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

Latest revision as of 21:49, 18 October 2025

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

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

~diagram uploaded by grogg007

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

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

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)
Preceded by Problem 16	Followed by Problem 18
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
CEMC Gauss (Grade 8)

Revision as of 00:39, 25 June 2025 (view source) Grogg007 (talk \| contribs) m (→‎Problem) ← Older edit		Latest revision as of 21:49, 18 October 2025 (view source) Anabel.disher (talk \| contribs)
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	~anabel.disher		~anabel.disher
		+	{{CEMC box\|year=2013\|competition=Gauss (Grade 8)\|num-b=16\|num-a=18}}