Difference between revisions of "2018 AMC 10A Problems/Problem 10"
(→Solution) |
Kempu33334 (talk | contribs) (→Solution 10 (Solution 1 but alternate)) |
||
(64 intermediate revisions by 30 users not shown) | |||
Line 1: | Line 1: | ||
− | Suppose that real number <math>x</math> satisfies <cmath>\sqrt{49-x^2}-\sqrt{25-x^2}=3</cmath> | + | ==Problem== |
+ | |||
+ | Suppose that real number <math>x</math> satisfies <cmath>\sqrt{49-x^2}-\sqrt{25-x^2}=3</cmath>What is the value of <math>\sqrt{49-x^2}+\sqrt{25-x^2}</math>? | ||
<math> | <math> | ||
− | \textbf{(A) }8 \qquad | + | \textbf{(A) }8\qquad |
\textbf{(B) }\sqrt{33}+8\qquad | \textbf{(B) }\sqrt{33}+8\qquad | ||
− | \textbf{(C) }9 \qquad | + | \textbf{(C) }9\qquad |
− | \textbf{(D) }2\sqrt{10}+4 \qquad | + | \textbf{(D) }2\sqrt{10}+4\qquad |
− | \textbf{(E) }12 \qquad | + | \textbf{(E) }12\qquad |
</math> | </math> | ||
− | == Solution == | + | ==Solution 10 (Solution 1 but alternate)== |
+ | |||
+ | We let <math>a=\sqrt{49-x^2}+\sqrt{25-x^2}</math>; in other words, we want to find <math>a</math>. We know that <math>a\cdot3=\left(\sqrt{49-x^2}+\sqrt{25-x^2}\right)\cdot\left(\sqrt{49-x^2}-\sqrt{25-x^2}\right)=\left(\sqrt{49-x^2}\right)^2-\left(\sqrt{25-x^2}\right)^2=\left(49-x^2\right)-\left(25-x^2\right)=24.</math> Thus, <math>a=\boxed{8}</math>. | ||
+ | |||
+ | ~Technodoggo | ||
+ | |||
+ | ==Video Solution (HOW TO THINK CREATIVELY!)== | ||
+ | https://youtu.be/P-atxiiTw2I | ||
+ | |||
+ | ~Education, the Study of Everything | ||
+ | |||
+ | |||
+ | |||
+ | |||
+ | == Video Solutions == | ||
+ | ===Video Solution 1=== | ||
+ | https://youtu.be/ba6w1OhXqOQ?t=1403 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
+ | ===Video Solution 2=== | ||
+ | https://youtu.be/zQG70XKAdeA ~ North America Math Contest Go Go Go | ||
− | + | ===Video Solution 3=== | |
+ | https://youtu.be/ZiZVIMmo260 | ||
− | + | ===Video Solution 4=== | |
+ | https://youtu.be/5cA87rbzFdw | ||
− | == See Also == | + | ~savannahsolver |
+ | |||
+ | ==See Also== | ||
{{AMC10 box|year=2018|ab=A|num-b=9|num-a=11}} | {{AMC10 box|year=2018|ab=A|num-b=9|num-a=11}} | ||
− | + | ||
+ | [[Category:Introductory Algebra Problems]] |
Latest revision as of 19:59, 20 April 2024
Contents
Problem
Suppose that real number satisfies What is the value of ?
Solution 10 (Solution 1 but alternate)
We let ; in other words, we want to find . We know that Thus, .
~Technodoggo
Video Solution (HOW TO THINK CREATIVELY!)
~Education, the Study of Everything
Video Solutions
Video Solution 1
https://youtu.be/ba6w1OhXqOQ?t=1403
~ pi_is_3.14
Video Solution 2
https://youtu.be/zQG70XKAdeA ~ North America Math Contest Go Go Go
Video Solution 3
Video Solution 4
~savannahsolver
See Also
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 |