

7,85061232




7,85061232










asked Dec 10 '18 at 2:38









Edward SeverinsenEdward Severinsen

18815




18815












  • $begingroup$
    I added the "algebra-precalculus" tag to your post. Cheers!
    $endgroup$
    – Robert Lewis
    Dec 10 '18 at 2:47










  • $begingroup$
    Draw a picture with the ladder. Label the sides of the triangle formed by the ladder, the wall, and the floor. If you call the base x, the ladder is x+4. Then use the Pythagorean theorem.
    $endgroup$
    – Joel Pereira
    Dec 10 '18 at 2:47


















  • $begingroup$
    I added the "algebra-precalculus" tag to your post. Cheers!
    $endgroup$
    – Robert Lewis
    Dec 10 '18 at 2:47










  • $begingroup$
    Draw a picture with the ladder. Label the sides of the triangle formed by the ladder, the wall, and the floor. If you call the base x, the ladder is x+4. Then use the Pythagorean theorem.
    $endgroup$
    – Joel Pereira
    Dec 10 '18 at 2:47
















$begingroup$
I added the "algebra-precalculus" tag to your post. Cheers!
$endgroup$
– Robert Lewis
Dec 10 '18 at 2:47




$begingroup$
I added the "algebra-precalculus" tag to your post. Cheers!
$endgroup$
– Robert Lewis
Dec 10 '18 at 2:47












$begingroup$
Draw a picture with the ladder. Label the sides of the triangle formed by the ladder, the wall, and the floor. If you call the base x, the ladder is x+4. Then use the Pythagorean theorem.
$endgroup$
– Joel Pereira
Dec 10 '18 at 2:47




$begingroup$
Draw a picture with the ladder. Label the sides of the triangle formed by the ladder, the wall, and the floor. If you call the base x, the ladder is x+4. Then use the Pythagorean theorem.
$endgroup$
– Joel Pereira
Dec 10 '18 at 2:47










2 Answers
2






active

oldest

votes


















6












$begingroup$

Let $d$ be the distance from the ladder to the wall, and $l$ the length of the ladder.
Then



$l = d + 4; tag 1$



since the wall is mos' likely perpendicular to the ground, we may deploy the Pythagorean theorem and write



$l^2 = (12)^2 + d^2; tag 2$



substituting (1) into (2) yields



$(d + 4)^2 = 144 + d^2, tag 3$



$d^2 + 8d + 16 = 144 + d^2, tag 4$



$8d + 16 = 144 Longrightarrow 8d = 128 Longrightarrow d = 16M Longrightarrow l = 20M. tag 5$






share|cite|improve this answer









$endgroup$









  • 2




    $begingroup$
    Oh my. I distributed the 2 exponent to d and 4 individually instead of multiplying the expression by itself. Not the first time this has gotten me.
    $endgroup$
    – Edward Severinsen
    Dec 10 '18 at 2:49






  • 1




    $begingroup$
    @EdwardSeverinsen: we're all learners, my friend!
    $endgroup$
    – Robert Lewis
    Dec 10 '18 at 2:50






  • 6




    $begingroup$
    No, the answer should be "at that angle, the ladder doesn't rest at the wall" ;-)
    $endgroup$
    – DonQuiKong
    Dec 10 '18 at 9:51






  • 1




    $begingroup$
    @DonQuiKong I wouldn't get on it, that's for sure!
    $endgroup$
    – JTPenguin
    Dec 10 '18 at 11:48



















13












$begingroup$

enter image description here



Given the length of the wall as $12$.



Take the length of the base as $x$.



Since the length of the ladder $l$ is $4$ meters greater than the base, we have $l = x+4$



Now according to the pythagorean theorem we have,



$begin{align}
(x+4)^2 &= 12^2 + x^2 \
x^2 + 16 + 8x &= 144 + x^2 \
8x &= 128 \
x & = 16
end{align}$



So, the length of the ladder $l = x+4 = 16+4 = 20$






share|cite|improve this answer











$endgroup$









  • 4




    $begingroup$
    Nice graphic, +1!
    $endgroup$
    – Robert Lewis
    Dec 10 '18 at 2:52






  • 2




    $begingroup$
    @RobertLewis Thanks!
    $endgroup$
    – Key Flex
    Dec 10 '18 at 2:57






  • 1




    $begingroup$
    Just as a note 4 times greater than the base is ambiguous and could imply it is x * 4 and not x + 4.
    $endgroup$
    – Felix Guo
    Dec 10 '18 at 7:36






  • 2




    $begingroup$
    @FelixGuo "4 times greater" isn't ambiguous at all: it can only mean $xtimes 4$ and never $x+4$. It was just a mistake, which has now been corrected.
    $endgroup$
    – David Richerby
    Dec 10 '18 at 14:47











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









6












$begingroup$

Let $d$ be the distance from the ladder to the wall, and $l$ the length of the ladder.
Then



$l = d + 4; tag 1$



since the wall is mos' likely perpendicular to the ground, we may deploy the Pythagorean theorem and write



$l^2 = (12)^2 + d^2; tag 2$



substituting (1) into (2) yields



$(d + 4)^2 = 144 + d^2, tag 3$



$d^2 + 8d + 16 = 144 + d^2, tag 4$



$8d + 16 = 144 Longrightarrow 8d = 128 Longrightarrow d = 16M Longrightarrow l = 20M. tag 5$






share|cite|improve this answer









$endgroup$









  • 2




    $begingroup$
    Oh my. I distributed the 2 exponent to d and 4 individually instead of multiplying the expression by itself. Not the first time this has gotten me.
    $endgroup$
    – Edward Severinsen
    Dec 10 '18 at 2:49






  • 1




    $begingroup$
    @EdwardSeverinsen: we're all learners, my friend!
    $endgroup$
    – Robert Lewis
    Dec 10 '18 at 2:50






  • 6




    $begingroup$
    No, the answer should be "at that angle, the ladder doesn't rest at the wall" ;-)
    $endgroup$
    – DonQuiKong
    Dec 10 '18 at 9:51






  • 1




    $begingroup$
    @DonQuiKong I wouldn't get on it, that's for sure!
    $endgroup$
    – JTPenguin
    Dec 10 '18 at 11:48
















6












$begingroup$

Let $d$ be the distance from the ladder to the wall, and $l$ the length of the ladder.
Then



$l = d + 4; tag 1$



since the wall is mos' likely perpendicular to the ground, we may deploy the Pythagorean theorem and write



$l^2 = (12)^2 + d^2; tag 2$



substituting (1) into (2) yields



$(d + 4)^2 = 144 + d^2, tag 3$



$d^2 + 8d + 16 = 144 + d^2, tag 4$



$8d + 16 = 144 Longrightarrow 8d = 128 Longrightarrow d = 16M Longrightarrow l = 20M. tag 5$






share|cite|improve this answer









$endgroup$









  • 2




    $begingroup$
    Oh my. I distributed the 2 exponent to d and 4 individually instead of multiplying the expression by itself. Not the first time this has gotten me.
    $endgroup$
    – Edward Severinsen
    Dec 10 '18 at 2:49






  • 1




    $begingroup$
    @EdwardSeverinsen: we're all learners, my friend!
    $endgroup$
    – Robert Lewis
    Dec 10 '18 at 2:50






  • 6




    $begingroup$
    No, the answer should be "at that angle, the ladder doesn't rest at the wall" ;-)
    $endgroup$
    – DonQuiKong
    Dec 10 '18 at 9:51






  • 1




    $begingroup$
    @DonQuiKong I wouldn't get on it, that's for sure!
    $endgroup$
    – JTPenguin
    Dec 10 '18 at 11:48














6












6








6





$begingroup$

Let $d$ be the distance from the ladder to the wall, and $l$ the length of the ladder.
Then



$l = d + 4; tag 1$



since the wall is mos' likely perpendicular to the ground, we may deploy the Pythagorean theorem and write



$l^2 = (12)^2 + d^2; tag 2$



substituting (1) into (2) yields



$(d + 4)^2 = 144 + d^2, tag 3$



$d^2 + 8d + 16 = 144 + d^2, tag 4$



$8d + 16 = 144 Longrightarrow 8d = 128 Longrightarrow d = 16M Longrightarrow l = 20M. tag 5$






share|cite|improve this answer









$endgroup$



Let $d$ be the distance from the ladder to the wall, and $l$ the length of the ladder.
Then



$l = d + 4; tag 1$



since the wall is mos' likely perpendicular to the ground, we may deploy the Pythagorean theorem and write



$l^2 = (12)^2 + d^2; tag 2$



substituting (1) into (2) yields



$(d + 4)^2 = 144 + d^2, tag 3$



$d^2 + 8d + 16 = 144 + d^2, tag 4$



$8d + 16 = 144 Longrightarrow 8d = 128 Longrightarrow d = 16M Longrightarrow l = 20M. tag 5$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 10 '18 at 2:46









Robert LewisRobert Lewis

45.4k23065




45.4k23065








  • 2




    $begingroup$
    Oh my. I distributed the 2 exponent to d and 4 individually instead of multiplying the expression by itself. Not the first time this has gotten me.
    $endgroup$
    – Edward Severinsen
    Dec 10 '18 at 2:49






  • 1




    $begingroup$
    @EdwardSeverinsen: we're all learners, my friend!
    $endgroup$
    – Robert Lewis
    Dec 10 '18 at 2:50






  • 6




    $begingroup$
    No, the answer should be "at that angle, the ladder doesn't rest at the wall" ;-)
    $endgroup$
    – DonQuiKong
    Dec 10 '18 at 9:51






  • 1




    $begingroup$
    @DonQuiKong I wouldn't get on it, that's for sure!
    $endgroup$
    – JTPenguin
    Dec 10 '18 at 11:48














  • 2




    $begingroup$
    Oh my. I distributed the 2 exponent to d and 4 individually instead of multiplying the expression by itself. Not the first time this has gotten me.
    $endgroup$
    – Edward Severinsen
    Dec 10 '18 at 2:49






  • 1




    $begingroup$
    @EdwardSeverinsen: we're all learners, my friend!
    $endgroup$
    – Robert Lewis
    Dec 10 '18 at 2:50






  • 6




    $begingroup$
    No, the answer should be "at that angle, the ladder doesn't rest at the wall" ;-)
    $endgroup$
    – DonQuiKong
    Dec 10 '18 at 9:51






  • 1




    $begingroup$
    @DonQuiKong I wouldn't get on it, that's for sure!
    $endgroup$
    – JTPenguin
    Dec 10 '18 at 11:48








2




2




$begingroup$
Oh my. I distributed the 2 exponent to d and 4 individually instead of multiplying the expression by itself. Not the first time this has gotten me.
$endgroup$
– Edward Severinsen
Dec 10 '18 at 2:49




$begingroup$
Oh my. I distributed the 2 exponent to d and 4 individually instead of multiplying the expression by itself. Not the first time this has gotten me.
$endgroup$
– Edward Severinsen
Dec 10 '18 at 2:49




1




1




$begingroup$
@EdwardSeverinsen: we're all learners, my friend!
$endgroup$
– Robert Lewis
Dec 10 '18 at 2:50




$begingroup$
@EdwardSeverinsen: we're all learners, my friend!
$endgroup$
– Robert Lewis
Dec 10 '18 at 2:50




6




6




$begingroup$
No, the answer should be "at that angle, the ladder doesn't rest at the wall" ;-)
$endgroup$
– DonQuiKong
Dec 10 '18 at 9:51




$begingroup$
No, the answer should be "at that angle, the ladder doesn't rest at the wall" ;-)
$endgroup$
– DonQuiKong
Dec 10 '18 at 9:51




1




1




$begingroup$
@DonQuiKong I wouldn't get on it, that's for sure!
$endgroup$
– JTPenguin
Dec 10 '18 at 11:48




$begingroup$
@DonQuiKong I wouldn't get on it, that's for sure!
$endgroup$
– JTPenguin
Dec 10 '18 at 11:48











13












$begingroup$

enter image description here



Given the length of the wall as $12$.



Take the length of the base as $x$.



Since the length of the ladder $l$ is $4$ meters greater than the base, we have $l = x+4$



Now according to the pythagorean theorem we have,



$begin{align}
(x+4)^2 &= 12^2 + x^2 \
x^2 + 16 + 8x &= 144 + x^2 \
8x &= 128 \
x & = 16
end{align}$



So, the length of the ladder $l = x+4 = 16+4 = 20$






share|cite|improve this answer











$endgroup$









  • 4




    $begingroup$
    Nice graphic, +1!
    $endgroup$
    – Robert Lewis
    Dec 10 '18 at 2:52






  • 2




    $begingroup$
    @RobertLewis Thanks!
    $endgroup$
    – Key Flex
    Dec 10 '18 at 2:57






  • 1




    $begingroup$
    Just as a note 4 times greater than the base is ambiguous and could imply it is x * 4 and not x + 4.
    $endgroup$
    – Felix Guo
    Dec 10 '18 at 7:36






  • 2




    $begingroup$
    @FelixGuo "4 times greater" isn't ambiguous at all: it can only mean $xtimes 4$ and never $x+4$. It was just a mistake, which has now been corrected.
    $endgroup$
    – David Richerby
    Dec 10 '18 at 14:47
















13












$begingroup$

enter image description here



Given the length of the wall as $12$.



Take the length of the base as $x$.



Since the length of the ladder $l$ is $4$ meters greater than the base, we have $l = x+4$



Now according to the pythagorean theorem we have,



$begin{align}
(x+4)^2 &= 12^2 + x^2 \
x^2 + 16 + 8x &= 144 + x^2 \
8x &= 128 \
x & = 16
end{align}$



So, the length of the ladder $l = x+4 = 16+4 = 20$






share|cite|improve this answer











$endgroup$









  • 4




    $begingroup$
    Nice graphic, +1!
    $endgroup$
    – Robert Lewis
    Dec 10 '18 at 2:52






  • 2




    $begingroup$
    @RobertLewis Thanks!
    $endgroup$
    – Key Flex
    Dec 10 '18 at 2:57






  • 1




    $begingroup$
    Just as a note 4 times greater than the base is ambiguous and could imply it is x * 4 and not x + 4.
    $endgroup$
    – Felix Guo
    Dec 10 '18 at 7:36






  • 2




    $begingroup$
    @FelixGuo "4 times greater" isn't ambiguous at all: it can only mean $xtimes 4$ and never $x+4$. It was just a mistake, which has now been corrected.
    $endgroup$
    – David Richerby
    Dec 10 '18 at 14:47














13












13








13





$begingroup$

enter image description here



Given the length of the wall as $12$.



Take the length of the base as $x$.



Since the length of the ladder $l$ is $4$ meters greater than the base, we have $l = x+4$



Now according to the pythagorean theorem we have,



$begin{align}
(x+4)^2 &= 12^2 + x^2 \
x^2 + 16 + 8x &= 144 + x^2 \
8x &= 128 \
x & = 16
end{align}$



So, the length of the ladder $l = x+4 = 16+4 = 20$






share|cite|improve this answer











$endgroup$



enter image description here



Given the length of the wall as $12$.



Take the length of the base as $x$.



Since the length of the ladder $l$ is $4$ meters greater than the base, we have $l = x+4$



Now according to the pythagorean theorem we have,



$begin{align}
(x+4)^2 &= 12^2 + x^2 \
x^2 + 16 + 8x &= 144 + x^2 \
8x &= 128 \
x & = 16
end{align}$



So, the length of the ladder $l = x+4 = 16+4 = 20$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 10 '18 at 8:17









SQB

1,74611026




1,74611026










answered Dec 10 '18 at 2:49









Key FlexKey Flex

7,85061232




7,85061232








  • 4




    $begingroup$
    Nice graphic, +1!
    $endgroup$
    – Robert Lewis
    Dec 10 '18 at 2:52






  • 2




    $begingroup$
    @RobertLewis Thanks!
    $endgroup$
    – Key Flex
    Dec 10 '18 at 2:57






  • 1




    $begingroup$
    Just as a note 4 times greater than the base is ambiguous and could imply it is x * 4 and not x + 4.
    $endgroup$
    – Felix Guo
    Dec 10 '18 at 7:36






  • 2




    $begingroup$
    @FelixGuo "4 times greater" isn't ambiguous at all: it can only mean $xtimes 4$ and never $x+4$. It was just a mistake, which has now been corrected.
    $endgroup$
    – David Richerby
    Dec 10 '18 at 14:47














  • 4




    $begingroup$
    Nice graphic, +1!
    $endgroup$
    – Robert Lewis
    Dec 10 '18 at 2:52






  • 2




    $begingroup$
    @RobertLewis Thanks!
    $endgroup$
    – Key Flex
    Dec 10 '18 at 2:57






  • 1




    $begingroup$
    Just as a note 4 times greater than the base is ambiguous and could imply it is x * 4 and not x + 4.
    $endgroup$
    – Felix Guo
    Dec 10 '18 at 7:36






  • 2




    $begingroup$
    @FelixGuo "4 times greater" isn't ambiguous at all: it can only mean $xtimes 4$ and never $x+4$. It was just a mistake, which has now been corrected.
    $endgroup$
    – David Richerby
    Dec 10 '18 at 14:47








4




4




$begingroup$
Nice graphic, +1!
$endgroup$
– Robert Lewis
Dec 10 '18 at 2:52




$begingroup$
Nice graphic, +1!
$endgroup$
– Robert Lewis
Dec 10 '18 at 2:52




2




2




$begingroup$
@RobertLewis Thanks!
$endgroup$
– Key Flex
Dec 10 '18 at 2:57




$begingroup$
@RobertLewis Thanks!
$endgroup$
– Key Flex
Dec 10 '18 at 2:57




1




1




$begingroup$
Just as a note 4 times greater than the base is ambiguous and could imply it is x * 4 and not x + 4.
$endgroup$
– Felix Guo
Dec 10 '18 at 7:36




$begingroup$
Just as a note 4 times greater than the base is ambiguous and could imply it is x * 4 and not x + 4.
$endgroup$
– Felix Guo
Dec 10 '18 at 7:36




2




2




$begingroup$
@FelixGuo "4 times greater" isn't ambiguous at all: it can only mean $xtimes 4$ and never $x+4$. It was just a mistake, which has now been corrected.
$endgroup$
– David Richerby
Dec 10 '18 at 14:47




$begingroup$
@FelixGuo "4 times greater" isn't ambiguous at all: it can only mean $xtimes 4$ and never $x+4$. It was just a mistake, which has now been corrected.
$endgroup$
– David Richerby
Dec 10 '18 at 14:47


















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Wiesbaden

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Wappen Deutschlandkarte 50.0825 8.24 117 Koordinaten: 50° 5′  N , 8° 14′  O Basisdaten Bundesland: Hessen Regierungsbezirk: Darmstadt Höhe: 117 m ü. NHN Fläche: 203,93 km 2 Einwohner: 278.654 (31. Dez. 2017) [1] Bevölkerungsdichte: 1366 Einwohner je km 2 Postleitzahlen: 65183–65207, 55246, 55252 Vorlage:Infobox Gemeinde in Deutschland/Wartung/PLZ enthält Text Vorwahlen: 0611, 06122, 06127, 06134 Kfz-Kennzeichen: WI Gemeindeschlüssel: 06 4 14 000 LOCODE: DE WIB NUTS: DE714 Stadtgliederung: 26 Ortsbezirke Adresse der Stadtverwaltung: Schlossplatz 6 65183 Wiesbaden Webpräsenz: www.wiesbaden.de Oberbürgermeister: Sven Gerich (SPD) Lage der Landeshauptstadt Wiesbaden in Hessen und im Regierungsbezirk Darmstadt Offizielles Logo der Landeshauptstadt Wiesbaden Flagge der Landeshauptstadt Wiesbaden Wiesbaden is
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Marschland

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Kilometerweite Marsch in den Vier- und Marschlanden am Elbdeich Die Wedeler Marschlandschaft in Schleswig-Holstein Der Hadelner Kanal zur Entwässerung der Marsch Die tief gelegene Marsch und ein Entwässerungskanal Das Schöpfwerk Otterndorf im Land Hadeln zur Entwässerung der Medem verfügte damals bereits über die größte Kreiselpumpe Europas. Das Schöpfwerk in Neuhaus für die Aue und des Neuhaus-Bülkauer Kanals Als Marsch(land) (v. niederdt., altsächs. mersc ) – auch Masch , Mersch oder Schwemmland genannt – bezeichnet man eine nacheiszeitlich entstandene geomorphologische Landform im Gebiet der nordwestdeutschen Küsten und Flüsse sowie vergleichbare Landformen weltweit. Inhaltsverzeichnis 1 Landschaft 2 Entstehung 3 Marschböden 4 Nutzung 5 Marschgebiete in Deutschland 6 Siehe auch 7 Literatur 8 Weblinks 9 Einzelnachweise Landschaft | Marschen sind generell flache Landstriche ohne natürliche Erhebungen. Si
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Dieringhausen

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Dieringhausen Stadt Gummersbach 50.982777777778 7.5316666666667 165 Koordinaten: 50° 58′ 58″  N , 7° 31′ 54″  O Höhe: 165  (160–245)  m Einwohner: 5055  (30. Jun. 2016) Postleitzahl: 51645 Vorwahl: 02261 Lage von Dieringhausen in Gummersbach Im Hohl ca. 1927–1930 Ansicht von Südwesten bei Bünghausen Dieringhausen (im örtlichen Dialekt Dierkesen ) ist ein Ortsteil von Gummersbach im Oberbergischen Kreis, Regierungsbezirk Köln, Nordrhein-Westfalen (Deutschland). Inhaltsverzeichnis 1 Geographie 2 Geschichte 3 Kultur 3.1 Sehenswürdigkeiten 3.2 Regelmäßige Veranstaltungen 4 Infrastruktur und Wirtschaft 4.1 Verkehr 4.1.1 Schienen- und Busverkehr 4.1.2 Straßen 4.2 Öffentliche Einrichtungen 4.2.1 Schulen und Bildungseinrichtungen 4.2.2 Kirchliche Einrichtungen 5 Persönlichkeiten 5.1 In Dieringhausen geboren 5.2 In Dieringhausen gelebt 6 Einzelnachweise 7
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