GRE Prep Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Area of the triangle AOB
[#permalink]
14 Jan 2025, 17:05
14 Jan 2025, 17:05
1
Expert Reply
1
Bookmarks
00:00
Question Stats:
55% (01:06) correct
44% (02:07) wrong based on 9 sessions
Hide Show timer Statistics
Attachment:
GRE Area of the triangle AOB.png [ 45.49 KiB | Viewed 1640 times ]
Quantity A |
Quantity B |
Area of the triangle AOB |
Area of the shaded region |
- Part of the project: GRE Quant & Verbal ADVANCED Daily Challenge 2024 Edition - Gain 20 Kudos & Get FREE Access to GRE Prep Club TESTS
- Also replying to the unanswered questions
Area of the triangle AOB
[#permalink]
22 Feb 2025, 13:51
22 Feb 2025, 13:51
Expert Reply
OFFICIAL EXPLANATION
GRE circle radius.png [ 95.04 KiB | Viewed 1459 times ]
\(\text { Let the radius of the circle above be ' } r \text { ' }\)
The area of the right triangle $\(A O B\)$ is $\(\frac{1}{2} \times\)$ Base $\(\times\)$ Height $\(=\frac{1}{2} \times A O \times O B=\frac{1}{2} \times r \times r=\frac{1}{2}\left(r^2\right)\)($ We have $\(\mathrm{AO}=\mathrm{OB}=\)$ radii $\(=\mathrm{r}\)$ )
Now, the area of the shaded portion $=$ area of sector $\(\operatorname{COD}=\frac{\theta}{360} \times \pi r^2=\frac{60}{360} \times \frac{22}{7} \times \mathrm{r}^2=\frac{11}{21}\left(\mathrm{r}^2\right)\)$
As $\(\frac{1}{2}\)$ is less than $\(\frac{11}{21}\)$, we get $\(\frac{1}{2}\left(\mathrm{r}^2\right)<\frac{11}{21}\left(\mathrm{r}^2\right)\)$
Hence column $\(B\)$ has higher quantity when compared with column $\(A\)$, so the answer is ( $\(B\)$ ).
Attachment:
GRE circle radius.png [ 95.04 KiB | Viewed 1459 times ]
\(\text { Let the radius of the circle above be ' } r \text { ' }\)
The area of the right triangle $\(A O B\)$ is $\(\frac{1}{2} \times\)$ Base $\(\times\)$ Height $\(=\frac{1}{2} \times A O \times O B=\frac{1}{2} \times r \times r=\frac{1}{2}\left(r^2\right)\)($ We have $\(\mathrm{AO}=\mathrm{OB}=\)$ radii $\(=\mathrm{r}\)$ )
Now, the area of the shaded portion $=$ area of sector $\(\operatorname{COD}=\frac{\theta}{360} \times \pi r^2=\frac{60}{360} \times \frac{22}{7} \times \mathrm{r}^2=\frac{11}{21}\left(\mathrm{r}^2\right)\)$
As $\(\frac{1}{2}\)$ is less than $\(\frac{11}{21}\)$, we get $\(\frac{1}{2}\left(\mathrm{r}^2\right)<\frac{11}{21}\left(\mathrm{r}^2\right)\)$
Hence column $\(B\)$ has higher quantity when compared with column $\(A\)$, so the answer is ( $\(B\)$ ).
gmatclubot
Area of the triangle AOB [#permalink]
22 Feb 2025, 13:51
Moderators:
