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Difference between revisions of "2013 CEMC Gauss (Grade 8) Problems/Problem 17"

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== Problem ==
 
== Problem ==
 
<math>PQRS</math> is a rectangle with diagonals <math>PR</math> and <math>QS</math>, as shown.
 
<math>PQRS</math> is a rectangle with diagonals <math>PR</math> and <math>QS</math>, as shown.
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{{Template:Image needed}}
 
 
The value of y is
 
The value of y is
  
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\textbf{(E)}\ 60
 
\textbf{(E)}\ 60
 
</math>
 
</math>
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[[File:2013CEMCGauss8P17diagram.png]]
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~diagram uploaded by [https://artofproblemsolving.com/wiki/index.php/User:grogg007 grogg007]
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== Solution 1==
 
== Solution 1==
 
The interior angles of a [[rectangle]] are all [[right angle]]s, and the [[acute angle]]s of a [[right triangle]] sum up to <math>90^{\circ}</math>. Thus, we have the following equations:
 
The interior angles of a [[rectangle]] are all [[right angle]]s, and the [[acute angle]]s of a [[right triangle]] sum up to <math>90^{\circ}</math>. Thus, we have the following equations:
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~anabel.disher
 
~anabel.disher
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{{CEMC box|year=2013|competition=Gauss (Grade 8)|num-b=16|num-a=18}}

Latest revision as of 21:49, 18 October 2025

Problem

$PQRS$ is a rectangle with diagonals $PR$ and $QS$, as shown. The value of y is

$\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 40 \qquad \textbf{(C)}\ 45 \qquad \textbf{(D)}\ 50 \qquad \textbf{(E)}\ 60$


2013CEMCGauss8P17diagram.png

~diagram uploaded by grogg007

Solution 1

The interior angles of a rectangle are all right angles, and the acute angles of a right triangle sum up to $90^{\circ}$. Thus, we have the following equations:

$5x^{\circ} + 4x^{\circ} = 90^{\circ}$

$4x^{\circ} + y^{\circ} = 90^{\circ}$

Solving the first equation for $x$, we get:

$9x^{\circ} = 90^{\circ}$

$x = 10$

Plugging $x$ into the second equation, we have:

$4 \times 10^{\circ} + y^{\circ} = 90^{\circ}$

$40^{\circ} + y^{\circ} = 90^{\circ}$

$y = \boxed {\textbf {(D) } 50}$

~anabel.disher

Solution 2

We can use the above process to find $x$, and then notice $y$ and $5x$ would be alternate interior angles. Thus,

$y = 5x = 5 \times 10 = \boxed {\textbf {(D) } 50}$

~anabel.disher

Solution 2.5

We can also get to the conclusion that $y = 5x$ by using the equations:

$4x = 90 - 5x$

$4x + y = 90 - 5x + y = 90$

$y = 90 - 90 + 5x = 5x = 5 \times 10 = \boxed {\textbf {(D) } 50}$

~anabel.disher

2013 CEMC Gauss (Grade 8) (Problems • Answer Key • Resources)
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Problem 16 		Followed by
Problem 18
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CEMC Gauss (Grade 8)
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