Difference between revisions of "2012 AMC 8 Problems/Problem 25"
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==Problem== | ==Problem== | ||
− | A square with area 4 is inscribed in a square with area 5, with | + | 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 | + | 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> | + | 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. | + | 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> 4 + 2ab = 5</math> |
+ | |||
+ | which gives us the value of <math>a \cdot b</math>, which is <math> \boxed{\textbf{(C)}\ \frac{1}2} </math>. | ||
+ | |||
+ | ==Solution 4== | ||
+ | |||
+ | First, observe that the given squares have areas <math>4</math> and <math>5</math>. | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | 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>. | ||
− | <math> | + | Since <math>2ab=1</math>, we divide by two getting <math>a \cdot b=\boxed{\textbf{(C) }\frac{1}{2}}</math> |
+ | |||
+ | ==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== | ==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
Contents
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 , and the other of length . What is the value of ?
Solution 1
The total area of the four congruent triangles formed by the squares is . Therefore, the area of one of these triangles is . The height of one of these triangles is and the base is . Using the formula for area of the triangle, we have . Multiply by on both sides to find that the value of is .
Solution 2
To solve this problem you could also use algebraic manipulation.
Since the area of the large square is , the side length is .
We then have the equation .
We also know that the side length of the smaller square is , since its area is . Then, the segment of length and segment of length form a right triangle whose hypotenuse would have length .
So our second equation is .
Square both equations.
Now, subtract, and obtain the equation . We can deduce that the value of is .
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.
which gives us the value of , which is .
Solution 4
First, observe that the given squares have areas and .
Then, observe that the 4 triangles with side lengths and have a combined area of .
We have, that is the total area of the 4 triangles in terms of and .
Since , we divide by two getting
Video Solution by Punxsutawney Phil
~sugar_rush
https://www.youtube.com/watch?v=QEwZ_17PQ6Q
Video Solution 2
https://youtu.be/MhxGq1sSA6U ~savannahsolver
See Also
2012 AMC 8 (Problems • Answer Key • Resources) | ||
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.