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

(Created page with "== Problem == <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 <math> \textbf{(A)}\...")
 
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<math>4x + y = 90 - 5x + y = 90</math>
 
<math>4x + y = 90 - 5x + y = 90</math>
  
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<math>y = 90 - 90 + 5x = 90 - 90 + 5 \times 10 = \boxed {\textbf {(D) } 50}</math>
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<math>y = 90 - 90 + 5x = 5x = 5 \times 10 = \boxed {\textbf {(D) } 50}</math>
  
 
~anabel.disher
 
~anabel.disher

Revision as of 14:50, 6 May 2025

Problem

$PQRS$ is a rectangle with diagonals $PR$ and $QS$, as shown.

An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.

The value of y is

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

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,

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

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

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