Difference between revisions of "2024 AMC 10A Problems/Problem 22"

(Problem)
(Solution)
Line 10: Line 10:
 
Drawing the altitude from N to point Q on line MO, we know NQ is <math>\sqrt3/2</math>, MQ is <math>3/2</math>, and QO is <math>1/2</math>.
 
Drawing the altitude from N to point Q on line MO, we know NQ is <math>\sqrt3/2</math>, MQ is <math>3/2</math>, and QO is <math>1/2</math>.
  
[[File:Drawing.sketchpad.png]]
+
[[File:Screenshot 2024-11-08 2.33.52 PM.png]]
  
 
==See also==
 
==See also==
 
{{AMC10 box|year=2024|ab=A|num-b=21|num-a=23}}
 
{{AMC10 box|year=2024|ab=A|num-b=21|num-a=23}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 17:35, 8 November 2024

Problem

Let $\mathcal K$ be the kite formed by joining two right triangles with legs $1$ and $\sqrt3$ along a common hypotenuse. Eight copies of $\mathcal K$ are used to form the polygon shown below. What is the area of triangle $\Delta ABC$?

03abbd4df2932f4a1d16a34c2b9e15b683dedb.png

$\textbf{(A) }2+3\sqrt3\qquad\textbf{(B) }\dfrac92\sqrt3\qquad\textbf{(C) }\dfrac{10+8\sqrt3}{3}\qquad\textbf{(D) }8\qquad\textbf{(E) }5\sqrt3$

Solution

Let $\mathcal K$ be quadrilateral MNOP. Drawing line MO splits the triangle into $\Delta MNO$. Drawing the altitude from N to point Q on line MO, we know NQ is $\sqrt3/2$, MQ is $3/2$, and QO is $1/2$.

Screenshot 2024-11-08 2.33.52 PM.png

See also

2024 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 21
Followed by
Problem 23
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 AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png