Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m
Line 1: Line 1:
 
== 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.
−
{{Template:Image needed}}
 
 
The value of y is
 
The value of y is
  
Line 15: Line 14:
 
\textbf{(E)}\ 60
 
\textbf{(E)}\ 60
 
</math>
 
</math>
 +
 +
 +
 +
[[File:2013CEMCGauss8P17diagram.png]]
 +
 +
~diagram uploaded by [[sharmaguy]]
 +
 
== 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:

Revision as of 18:29, 17 June 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 sharmaguy

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2013_CEMC_Gauss_(Grade_8)_Problems/Problem_17&oldid=251362"