Difference between revisions of "2018 AMC 10A Problems/Problem 10"

Revision as of 15:32, 8 February 2018

Suppose that real number $x$ satisfies $\sqrt{49-x^2}-\sqrt{25-x^2}=3$ . What is the value of $\sqrt{49-x^2}+\sqrt{25-x^2}$ ?

$\textbf{(A) }8 \qquad \textbf{(B) }\sqrt{33}+8\qquad \textbf{(C) }9 \qquad \textbf{(D) }2\sqrt{10}+4 \qquad \textbf{(E) }12 \qquad$

$(\sqrt {49-x^2} + \sqrt {25-x^2}) * (\sqrt {49-x^2} - \sqrt {25-x^2}) = 49-x^2 - 25 +x^2 = 24$

Given that $(\sqrt {49-x^2} - \sqrt {25-x^2})$ = 3, $(\sqrt {49-x^2} + \sqrt {25-x^2}) = \frac {24} {8} = \boxed{3}$

2018 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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
All AMC 10 Problems and Solutions

@@ Line 13: / Line 13: @@
 <math>(\sqrt {49-x^2} + \sqrt {25-x^2}) * (\sqrt {49-x^2} - \sqrt {25-x^2}) = 49-x^2 - 25 +x^2 = 24</math>
-Given that <math>(\sqrt {49-x^2} - \sqrt {25-x^2})</math> = 3, <math>(\sqrt {49-x^2} + \sqrt {25-x^2}) = \frac {24} {8} = \box{3}</math>
+Given that <math>(\sqrt {49-x^2} - \sqrt {25-x^2})</math> = 3, <math>(\sqrt {49-x^2} + \sqrt {25-x^2}) = \frac {24} {8} = \boxed{3}</math>
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