2006 CEMC Pascal Problems/Problem 2

Problem

What is the value of $\sqrt{36 + 64} - \sqrt{25 - 16}$ ?

$\text{ (A) }\ 5 \qquad\text{ (B) }\ 7 \qquad\text{ (C) }\ 13 \qquad\text{ (D) }\ 11 \qquad\text{ (E) }\ 9$

Solution 1

$\sqrt{36 + 64} - \sqrt{25 - 16} = \sqrt{100} - \sqrt{9} = 10 - 3 = \boxed {\textbf {(B) } 7}$

~anabel.disher

Solution 2

We can use the fact that $6^2=36, 8^2 = 64, 5^2 = 25$ , and $4^2 = 16$ . $6$ , $8$ , and $10$ form a Pythagorean triple, as well as $3$ , $4$ , and $5$ , so we can see:

$\sqrt{36 + 64} - \sqrt{25 - 16} = 10 - 3 = \boxed {\textbf {(B) } 7}$ .

~anabel.disher

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2006 CEMC Pascal Problems/Problem 2

Problem

Solution 1

Solution 2