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 "2009 UNCO Math Contest II Problems/Problem 9"

(→‎Solution 2)
(→‎Solution 2 (Better Explained))
 
(10 intermediate revisions by the same user not shown)
Line 15: Line 15:
  
 
== Solution ==
 
== Solution ==
−
{{solution}}
 
 
Let x be the length of a side. Then the square has area <math>x^2</math> and each portion has area <math>x^2 \times\frac{1}{3}</math>
 
Let x be the length of a side. Then the square has area <math>x^2</math> and each portion has area <math>x^2 \times\frac{1}{3}</math>
 
If x is the base of one of the triangles, then the height will be <math>\frac{2x}{3}</math>.
 
If x is the base of one of the triangles, then the height will be <math>\frac{2x}{3}</math>.
Line 22: Line 21:
 
Thus, the area of the square is 64*13 = 832.
 
Thus, the area of the square is 64*13 = 832.
  
−
== Solution 2 ==
+
== Solution 2 (Better Explained) ==
 
<asy>
 
<asy>
 
unitsize(135);
 
unitsize(135);
Line 63: Line 62:
  
 
<math>3x</math> = <math>8\sqrt{13}</math>
 
<math>3x</math> = <math>8\sqrt{13}</math>
 +
 +
Area of square ABCD = <math>9x^{2}</math> = <math>(8\sqrt{13})^{2}</math> = 832
  
 
== See also ==
 
== See also ==

Latest revision as of 10:07, 28 September 2023

Problem

A square is divided into three pieces of equal area by two parallel lines as shown. If the distance between the two parallel lines is $8$ what is the area of the square?

[asy] draw((0,0)--(1,0)--(1,1)--(0,1)--cycle,black); draw((1,0)--(0,2/3),black); draw((1,1/3)--(0,1),black); [/asy]

Solution

Let x be the length of a side. Then the square has area $x^2$ and each portion has area $x^2 \times\frac{1}{3}$ If x is the base of one of the triangles, then the height will be $\frac{2x}{3}$. By the pythaogrean theorem, longer side of the parallelogram has length $\sqrt(x^2+(\frac{2x}{3})^2)$ Thus sqrt(13)*x/3*8 = x^2/3. Solving this gives x = 8*sqrt(13). Thus, the area of the square is 64*13 = 832.

Solution 2 (Better Explained)

[asy] unitsize(135); defaultpen(linewidth(.8pt)+fontsize(10pt)); pair A, B, C, D, E, F; A=(0, 0); B=(1, 0); C=(1, -1); D=(0,-1); E=(0,-1/3); F=(1,-2/3); draw(A--B--C--D--cycle); label("$A$",A,N); label("$B$",B,N); label("$C$",C,E); label("$D$",D,S); label("$E$",E,W); label("$F$",F,SE); draw(A--F); draw(C--E); draw(A--C); draw(E--F); [/asy]

In $\triangle$ADC, $\frac {AE}{ED}$ = $\frac {1}{2}$ because $\frac {[ACE]}{[CED]}$ = $\frac {1}{2}$ (According to question)

$[ECD]$ = $2[EFC]$ (AECF is a parallelogram and $2[EFC]$ = $[AECF]$)

$2[EFC]$ = $[EDC]$ (same reason as before)

$[EFC]$ = $\frac {1}{2}$ * 8 * EC

$[EDC]$ = $\frac{1}{2}$ * EC * (Perpendicular from D to EC)

2 * $\frac {1}{2}$ * 8 * EC = $\frac{1}{2}$ * EC * (Perpendicular from D to EC)

(Perpendicular from D to EC) = 16

Let AE = $x$, then ED = $2x$. So the side length of square is $3x$.

(Perpendicular from D to EC) = $\frac {ED * DC}{EC}$

EC = $\sqrt{13}x$ (By Pythagoras Theorem)

16 = $\frac {2x * 3x}{\sqrt{13}x}$

16 = $\frac {6x^{2}}{\sqrt{13}x}$

$x$ = $\frac {8\sqrt{13}}{3}$

$3x$ = $8\sqrt{13}$

Area of square ABCD = $9x^{2}$ = $(8\sqrt{13})^{2}$ = 832

See also

2009 UNCO Math Contest II (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10
All UNCO Math Contest Problems and Solutions
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2009_UNCO_Math_Contest_II_Problems/Problem_9&oldid=198753"