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

Revision as of 11:13, 12 November 2019

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

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

@@ Line 48: / Line 48: @@
 We have, that <math>4\frac{ab}{2}=2ab</math> is the total area of the 4 triangles in terms of <math>a</math> and <math>b</math>.
-Since <math>2ab=1</math>, we divide by two getting <math>ab=\boxed{\textbf{(C)}\frac{ab}{2}}</math>
+Since <math>2ab=1</math>, we divide by two getting <math>ab=\boxed{\textbf{(C) }\frac{ab}{2}}</math>
 ==See Also==
 {{AMC8 box|year=2012|num-b=24|after=Last Problem}}
 {{MAA Notice}}

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

Page

Toolbox

Search