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 "1978 AHSME Problems/Problem 29"

(→‎Solution)
(→‎Solution)
Line 5: Line 5:
 
==Solution==
 
==Solution==
  
−
Notice that the area of \triangle <math>DAB</math> is the same as that of \triangle <math>A'AB</math> (same base, same height). Thus, the area of \triangle   <math>A'AB</math> is twice that (same height, twice the base). Similarly, [\triangle <math>BB'C</math>] = 2 \cdot [\triangle <math>ABC</math>], and so on.
+
Notice that the area of <math>\triangle</math> <math>DAB</math> is the same as that of <math>\triangle</math> <math>A'AB</math> (same base, same height). Thus, the area of <math>\triangle</math> <math>A'AB</math> is twice that (same height, twice the base). Similarly, [<math>\triangle</math> <math>BB'C</math>] = 2 \cdot [<math>\triangle</math> <math>ABC</math>], and so on.
  
−
Adding all of these, we see that the area the four triangles around <math>ABCD</math> is twice [\triangle <math>DAB</math>] + [\triangle <math>ABC</math>] + [\triangle <math>BCD</math>] + [\triangle <math>CDA</math>], which is itself twice the area of the quadrilateral <math>ABCD</math>. Finally, [<math>A'B'C'D'</math>] = [<math>ABCD</math>] + 4 \cdot [<math>ABCD</math>] = 5 \cdot [<math>ABCD</math>] = \fbox{50}.
+
Adding all of these, we see that the area the four triangles around <math>ABCD</math> is twice [<math>\triangle</math> <math>DAB</math>] + [<math>\triangle</math> <math>ABC</math>] + [<math>\triangle</math> <math>BCD</math>] + [<math>\triangle</math> <math>CDA</math>], which is itself twice the area of the quadrilateral <math>ABCD</math>. Finally, [<math>A'B'C'D'</math>] = [<math>ABCD</math>] + 4 \cdot [<math>ABCD</math>] = 5 \cdot [<math>ABCD</math>] = \fbox{50}.
  
 
~ Mathavi
 
~ Mathavi
  
 
Note: Anyone with a diagram would be of great help (still new to LaTex).
 
Note: Anyone with a diagram would be of great help (still new to LaTex).

Revision as of 13:03, 23 August 2022

Problem

Sides $AB,~ BC, ~CD$ and $DA$, respectively, of convex quadrilateral $ABCD$ are extended past $B,~ C ,~ D$ and $A$ to points $B',~C',~ D'$ and $A'$. Also, $AB = BB' = 6,~ BC = CC' = 7, ~CD = DD' = 8$ and $DA = AA' = 9$; and the area of $ABCD$ is $10$. The area of $A 'B 'C'D'$ is

$\textbf{(A) }20\qquad \textbf{(B) }40\qquad \textbf{(C) }45\qquad \textbf{(D) }50\qquad \textbf{(E) }60$

Solution

Notice that the area of $\triangle$ $DAB$ is the same as that of $\triangle$ $A'AB$ (same base, same height). Thus, the area of $\triangle$ $A'AB$ is twice that (same height, twice the base). Similarly, [$\triangle$ $BB'C$] = 2 \cdot [$\triangle$ $ABC$], and so on.

Adding all of these, we see that the area the four triangles around $ABCD$ is twice [$\triangle$ $DAB$] + [$\triangle$ $ABC$] + [$\triangle$ $BCD$] + [$\triangle$ $CDA$], which is itself twice the area of the quadrilateral $ABCD$. Finally, [$A'B'C'D'$] = [$ABCD$] + 4 \cdot [$ABCD$] = 5 \cdot [$ABCD$] = \fbox{50}.

~ Mathavi

Note: Anyone with a diagram would be of great help (still new to LaTex).

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1978_AHSME_Problems/Problem_29&oldid=177400"