Difference between revisions of "2015 AMC 10A Problems/Problem 11"

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{{duplicate|[[2015 AMC 12A Problems|2015 AMC 12A #8]] and [[2015 AMC 10A Problems|2015 AMC 10A #11]]}}
 
==Problem 11==
 
==Problem 11==
  
 
The ratio of the length to the width of a rectangle is <math>4</math> : <math>3</math>. If the rectangle has diagonal of length <math>d</math>, then the area may be expressed as <math>kd^2</math> for some constant <math>k</math>. What is <math>k</math>?
 
The ratio of the length to the width of a rectangle is <math>4</math> : <math>3</math>. If the rectangle has diagonal of length <math>d</math>, then the area may be expressed as <math>kd^2</math> for some constant <math>k</math>. What is <math>k</math>?
  
<math>\textbf{(A)}\ \frac{2}{7}\qquad\textbf{(B)}\ \frac{3}{7}\qquad\textbf{(C)}\ \frac{12}{25}\qquad\textbf{(D)}}\ \frac{16}{25}\qquad\textbf{(E)}\ \frac{3}{4}</math>
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<math>\textbf{(A)}\ \frac{2}{7}\qquad\textbf{(B)}\ \frac{3}{7}\qquad\textbf{(C)}\ \frac{12}{25}\qquad\textbf{(D)}\ \frac{16}{25}\qquad\textbf{(E)}\ \frac{3}{4}</math>
  
 
==Solution==
 
==Solution==
Let the rectangle have length <math>4x</math> and width <math>3x</math>. Then by 3-4-5 triangles (or the Pythagorean Theorem), we have <math>d = 5x</math>, and so <math>d = \dfrac{x}{5}</math>. Hence, the area of the rectangle is <math>3x \cdot 4x = 12x^2 = \dfrac{12d^2}{25}</math>, so the answer is <math>\boxed{\textbf{(C) }\frac{12}{25}}</math>
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Let the rectangle have length <math>4x</math> and width <math>3x</math>. Then by <math>3-4-5</math> triangles (or the Pythagorean Theorem), we have <math>d = 5x</math>, and so <math>x = \dfrac{d}{5}</math>. Hence, the area of the rectangle is <math>3x \cdot 4x = 12x^2 = \dfrac{12d^2}{25}</math>, so the answer is <math>\boxed{\textbf{(C) }\frac{12}{25}}</math>
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==Video Solution==
 +
https://youtu.be/Ep-IQmOK-nU
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 +
~savannahsolver
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== See also ==
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{{AMC10 box|year=2015|ab=A|num-b=10|num-a=12}}
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{{AMC12 box|year=2015|ab=A|num-b=7|num-a=9}}
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[[Category:Introductory Geometry Problems]]
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{{MAA Notice}}

Latest revision as of 03:52, 13 January 2021

The following problem is from both the 2015 AMC 12A #8 and 2015 AMC 10A #11, so both problems redirect to this page.

Problem 11

The ratio of the length to the width of a rectangle is $4$ : $3$. If the rectangle has diagonal of length $d$, then the area may be expressed as $kd^2$ for some constant $k$. What is $k$?

$\textbf{(A)}\ \frac{2}{7}\qquad\textbf{(B)}\ \frac{3}{7}\qquad\textbf{(C)}\ \frac{12}{25}\qquad\textbf{(D)}\ \frac{16}{25}\qquad\textbf{(E)}\ \frac{3}{4}$

Solution

Let the rectangle have length $4x$ and width $3x$. Then by $3-4-5$ triangles (or the Pythagorean Theorem), we have $d = 5x$, and so $x = \dfrac{d}{5}$. Hence, the area of the rectangle is $3x \cdot 4x = 12x^2 = \dfrac{12d^2}{25}$, so the answer is $\boxed{\textbf{(C) }\frac{12}{25}}$

Video Solution

https://youtu.be/Ep-IQmOK-nU

~savannahsolver

See also

2015 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions
2015 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

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