Difference between revisions of "2012 AMC 8 Problems/Problem 25"

(Solution 3 (similar to solution 1))
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==Problem==
 
==Problem==
A square with area 4 is inscribed in a square with area 5, with one vertex of the smaller square on each side of the larger square. A vertex of the smaller square divides a side of the larger square into two segments, one of length <math> a </math>, and the other of length <math> b </math>. What is the value of <math> ab </math>?
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A square with area 4 is inscribed in a square with area 5, with each vertex of the smaller square on a side of the larger square. A vertex of the smaller square divides a side of the larger square into two segments, one of length <math> a </math>, and the other of length <math> b </math>. What is the value of <math> ab </math>?
  
 
<asy>
 
<asy>
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To solve this problem you could also use algebraic manipulation.
 
To solve this problem you could also use algebraic manipulation.
  
Since the area of the large square is <math> 5 </math>, the sidelength is <math> \sqrt{5} </math>.
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Since the area of the large square is <math> 5 </math>, the side length is <math> \sqrt{5} </math>.
  
 
We then have the equation <math> a + b = \sqrt{5} </math>.
 
We then have the equation <math> a + b = \sqrt{5} </math>.
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<math> a^2 + b^2 = 4 </math>
 
<math> a^2 + b^2 = 4 </math>
  
Now, subtract, and obtain the equation <math> 2ab = 1 </math>. We can deduce that the value of <math> ab </math> is <math> \boxed{\textbf{(C)}\ \frac{1}2} </math>.
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Now, subtract, and obtain the equation <math> 2ab = 1 </math>. We can deduce that the value of <math> a \cdot b </math> is <math> \boxed{\textbf{(C)}\ \frac{1}2} </math>.
  
 
==Solution 3 (similar to solution 1)==
 
==Solution 3 (similar to solution 1)==
  
Since we know 4 of the triangles both have side lengths a and b, we can create an equation.
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Since we know 4 of the triangles both have side lengths a and b, we can create an equation, which the area of the inner square plus the sum of the 4 triangles being the area of the outer square.
  
<math> Area(Inner square) + Area(triangles) = Area(Outer square)
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<math> 4 + 2ab = 5</math>
  
</math> 4 + 2ab = 5<math>
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which gives us the value of <math>a \cdot b</math>, which is  <math> \boxed{\textbf{(C)}\ \frac{1}2} </math>.
  
which gives us the value of ab, which is </math> \boxed{\textbf{(C)}\ \frac{1}2} $.
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==Solution 4==
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First, observe that the given squares have areas <math>4</math> and <math>5</math>.
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Then, observe that the 4 triangles with side lengths <math>a</math> and <math>b</math> have a combined area of <math>5-4=1</math>.
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We have, that <math>4\cdot\frac{ab}{2}=2ab</math> is the total area of the 4 triangles in terms of <math>a</math> and <math>b</math>.
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Since <math>2ab=1</math>, we divide by two getting <math>a \cdot b=\boxed{\textbf{(C) }\frac{1}{2}}</math>
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==Video Solution by Punxsutawney Phil==
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https://youtu.be/RyKWp2YDHJM
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~sugar_rush
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https://www.youtube.com/watch?v=QEwZ_17PQ6Q
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==Video Solution 2==
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https://youtu.be/MhxGq1sSA6U ~savannahsolver
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2012|num-b=24|after=Last Problem}}
 
{{AMC8 box|year=2012|num-b=24|after=Last Problem}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 09:38, 18 May 2022

Problem

A square with area 4 is inscribed in a square with area 5, with each vertex of the smaller square on a side of the larger square. A vertex of the smaller square divides a side of the larger square into two segments, one of length $a$, and the other of length $b$. What is the value of $ab$?

[asy] draw((0,2)--(2,2)--(2,0)--(0,0)--cycle); draw((0,0.3)--(0.3,2)--(2,1.7)--(1.7,0)--cycle); label("$a$",(-0.1,0.15)); label("$b$",(-0.1,1.15));[/asy]

$\textbf{(A)}\hspace{.05in}\frac{1}5\qquad\textbf{(B)}\hspace{.05in}\frac{2}5\qquad\textbf{(C)}\hspace{.05in}\frac{1}2\qquad\textbf{(D)}\hspace{.05in}1\qquad\textbf{(E)}\hspace{.05in}4$

Solution 1

The total area of the four congruent triangles formed by the squares is $5-4 = 1$. Therefore, the area of one of these triangles is $\frac{1}{4}$. The height of one of these triangles is $a$ and the base is $b$. Using the formula for area of the triangle, we have $\frac{ab}{2} = \frac{1}{4}$. Multiply by $2$ on both sides to find that the value of $ab$ is $\boxed{\textbf{(C)}\ \frac{1}2}$.

Solution 2

To solve this problem you could also use algebraic manipulation.

Since the area of the large square is $5$, the side length is $\sqrt{5}$.

We then have the equation $a + b = \sqrt{5}$.

We also know that the side length of the smaller square is $2$, since its area is $4$. Then, the segment of length $a$ and segment of length $b$ form a right triangle whose hypotenuse would have length $2$.

So our second equation is $\sqrt{{a^2}+{b^2}} = 2$.

Square both equations.

$a^2 + 2ab + b^2 = 5$

$a^2 + b^2 = 4$

Now, subtract, and obtain the equation $2ab = 1$. We can deduce that the value of $a \cdot b$ is $\boxed{\textbf{(C)}\ \frac{1}2}$.

Solution 3 (similar to solution 1)

Since we know 4 of the triangles both have side lengths a and b, we can create an equation, which the area of the inner square plus the sum of the 4 triangles being the area of the outer square.

$4 + 2ab = 5$

which gives us the value of $a \cdot b$, which is $\boxed{\textbf{(C)}\ \frac{1}2}$.

Solution 4

First, observe that the given squares have areas $4$ and $5$.

Then, observe that the 4 triangles with side lengths $a$ and $b$ have a combined area of $5-4=1$.

We have, that $4\cdot\frac{ab}{2}=2ab$ is the total area of the 4 triangles in terms of $a$ and $b$.

Since $2ab=1$, we divide by two getting $a \cdot b=\boxed{\textbf{(C) }\frac{1}{2}}$

Video Solution by Punxsutawney Phil

https://youtu.be/RyKWp2YDHJM

~sugar_rush

https://www.youtube.com/watch?v=QEwZ_17PQ6Q

Video Solution 2

https://youtu.be/MhxGq1sSA6U ~savannahsolver

See Also

2012 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 24
Followed by
Last Problem
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 AJHSME/AMC 8 Problems and Solutions

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