2001 CEMC Gauss (Grade 8) Problems/Problem 16

Problem

In the diagram, the value of $x$ is


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

Solution 1

To make the solution easier to understand, we can name different points.


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Since all of the sides of triangle $ABC$ are equal, the triangle is equilateral. Thus, all of its angles are $60^{\circ}$ in measure.


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We can also notice that triangle $CAD$ is isosceles, with $CA = CD$. Thus, we can let $m \angle CAD = m \angle CDA = y$.


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The sum of the interior angles of a triangle is always $180^{\circ}$, so we have:

$y + y + 30^{\circ} = 180^{\circ}$

$2y + 30^{\circ} = 180^{\circ}$

$2y = 150^{\circ}$

$y = 75^{\circ}$


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We can now see that $EAB$ is a straight line. Thus, we have:

$x^{\circ} + 75^{\circ} + 60^{\circ} = 180^{\circ}$

$x^{\circ} + 135^{\circ} = 180^{\circ}$

$x = \boxed {\textbf {(E) } 45}$

~anabel.disher

2001 CEMC Gauss (Grade 8) (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
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CEMC Gauss (Grade 8)