2009 CEMC Gauss (Grade 8) Problems/Problem 4

Problem

In the diagram, $AB$ is a line segment.

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The value of $x$ is

$\text{ (A) }\ 128 \qquad\text{ (B) }\ 38 \qquad\text{ (C) }\ 48 \qquad\text{ (D) }\ 142 \qquad\text{ (E) }\ 308$

Solution 1

Since $AB$ is a line segment, we know that the right angle, $x$ , and $52^{\circ}$ form a straight angle.

This means that:

$90 + x + 52 = 180$

$x + 142 = 180$

$x = \boxed {\textbf {(B) } 38}$

~anabel.disher

Solution 2 (answer choices)

Trying answer choice A, we can see that the angle is too big because $128 + 52 = 180$ , so adding $90$ would lead to the number being $90$ . This means that the answer must be B or C.

$38 + 52 + 90 = 90 + 90 = 180$ , which works, so the answer is $\boxed {\textbf {(B) } 38}$ .

~anabel.disher

Solution 2.5 (answer choices)

We can use the first part of solution 2 where we try answer choice A.

Since answer choice A gives an answer that is $90$ degrees higher than it is supposed to, we can simply see what answer choice is $90$ degrees less than answer choice A.

$128 - 90 = 38$ , so the answer is $\boxed {\textbf {(B) } 38}$ .

~anabel.disher

2009 CEMC Gauss (Grade 8) (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
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CEMC Gauss (Grade 8)

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2009 CEMC Gauss (Grade 8) Problems/Problem 4

Contents

Problem

Solution 1

Solution 2 (answer choices)

Solution 2.5 (answer choices)