Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2012 AMC 8 Problems/Problem 25"
Line 8: Line 8:
  
 
<math> \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 </math>
 
<math> \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 </math>
 +
 +
==Solution 1==
 +
The total area of the four congruent triangles formed by the squares is <math> 1 </math>. Therefore, the area of one of these triangles is  <math> \frac{1}{4} </math>. The height of one of these triangles is <math> a </math> and the base is <math> b </math>. Using the formula for area of the triangle, we have <math> \frac{ab}{2} = \frac{1}{4} </math>. Multiply by <math> 2 </math> on both sides to find that the value of <math> ab </math> is <math> \boxed{\textbf{(C)}\ \frac{1}2} </math>.
 +
 +
==Solution 2==
 +
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>.
 +
 +
We then have the equation <math> a + b = \sqrt{5} </math>.
 +
 +
We also know that the sidelength of the smaller square is  <math> 2 </math>, since its area is <math> 4 </math>. Then, the segment of length <math> a </math> and segment of length <math> b </math> form a right triangle whose hypotenuse would have length <math> 2 </math>.
 +
 +
So our second equation is <math> \sqrt{{a^2}+{b^2}} = 2 </math>.
 +
 +
Square both equations.
 +
 +
<math> a^2 + 2ab + b^2 = 5 </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>.
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2012|num-b=24|after=Last Problem}}
 
{{AMC8 box|year=2012|num-b=24|after=Last Problem}}

Revision as of 12:38, 24 November 2012

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 $a$, 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 $1$. 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 sidelength is $\sqrt{5}$.

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

We also know that the sidelength 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 $ab$ is $\boxed{\textbf{(C)}\ \frac{1}2}$.

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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_8_Problems/Problem_25&oldid=49745"