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

Revision as of 08:29, 16 July 2024

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 $\boxed{\textbf{(C)}\ \frac{1}2}$ .

Video Solution 2

https://youtu.be/MhxGq1sSA6U ~savannahsolver

Video Solution by OmegaLearn

https://youtu.be/j3QSD5eDpzU?t=2

~ pi_is_3.14

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.

@@ Line 1: / Line 1: @@
 ==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>?
+A square with area <math>4</math> is inscribed in a square with area <math>5</math>, 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>
@@ Line 13: / Line 13: @@
 The total area of the four congruent triangles formed by the squares is <math>5-4 = 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==
+==Video Solution 2==
-To solve this problem you could also use algebraic manipulation.
+https://youtu.be/MhxGq1sSA6U ~savannahsolver
-Since the area of the large square is <math> 5 </math>, the sidelength is <math> \sqrt{5} </math>.
+==Video Solution by OmegaLearn==
+https://youtu.be/j3QSD5eDpzU?t=2
-We then have the equation <math> a + b = \sqrt{5} </math>.
+~ pi_is_3.14
-We also know that the side length 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>.
-==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.
-<math> 4 + 2ab = 5</math>
-which gives us the value of ab, 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\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>
 ==See Also==
 {{AMC8 box|year=2012|num-b=24|after=Last Problem}}
 {{MAA Notice}}

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