Finding the diagonals of a rhombus with side length $13$, where the sum of the diagonals is $34$ [closed]
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How can we find diagonals of the rhombus with side length $a=13$ cm and sum of diagonals $d_1+d_2=34$ cm?
Anything doesn't seem to work...
I would really appreciate it, if anyone could help me / solve it :)
Since I have only basic knowledge in geometry, please post your EASIEST answer possible.
geometry
asked Jan 11 at 19:27
Jaanis SoosaarJaanis Soosaar
84
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closed as off-topic by Henrik, Abcd, metamorphy, Cesareo, Iuli Jan 12 at 9:25
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Henrik, Abcd, metamorphy, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
How can we find diagonals of the rhombus with side length $a=13$ cm and sum of diagonals $d_1+d_2=34$ cm?
Anything doesn't seem to work...
I would really appreciate it, if anyone could help me / solve it :)
Since I have only basic knowledge in geometry, please post your EASIEST answer possible.
geometry
asked Jan 11 at 19:27
Jaanis SoosaarJaanis Soosaar
84
$endgroup$
closed as off-topic by Henrik, Abcd, metamorphy, Cesareo, Iuli Jan 12 at 9:25
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Henrik, Abcd, metamorphy, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
How can we find diagonals of the rhombus with side length $a=13$ cm and sum of diagonals $d_1+d_2=34$ cm?
Anything doesn't seem to work...
I would really appreciate it, if anyone could help me / solve it :)
Since I have only basic knowledge in geometry, please post your EASIEST answer possible.
geometry
asked Jan 11 at 19:27
Jaanis SoosaarJaanis Soosaar
84
$endgroup$
How can we find diagonals of the rhombus with side length $a=13$ cm and sum of diagonals $d_1+d_2=34$ cm?
Anything doesn't seem to work...
I would really appreciate it, if anyone could help me / solve it :)
Since I have only basic knowledge in geometry, please post your EASIEST answer possible.
geometry
geometry
asked Jan 11 at 19:27
Jaanis SoosaarJaanis Soosaar
84
asked Jan 11 at 19:27
Jaanis SoosaarJaanis Soosaar
84
edited Jan 11 at 19:59
Jaanis Soosaar
asked Jan 11 at 19:27
Jaanis SoosaarJaanis Soosaar
84
asked Jan 11 at 19:27
Jaanis SoosaarJaanis Soosaar
84
asked Jan 11 at 19:27
Jaanis SoosaarJaanis Soosaar
84
84
closed as off-topic by Henrik, Abcd, metamorphy, Cesareo, Iuli Jan 12 at 9:25
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Henrik, Abcd, metamorphy, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Henrik, Abcd, metamorphy, Cesareo, Iuli Jan 12 at 9:25
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Henrik, Abcd, metamorphy, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
Results on quadratic equations make this problem easy to solve:
- As the diagonals in a rhombus are perpendicular, applying Pythagoras results in
$$d_1^2+d_2^2=4cdot13^2. $$
- On the other, the hypothesis $;d_1+d_2=34$ implies
$$4cdot 17^2=(d_1+d_2)^2=d_1^2+d_2^2+2d_1d_2=4cdot13^2+2d_1d_2,$$
so that $;d_1d_2=240$.
Therefore, it comes down to the standard problem of finding two numbers , given their sum $s$ and their product $p$. We know they're the roots (if any) of the quadratic equation
$$x^2-sx+p=0.$$
answered Jan 11 at 20:01
BernardBernard
124k741117
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$begingroup$
Yes! Thank you for "dumbing" it down for me :)
$endgroup$
– Jaanis Soosaar
Jan 11 at 20:10
add a comment |
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Hint: The diagonals of a rhombus are perpendicular. Set up a system of equations using the given information and the Pythagorean theorem.
answered Jan 11 at 19:30
jmerryjmerry
17.1k11633
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add a comment |
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You can also use the theorem of cosines:
$$d_1^2=13^2+13^2-2cdot 13^2cos(180^{circ}-alpha)$$
$$d_2^2=13^2+13^2-2cdot 13^2cos(alpha)$$ and use that $$d_1+d_2=34$$ and note that $$cos(pi-x)=-cos(x)$$
answered Jan 11 at 19:33
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.9k42867
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add a comment |
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Here's how to do this problem the wrong way. Be suspicious of the number $13$ and remember that "$5$-$12$-$13$" forms a right triangle. Guess that the diagonals split the rhombus into $4$ seperate "$5$-$12$-$13$" right triangles, which implies diagonals of lengths $10$ and $24$, which do actually add to $34$, so that must be the correct answer.
answered Jan 11 at 23:38
Josh B.Josh B.
2,55511425
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add a comment |
$begingroup$
Perpendicular lengths of rhombus semi diagonals $(x,y)$ can be mentally solved
$$ 2x+2y=34quad x+y=17 tag1$$
Perimeter
$$ 4 sqrt{x^2+y^2} = 4 (13)quad rightarrow {x^2+y^2}=13^2 $$
Using identity
$$ 2xy= (x+y)^2-(x^2+y^2) =17^2-13^2= 30(4)=120 rightarrow xy=60 tag2 $$
Since the sum and product of $x$ and $y$ are known we can factorize
$$ (x-5)(y-12)=0 rightarrow d_1= 2x =10, d_2=2y= 24. tag3 $$
answered Jan 12 at 0:31
NarasimhamNarasimham
21.2k62258
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add a comment |
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Results on quadratic equations make this problem easy to solve:
- As the diagonals in a rhombus are perpendicular, applying Pythagoras results in
$$d_1^2+d_2^2=4cdot13^2. $$
- On the other, the hypothesis $;d_1+d_2=34$ implies
$$4cdot 17^2=(d_1+d_2)^2=d_1^2+d_2^2+2d_1d_2=4cdot13^2+2d_1d_2,$$
so that $;d_1d_2=240$.
Therefore, it comes down to the standard problem of finding two numbers , given their sum $s$ and their product $p$. We know they're the roots (if any) of the quadratic equation
$$x^2-sx+p=0.$$
answered Jan 11 at 20:01
BernardBernard
124k741117
$endgroup$
$begingroup$
Yes! Thank you for "dumbing" it down for me :)
$endgroup$
– Jaanis Soosaar
Jan 11 at 20:10
add a comment |
$begingroup$
Results on quadratic equations make this problem easy to solve:
- As the diagonals in a rhombus are perpendicular, applying Pythagoras results in
$$d_1^2+d_2^2=4cdot13^2. $$
- On the other, the hypothesis $;d_1+d_2=34$ implies
$$4cdot 17^2=(d_1+d_2)^2=d_1^2+d_2^2+2d_1d_2=4cdot13^2+2d_1d_2,$$
so that $;d_1d_2=240$.
Therefore, it comes down to the standard problem of finding two numbers , given their sum $s$ and their product $p$. We know they're the roots (if any) of the quadratic equation
$$x^2-sx+p=0.$$
answered Jan 11 at 20:01
BernardBernard
124k741117
$endgroup$
$begingroup$
Yes! Thank you for "dumbing" it down for me :)
$endgroup$
– Jaanis Soosaar
Jan 11 at 20:10
add a comment |
$begingroup$
Results on quadratic equations make this problem easy to solve:
- As the diagonals in a rhombus are perpendicular, applying Pythagoras results in
$$d_1^2+d_2^2=4cdot13^2. $$
- On the other, the hypothesis $;d_1+d_2=34$ implies
$$4cdot 17^2=(d_1+d_2)^2=d_1^2+d_2^2+2d_1d_2=4cdot13^2+2d_1d_2,$$
so that $;d_1d_2=240$.
Therefore, it comes down to the standard problem of finding two numbers , given their sum $s$ and their product $p$. We know they're the roots (if any) of the quadratic equation
$$x^2-sx+p=0.$$
answered Jan 11 at 20:01
BernardBernard
124k741117
$endgroup$
Results on quadratic equations make this problem easy to solve:
- As the diagonals in a rhombus are perpendicular, applying Pythagoras results in
$$d_1^2+d_2^2=4cdot13^2. $$
- On the other, the hypothesis $;d_1+d_2=34$ implies
$$4cdot 17^2=(d_1+d_2)^2=d_1^2+d_2^2+2d_1d_2=4cdot13^2+2d_1d_2,$$
so that $;d_1d_2=240$.
Therefore, it comes down to the standard problem of finding two numbers , given their sum $s$ and their product $p$. We know they're the roots (if any) of the quadratic equation
$$x^2-sx+p=0.$$
answered Jan 11 at 20:01
BernardBernard
124k741117
answered Jan 11 at 20:01
BernardBernard
124k741117
answered Jan 11 at 20:01
BernardBernard
124k741117
answered Jan 11 at 20:01
BernardBernard
124k741117
124k741117
$begingroup$
Yes! Thank you for "dumbing" it down for me :)
$endgroup$
– Jaanis Soosaar
Jan 11 at 20:10
add a comment |
$begingroup$
Yes! Thank you for "dumbing" it down for me :)
$endgroup$
– Jaanis Soosaar
Jan 11 at 20:10
$begingroup$
Yes! Thank you for "dumbing" it down for me :)
$endgroup$
– Jaanis Soosaar
Jan 11 at 20:10
$begingroup$
Yes! Thank you for "dumbing" it down for me :)
$endgroup$
– Jaanis Soosaar
Jan 11 at 20:10
add a comment |
$begingroup$
Hint: The diagonals of a rhombus are perpendicular. Set up a system of equations using the given information and the Pythagorean theorem.
answered Jan 11 at 19:30
jmerryjmerry
17.1k11633
$endgroup$
add a comment |
$begingroup$
Hint: The diagonals of a rhombus are perpendicular. Set up a system of equations using the given information and the Pythagorean theorem.
answered Jan 11 at 19:30
jmerryjmerry
17.1k11633
$endgroup$
add a comment |
$begingroup$
Hint: The diagonals of a rhombus are perpendicular. Set up a system of equations using the given information and the Pythagorean theorem.
answered Jan 11 at 19:30
jmerryjmerry
17.1k11633
$endgroup$
Hint: The diagonals of a rhombus are perpendicular. Set up a system of equations using the given information and the Pythagorean theorem.
answered Jan 11 at 19:30
jmerryjmerry
17.1k11633
answered Jan 11 at 19:30
jmerryjmerry
17.1k11633
answered Jan 11 at 19:30
jmerryjmerry
17.1k11633
answered Jan 11 at 19:30
jmerryjmerry
17.1k11633
17.1k11633
add a comment |
add a comment |
$begingroup$
You can also use the theorem of cosines:
$$d_1^2=13^2+13^2-2cdot 13^2cos(180^{circ}-alpha)$$
$$d_2^2=13^2+13^2-2cdot 13^2cos(alpha)$$ and use that $$d_1+d_2=34$$ and note that $$cos(pi-x)=-cos(x)$$
answered Jan 11 at 19:33
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.9k42867
$endgroup$
add a comment |
$begingroup$
You can also use the theorem of cosines:
$$d_1^2=13^2+13^2-2cdot 13^2cos(180^{circ}-alpha)$$
$$d_2^2=13^2+13^2-2cdot 13^2cos(alpha)$$ and use that $$d_1+d_2=34$$ and note that $$cos(pi-x)=-cos(x)$$
answered Jan 11 at 19:33
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.9k42867
$endgroup$
add a comment |
$begingroup$
You can also use the theorem of cosines:
$$d_1^2=13^2+13^2-2cdot 13^2cos(180^{circ}-alpha)$$
$$d_2^2=13^2+13^2-2cdot 13^2cos(alpha)$$ and use that $$d_1+d_2=34$$ and note that $$cos(pi-x)=-cos(x)$$
answered Jan 11 at 19:33
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.9k42867
$endgroup$
You can also use the theorem of cosines:
$$d_1^2=13^2+13^2-2cdot 13^2cos(180^{circ}-alpha)$$
$$d_2^2=13^2+13^2-2cdot 13^2cos(alpha)$$ and use that $$d_1+d_2=34$$ and note that $$cos(pi-x)=-cos(x)$$
answered Jan 11 at 19:33
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.9k42867
answered Jan 11 at 19:33
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.9k42867
answered Jan 11 at 19:33
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.9k42867
answered Jan 11 at 19:33
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.9k42867
78.9k42867
add a comment |
add a comment |
$begingroup$
Here's how to do this problem the wrong way. Be suspicious of the number $13$ and remember that "$5$-$12$-$13$" forms a right triangle. Guess that the diagonals split the rhombus into $4$ seperate "$5$-$12$-$13$" right triangles, which implies diagonals of lengths $10$ and $24$, which do actually add to $34$, so that must be the correct answer.
answered Jan 11 at 23:38
Josh B.Josh B.
2,55511425
$endgroup$
add a comment |
$begingroup$
Here's how to do this problem the wrong way. Be suspicious of the number $13$ and remember that "$5$-$12$-$13$" forms a right triangle. Guess that the diagonals split the rhombus into $4$ seperate "$5$-$12$-$13$" right triangles, which implies diagonals of lengths $10$ and $24$, which do actually add to $34$, so that must be the correct answer.
answered Jan 11 at 23:38
Josh B.Josh B.
2,55511425
$endgroup$
add a comment |
$begingroup$
Here's how to do this problem the wrong way. Be suspicious of the number $13$ and remember that "$5$-$12$-$13$" forms a right triangle. Guess that the diagonals split the rhombus into $4$ seperate "$5$-$12$-$13$" right triangles, which implies diagonals of lengths $10$ and $24$, which do actually add to $34$, so that must be the correct answer.
answered Jan 11 at 23:38
Josh B.Josh B.
2,55511425
$endgroup$
Here's how to do this problem the wrong way. Be suspicious of the number $13$ and remember that "$5$-$12$-$13$" forms a right triangle. Guess that the diagonals split the rhombus into $4$ seperate "$5$-$12$-$13$" right triangles, which implies diagonals of lengths $10$ and $24$, which do actually add to $34$, so that must be the correct answer.
answered Jan 11 at 23:38
Josh B.Josh B.
2,55511425
answered Jan 11 at 23:38
Josh B.Josh B.
2,55511425
answered Jan 11 at 23:38
Josh B.Josh B.
2,55511425
answered Jan 11 at 23:38
Josh B.Josh B.
2,55511425
2,55511425
add a comment |
add a comment |
$begingroup$
Perpendicular lengths of rhombus semi diagonals $(x,y)$ can be mentally solved
$$ 2x+2y=34quad x+y=17 tag1$$
Perimeter
$$ 4 sqrt{x^2+y^2} = 4 (13)quad rightarrow {x^2+y^2}=13^2 $$
Using identity
$$ 2xy= (x+y)^2-(x^2+y^2) =17^2-13^2= 30(4)=120 rightarrow xy=60 tag2 $$
Since the sum and product of $x$ and $y$ are known we can factorize
$$ (x-5)(y-12)=0 rightarrow d_1= 2x =10, d_2=2y= 24. tag3 $$
answered Jan 12 at 0:31
NarasimhamNarasimham
21.2k62258
$endgroup$
add a comment |
$begingroup$
Perpendicular lengths of rhombus semi diagonals $(x,y)$ can be mentally solved
$$ 2x+2y=34quad x+y=17 tag1$$
Perimeter
$$ 4 sqrt{x^2+y^2} = 4 (13)quad rightarrow {x^2+y^2}=13^2 $$
Using identity
$$ 2xy= (x+y)^2-(x^2+y^2) =17^2-13^2= 30(4)=120 rightarrow xy=60 tag2 $$
Since the sum and product of $x$ and $y$ are known we can factorize
$$ (x-5)(y-12)=0 rightarrow d_1= 2x =10, d_2=2y= 24. tag3 $$
answered Jan 12 at 0:31
NarasimhamNarasimham
21.2k62258
$endgroup$
add a comment |
$begingroup$
Perpendicular lengths of rhombus semi diagonals $(x,y)$ can be mentally solved
$$ 2x+2y=34quad x+y=17 tag1$$
Perimeter
$$ 4 sqrt{x^2+y^2} = 4 (13)quad rightarrow {x^2+y^2}=13^2 $$
Using identity
$$ 2xy= (x+y)^2-(x^2+y^2) =17^2-13^2= 30(4)=120 rightarrow xy=60 tag2 $$
Since the sum and product of $x$ and $y$ are known we can factorize
$$ (x-5)(y-12)=0 rightarrow d_1= 2x =10, d_2=2y= 24. tag3 $$
answered Jan 12 at 0:31
NarasimhamNarasimham
21.2k62258
$endgroup$
Perpendicular lengths of rhombus semi diagonals $(x,y)$ can be mentally solved
$$ 2x+2y=34quad x+y=17 tag1$$
Perimeter
$$ 4 sqrt{x^2+y^2} = 4 (13)quad rightarrow {x^2+y^2}=13^2 $$
Using identity
$$ 2xy= (x+y)^2-(x^2+y^2) =17^2-13^2= 30(4)=120 rightarrow xy=60 tag2 $$
Since the sum and product of $x$ and $y$ are known we can factorize
$$ (x-5)(y-12)=0 rightarrow d_1= 2x =10, d_2=2y= 24. tag3 $$
answered Jan 12 at 0:31
NarasimhamNarasimham
21.2k62258
answered Jan 12 at 0:31
NarasimhamNarasimham
21.2k62258
answered Jan 12 at 0:31
NarasimhamNarasimham
21.2k62258
answered Jan 12 at 0:31
NarasimhamNarasimham
21.2k62258
21.2k62258
add a comment |
add a comment |
