How can I deduce the hypotenuse from the information given?
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
edited Dec 12 '18 at 4:12
Key Flex
7,52441232
asked Dec 10 '18 at 2:38
Edward Severinsen
18815
add a comment |
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
edited Dec 12 '18 at 4:12
Key Flex
7,52441232
asked Dec 10 '18 at 2:38
Edward Severinsen
18815
I added the "algebra-precalculus" tag to your post. Cheers!
– Robert Lewis
Dec 10 '18 at 2:47
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.
– Joel Pereira
Dec 10 '18 at 2:47
add a comment |
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
edited Dec 12 '18 at 4:12
Key Flex
7,52441232
asked Dec 10 '18 at 2:38
Edward Severinsen
18815
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
edited Dec 12 '18 at 4:12
Key Flex
7,52441232
asked Dec 10 '18 at 2:38
Edward Severinsen
18815
edited Dec 12 '18 at 4:12
Key Flex
7,52441232
asked Dec 10 '18 at 2:38
Edward Severinsen
18815
edited Dec 12 '18 at 4:12
Key Flex
7,52441232
edited Dec 12 '18 at 4:12
Key Flex
7,52441232
edited Dec 12 '18 at 4:12
Key Flex
7,52441232
7,52441232
asked Dec 10 '18 at 2:38
Edward Severinsen
18815
asked Dec 10 '18 at 2:38
Edward Severinsen
18815
asked Dec 10 '18 at 2:38
Edward Severinsen
18815
18815
I added the "algebra-precalculus" tag to your post. Cheers!
– Robert Lewis
Dec 10 '18 at 2:47
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.
– Joel Pereira
Dec 10 '18 at 2:47
add a comment |
I added the "algebra-precalculus" tag to your post. Cheers!
– Robert Lewis
Dec 10 '18 at 2:47
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.
– Joel Pereira
Dec 10 '18 at 2:47
I added the "algebra-precalculus" tag to your post. Cheers!
– Robert Lewis
Dec 10 '18 at 2:47
I added the "algebra-precalculus" tag to your post. Cheers!
– Robert Lewis
Dec 10 '18 at 2:47
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.
– Joel Pereira
Dec 10 '18 at 2:47
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.
– Joel Pereira
Dec 10 '18 at 2:47
add a comment |
2 Answers
2
active
oldest
votes
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
$(d + 4)^2 = 144 + d^2, tag 3$
$d^2 + 8d + 16 = 144 + d^2, tag 4$
$8d + 16 = 144 Longrightarrow 8d = 128 Longrightarrow d = 16M Longrightarrow l = 20M. tag 5$
answered Dec 10 '18 at 2:46
Robert Lewis
43.7k22963
2
Oh my. I distributed the 2 exponent todand4individually instead of multiplying the expression by itself. Not the first time this has gotten me.
– Edward Severinsen
Dec 10 '18 at 2:49
1
@EdwardSeverinsen: we're all learners, my friend!
– Robert Lewis
Dec 10 '18 at 2:50
6
No, the answer should be "at that angle, the ladder doesn't rest at the wall" ;-)
– DonQuiKong
Dec 10 '18 at 9:51
1
@DonQuiKong I wouldn't get on it, that's for sure!
– JTPenguin
Dec 10 '18 at 11:48
add a comment |
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 $l = x+4 = 16+4 = 20$
edited Dec 10 '18 at 8:17
SQB
1,73211026
answered Dec 10 '18 at 2:49
Key Flex
7,52441232
4
Nice graphic, +1!
– Robert Lewis
Dec 10 '18 at 2:52
2
@RobertLewis Thanks!
– Key Flex
Dec 10 '18 at 2:57
1
Just as a note4 times greater than the baseis ambiguous and could imply it isx * 4and notx + 4.
– Felix Guo
Dec 10 '18 at 7:36
2
@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.
– David Richerby
Dec 10 '18 at 14:47
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
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
$(d + 4)^2 = 144 + d^2, tag 3$
$d^2 + 8d + 16 = 144 + d^2, tag 4$
$8d + 16 = 144 Longrightarrow 8d = 128 Longrightarrow d = 16M Longrightarrow l = 20M. tag 5$
answered Dec 10 '18 at 2:46
Robert Lewis
43.7k22963
2
Oh my. I distributed the 2 exponent todand4individually instead of multiplying the expression by itself. Not the first time this has gotten me.
– Edward Severinsen
Dec 10 '18 at 2:49
1
@EdwardSeverinsen: we're all learners, my friend!
– Robert Lewis
Dec 10 '18 at 2:50
6
No, the answer should be "at that angle, the ladder doesn't rest at the wall" ;-)
– DonQuiKong
Dec 10 '18 at 9:51
1
@DonQuiKong I wouldn't get on it, that's for sure!
– JTPenguin
Dec 10 '18 at 11:48
add a comment |
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
$(d + 4)^2 = 144 + d^2, tag 3$
$d^2 + 8d + 16 = 144 + d^2, tag 4$
$8d + 16 = 144 Longrightarrow 8d = 128 Longrightarrow d = 16M Longrightarrow l = 20M. tag 5$
answered Dec 10 '18 at 2:46
Robert Lewis
43.7k22963
2
Oh my. I distributed the 2 exponent todand4individually instead of multiplying the expression by itself. Not the first time this has gotten me.
– Edward Severinsen
Dec 10 '18 at 2:49
1
@EdwardSeverinsen: we're all learners, my friend!
– Robert Lewis
Dec 10 '18 at 2:50
6
No, the answer should be "at that angle, the ladder doesn't rest at the wall" ;-)
– DonQuiKong
Dec 10 '18 at 9:51
1
@DonQuiKong I wouldn't get on it, that's for sure!
– JTPenguin
Dec 10 '18 at 11:48
add a comment |
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
$(d + 4)^2 = 144 + d^2, tag 3$
$d^2 + 8d + 16 = 144 + d^2, tag 4$
$8d + 16 = 144 Longrightarrow 8d = 128 Longrightarrow d = 16M Longrightarrow l = 20M. tag 5$
answered Dec 10 '18 at 2:46
Robert Lewis
43.7k22963
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
$(d + 4)^2 = 144 + d^2, tag 3$
$d^2 + 8d + 16 = 144 + d^2, tag 4$
$8d + 16 = 144 Longrightarrow 8d = 128 Longrightarrow d = 16M Longrightarrow l = 20M. tag 5$
answered Dec 10 '18 at 2:46
Robert Lewis
43.7k22963
answered Dec 10 '18 at 2:46
Robert Lewis
43.7k22963
answered Dec 10 '18 at 2:46
Robert Lewis
43.7k22963
answered Dec 10 '18 at 2:46
Robert Lewis
43.7k22963
43.7k22963
2
Oh my. I distributed the 2 exponent todand4individually instead of multiplying the expression by itself. Not the first time this has gotten me.
– Edward Severinsen
Dec 10 '18 at 2:49
1
@EdwardSeverinsen: we're all learners, my friend!
– Robert Lewis
Dec 10 '18 at 2:50
6
No, the answer should be "at that angle, the ladder doesn't rest at the wall" ;-)
– DonQuiKong
Dec 10 '18 at 9:51
1
@DonQuiKong I wouldn't get on it, that's for sure!
– JTPenguin
Dec 10 '18 at 11:48
add a comment |
2
Oh my. I distributed the 2 exponent todand4individually instead of multiplying the expression by itself. Not the first time this has gotten me.
– Edward Severinsen
Dec 10 '18 at 2:49
1
@EdwardSeverinsen: we're all learners, my friend!
– Robert Lewis
Dec 10 '18 at 2:50
6
No, the answer should be "at that angle, the ladder doesn't rest at the wall" ;-)
– DonQuiKong
Dec 10 '18 at 9:51
1
@DonQuiKong I wouldn't get on it, that's for sure!
– JTPenguin
Dec 10 '18 at 11:48
2
2
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.– Edward Severinsen
Dec 10 '18 at 2:49
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.– Edward Severinsen
Dec 10 '18 at 2:49
1
1
@EdwardSeverinsen: we're all learners, my friend!
– Robert Lewis
Dec 10 '18 at 2:50
@EdwardSeverinsen: we're all learners, my friend!
– Robert Lewis
Dec 10 '18 at 2:50
6
6
No, the answer should be "at that angle, the ladder doesn't rest at the wall" ;-)
– DonQuiKong
Dec 10 '18 at 9:51
No, the answer should be "at that angle, the ladder doesn't rest at the wall" ;-)
– DonQuiKong
Dec 10 '18 at 9:51
1
1
@DonQuiKong I wouldn't get on it, that's for sure!
– JTPenguin
Dec 10 '18 at 11:48
@DonQuiKong I wouldn't get on it, that's for sure!
– JTPenguin
Dec 10 '18 at 11:48
add a comment |
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 $l = x+4 = 16+4 = 20$
edited Dec 10 '18 at 8:17
SQB
1,73211026
answered Dec 10 '18 at 2:49
Key Flex
7,52441232
4
Nice graphic, +1!
– Robert Lewis
Dec 10 '18 at 2:52
2
@RobertLewis Thanks!
– Key Flex
Dec 10 '18 at 2:57
1
Just as a note4 times greater than the baseis ambiguous and could imply it isx * 4and notx + 4.
– Felix Guo
Dec 10 '18 at 7:36
2
@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.
– David Richerby
Dec 10 '18 at 14:47
add a comment |
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 $l = x+4 = 16+4 = 20$
edited Dec 10 '18 at 8:17
SQB
1,73211026
answered Dec 10 '18 at 2:49
Key Flex
7,52441232
4
Nice graphic, +1!
– Robert Lewis
Dec 10 '18 at 2:52
2
@RobertLewis Thanks!
– Key Flex
Dec 10 '18 at 2:57
1
Just as a note4 times greater than the baseis ambiguous and could imply it isx * 4and notx + 4.
– Felix Guo
Dec 10 '18 at 7:36
2
@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.
– David Richerby
Dec 10 '18 at 14:47
add a comment |
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 $l = x+4 = 16+4 = 20$
edited Dec 10 '18 at 8:17
SQB
1,73211026
answered Dec 10 '18 at 2:49
Key Flex
7,52441232
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 $l = x+4 = 16+4 = 20$
edited Dec 10 '18 at 8:17
SQB
1,73211026
answered Dec 10 '18 at 2:49
Key Flex
7,52441232
edited Dec 10 '18 at 8:17
SQB
1,73211026
edited Dec 10 '18 at 8:17
SQB
1,73211026
edited Dec 10 '18 at 8:17
SQB
1,73211026
1,73211026
answered Dec 10 '18 at 2:49
Key Flex
7,52441232
answered Dec 10 '18 at 2:49
Key Flex
7,52441232
answered Dec 10 '18 at 2:49
Key Flex
7,52441232
7,52441232
4
Nice graphic, +1!
– Robert Lewis
Dec 10 '18 at 2:52
2
@RobertLewis Thanks!
– Key Flex
Dec 10 '18 at 2:57
1
Just as a note4 times greater than the baseis ambiguous and could imply it isx * 4and notx + 4.
– Felix Guo
Dec 10 '18 at 7:36
2
@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.
– David Richerby
Dec 10 '18 at 14:47
add a comment |
4
Nice graphic, +1!
– Robert Lewis
Dec 10 '18 at 2:52
2
@RobertLewis Thanks!
– Key Flex
Dec 10 '18 at 2:57
1
Just as a note4 times greater than the baseis ambiguous and could imply it isx * 4and notx + 4.
– Felix Guo
Dec 10 '18 at 7:36
2
@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.
– David Richerby
Dec 10 '18 at 14:47
4
4
Nice graphic, +1!
– Robert Lewis
Dec 10 '18 at 2:52
Nice graphic, +1!
– Robert Lewis
Dec 10 '18 at 2:52
2
2
@RobertLewis Thanks!
– Key Flex
Dec 10 '18 at 2:57
@RobertLewis Thanks!
– Key Flex
Dec 10 '18 at 2:57
1
1
Just as a note
4 times greater than the base is ambiguous and could imply it is x * 4 and not x + 4.– Felix Guo
Dec 10 '18 at 7:36
Just as a note
4 times greater than the base is ambiguous and could imply it is x * 4 and not x + 4.– Felix Guo
Dec 10 '18 at 7:36
2
2
@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.
– David Richerby
Dec 10 '18 at 14:47
@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.
– David Richerby
Dec 10 '18 at 14:47
add a comment |
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I added the "algebra-precalculus" tag to your post. Cheers!
– Robert Lewis
Dec 10 '18 at 2:47
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.
– Joel Pereira
Dec 10 '18 at 2:47