Difference between revisions of "2005 AMC 8 Problems/Problem 23"

Revision as of 12:50, 11 October 2024

Problem

Isosceles right triangle $ABC$ encloses a semicircle of area $2\pi$ . The circle has its center $O$ on hypotenuse $\overline{AB}$ and is tangent to sides $\overline{AC}$ and $\overline{BC}$ . What is the area of triangle $ABC$ ?

$[asy]pair a=(4,4), b=(0,0), c=(0,4), d=(4,0), o=(2,2); draw(circle(o, 2)); clip(a--b--c--cycle); draw(a--b--c--cycle); dot(o); label("$C$", c, NW); label("$A$", a, NE); label("$B$", b, SW);[/asy]$

$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 3\pi\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 4\pi$

Solution

First, we notice half a square so first let's create a square. Once we have a square, we will have a full circle. This circle has a diameter of 4 which will be the side of the square. The area would be $4\cdot 4 = 16.$ Divide 16 by 2 to get the original shape and you get $\boxed{8}$

Solution 2

We can figure out the radius of the semicircle because the question states that the area of the semicircle is $2\pi$ and we can multiply it by 2 to get $4\pi$ which we can see it is 2 from the formula. Draw line segment OD such that it is the midsegment of triangle ABC, using the midsegment theorem we can see that line segment AC = 2*2=4. Since triangle ABC is an isosceles right triangle we can calculate the area to be $\frac{4^2}{2}$ = $\boxed{8}$

Video Solution

https://www.youtube.com/watch?v=cNbXCQXUc6E ~David

Video Solution by OmegaLearn

https://youtu.be/j3QSD5eDpzU?t=1116

~ pi_is_3.14

2005 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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 14: / Line 14: @@
 ==Solution==
-The semi circle has an area of <math>\pi r^2 /2 = 2\pi</math> and a radius of <math>2</math>.
+First, we notice half a square so first let's create a square. Once we have a square, we will have a full circle. This circle has a diameter of 4 which will be the side of the square. The area would be <math>4\cdot 4 = 16.</math> Divide 16 by 2 to get the original shape and you get <math>\boxed{8}</math>
-Because this is an isosceles right triangle, the center is the midpoint of the hypotenuse. Radii drawn to the tangent points of the semicircle and the radii also divide the legs into two equal segments. They also create a square in the top left corner. From this, we can conclude the legs of the triangle are twice the length of the radii, <math>4</math>. The area of the triangle is <math>(4)(4)/2 = \boxed{\textbf{(B)}\ 8}</math>.
+==Solution 2==
+We can figure out the radius of the semicircle because the question states that the area of the semicircle is <math> 2\pi</math> and we can multiply it by 2 to get <math> 4\pi </math> which we can see it is 2 from the formula. Draw line segment OD such that it is the midsegment of triangle ABC, using the midsegment theorem we can see that line segment AC = 2*2=4. Since triangle ABC is an isosceles right triangle we can calculate the area to be <math>\frac{4^2}{2}</math> = <math>\boxed{8}</math>
+==Video Solution==
+https://www.youtube.com/watch?v=cNbXCQXUc6E   ~David
+==Video Solution by OmegaLearn==
+https://youtu.be/j3QSD5eDpzU?t=1116
+~ pi_is_3.14
-==Easier More Logical Solution==
-We see half a square so first let's create a square. Once we have a square, we will have to full circle. This circle has a diameter of 4 which will be the side of the square. The area would be 4*4 = 16. Divide 16 by 2 to get he original shape and you get [b]8[/b].
 ==See Also==
 {{AMC8 box|year=2005|num-b=22|num-a=24}}
 {{MAA Notice}}

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