Difference between revisions of "2021 Fall AMC 10B Problems/Problem 2"

Revision as of 18:29, 29 November 2021

Problem

What is the area of the shaded figure shown below? $[asy] size(200); defaultpen(linewidth(0.4)+fontsize(12)); pen s = linewidth(0.8)+fontsize(8); pair O,X,Y; O = origin; X = (6,0); Y = (0,5); fill((1,0)--(3,5)--(5,0)--(3,2)--cycle, palegray+opacity(0.2)); for (int i=1; i<7; ++i) { draw((i,0)--(i,5), gray+dashed); label("${"+string(i)+"}$", (i,0), 2*S); if (i<6) { draw((0,i)--(6,i), gray+dashed); label("${"+string(i)+"}$", (0,i), 2*W); } } label("$0$", O, 2*SW); draw(O--X+(0.15,0), EndArrow); draw(O--Y+(0,0.15), EndArrow); draw((1,0)--(3,5)--(5,0)--(3,2)--(1,0), black+1.5); [/asy]$

$(\textbf{A})\: 4\qquad(\textbf{B}) \: 6\qquad(\textbf{C}) \: 8\qquad(\textbf{D}) \: 10\qquad(\textbf{E}) \: 12$

Solution #1

We have $2$ isosceles triangles. Thus, the area of the shaded region is $\frac{1}{2} \cdot 5 \cdot 4 - \left(\frac{1}{2} \cdot 4 \cdot 2\right) = 10 - 4 = 6.$ Thus our answer is $\boxed{(\textbf{B}.)}.$

~NH14

Solution #2

As we can see, the shape is symmetrical, so it will be equally valid to simply calculate one of the half's area and multiply by 2. One half's area is $\frac{bh}2$ , so two halves would be $bh=3\cdot2=6$ . Thus our answer is $\boxed{(\textbf{B}.)}.$

~Hefei417, or 陆畅 Sunny from China

Solution #3 (Overkill)

We start by finding the points. The outlined shape is made up of $(1,0),(3,5),(5,0),(3,2)$ . By the Shoelace Theorem, we find the area to be $6$ , or $\boxed{B}$ .

https://artofproblemsolving.com/wiki/index.php/Shoelace_Theorem

~Taco12

~I-AM-DA-KING for the link

Solution #4 (Pick's Theorem)

We can use Pick's Theorem. We have $4$ interior points and $6$ boundary points. By Pick's Theorem, we get $4+\frac{6}{2}-1 = 4+3-1 = 6.$ Checking our answer choices, we find our answer to be $\boxed{B}$ .

~danprathab

Solution 5

The area is $\begin{align*} \frac{1}{2} \left( 5 - 1 \right) 5 - \frac{1}{2} \left( 5 - 1 \right) 2 & = 6 . \end{align*}$

Therefore, the answer is $\boxed{\textbf{(B) }6}$ .

~Steven Chen (www.professorchenedu.com)

Solution 6

We have that the area of the shaded region is the difference between the area of 2 triangles. So, we have the area is $\dfrac{1}{2}(4\cdot 5) - \dfrac{1}{2}(4\cdot 2) = \boxed{\textbf{(B) or 6}}$ ~~stjwyl

Video Solution by Interstigation

https://youtu.be/p9_RH4s-kBA?t=110

@@ Line 73: / Line 73: @@
 ~Steven Chen (www.professorchenedu.com)
+== Solution 6 ==
+We have that the area of the shaded region is the difference between the area of 2 triangles. So, we have the area is <math>\dfrac{1}{2}(4\cdot 5) - \dfrac{1}{2}(4\cdot 2) = \boxed{\textbf{(B) or 6}}</math> ~~stjwyl
 ==Video Solution by Interstigation==
 https://youtu.be/p9_RH4s-kBA?t=110

2021 Fall AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
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

Page

Toolbox

Search