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

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==Solution 10 (Solution 1 but alternate)==
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==Solution 1==
  
 
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>.  
 
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
 
~Technodoggo
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==Solution 2==
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Let <math>a = \sqrt{49-x^2}</math>, and <math>b = \sqrt{25-x^2}</math>. Solving for the constants in terms of x, a , and b, we get <math>a^2 + x^2 = 49</math>, and <math>b^2 + x^2 = 25</math>. Subtracting the second equation from the first gives us <math>a^2 - b^2 = 24</math>. Difference of squares gives us <math>(a+b)(a-b) = 24</math>. Since we want to find <math>a+b = \sqrt{49-x^2}+\sqrt{25-x^2}</math>, and we know <math>a-b = 3</math>, we get <math>3(a+b) = 24</math>, so <math>a+b = \boxed{\textbf{(A) }8}</math>
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~idk12345678
  
 
==Video Solution (HOW TO THINK CREATIVELY!)==
 
==Video Solution (HOW TO THINK CREATIVELY!)==

Latest revision as of 19:20, 16 May 2024

Problem

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

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

Solution 1

We let $a=\sqrt{49-x^2}+\sqrt{25-x^2}$; in other words, we want to find $a$. We know that $a\cdot3=\left(\sqrt{49-x^2}+\sqrt{25-x^2}\right)\cdot\left(\sqrt{49-x^2}-\sqrt{25-x^2}\right)=\left(\sqrt{49-x^2}\right)^2-\left(\sqrt{25-x^2}\right)^2=\left(49-x^2\right)-\left(25-x^2\right)=24.$ Thus, $a=\boxed{8}$.

~Technodoggo

Solution 2

Let $a = \sqrt{49-x^2}$, and $b = \sqrt{25-x^2}$. Solving for the constants in terms of x, a , and b, we get $a^2 + x^2 = 49$, and $b^2 + x^2 = 25$. Subtracting the second equation from the first gives us $a^2 - b^2 = 24$. Difference of squares gives us $(a+b)(a-b) = 24$. Since we want to find $a+b = \sqrt{49-x^2}+\sqrt{25-x^2}$, and we know $a-b = 3$, we get $3(a+b) = 24$, so $a+b = \boxed{\textbf{(A) }8}$


~idk12345678

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

~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
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