Events & Promotions
|
|
GMAT 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.
Jul 0901:00 PM EDT
-02:00 PM EDT
More Target Test Prep students are earning 100th percentile GMAT scores. Donât settle for less when the Target Test Prep course gives you everything you need to score high on the GMAT. Start your 5-day free trial today.
Jul 1206:30 AM PDT
-08:30 AM PDT
Join our webinar to master GMAT RC Main Point Questions! Learn effective strategies to decipher passage themes and purposes. Senior Verbal Expert- Sunita Singhvi will equip you with tools to navigate complexities and avoid common traps set by the GMAT.
Jul 1311:00 AM EDT
-12:30 PM EDT
Looking for your GMAT motivation to break through the score plateau? Pragati improved her score by massive 160 points with strategic guidance and hard-work! Find out how personalized mentorship and a strong mindset can turn GMAT struggles into success.
Jul 1508:00 PM PDT
-09:00 PM PDT
Full-length FE mock with insightful analytics, weakness diagnosis, and video explanations!
Jul 1611:00 AM EDT
-12:00 PM EDT
Prefer video-based learning? The Target Test Prep OnDemand course is a one-of-a-kind video masterclass featuring 400 hours of lecture-style teaching by Scott Woodbury-Stewart, founder of Target Test Prep and one of the most accomplished GMAT instructors.
Jul 1911:00 AM IST
-01:00 PM IST
Do RC/MSR passages scare you? e-GMAT is conducting a masterclass to help you learn â Learn effective reading strategies Tackle difficult RC & MSR with confidence Excel in timed test environment
09 Oct 2022, 09:46
Kudos
Bookmarks
B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
46% (01:54) correct
54%
(03:06)
wrong
based on 68
sessions
History
Date
Time
Result
Not Attempted Yet
On the coordinate plane there are 3 points đŽ(â1, 1), đ”(1, â3), and đ¶(3, 4). What is the area of the triangle connecting the 3 points?
(A) 10
(B) 11
(C) 12
(D) 14
(E) 13
(A) 10
(B) 11
(C) 12
(D) 14
(E) 13
gmatprepper94111
Joined: 17 Jul 2022
Last visit: 23 Nov 2022
Posts: 24
Own Kudos:
Location: United States (CA)
Concentration: Finance, Strategy
Schools: Stanford '25
GPA: 3.9
WE:Corporate Finance (Finance: Venture Capital)
13 Oct 2022, 01:14
Kudos
Bookmarks
=================================
On the coordinate plane there are 3 points đŽ(â1, 1), đ”(1, â3), and đ¶(3, 4). What is the area of the triangle connecting the 3 points?
(A) 10
(B) 11
(C) 12
(D) 14
(E) 13
=================================
We want to take the area. Area is 0.5 * base * height -- where height is the altitude of the triangle.
I want AB to be the base of my triangle. This means that we want the altitude dropped from C to be our height.
To simplify things, I'm going to move the triangle such that B sits on the origin; this moves the triangle -1 in the x-direction and +3 in the y direction.
A -> (-2, 4)
B -> (0, 0)
C -> (2, 7)
Let's check if we're lucky and it's a right triangle.
-> Slope of AB is -4 / 2 = -2
-> Slope of BC is 7 / 2 = 3.5
-> Nope!
So we have to tackle the altitude in two steps:
(1) Figure out the point D where the continuation of AB intersects a perpendicular line that runs through point C
(2) Take the distance between C and D
Because it runs through the origin, we can just write the equation of AB as y = -2x --> we know this describes our point D
We then can write the equation of CD.
--> The slope of CD is the negative reciprocal of the slope of AB
--> CD runs through C
So we get y - 7 = 0.5(x -2); we can plug in our equation from AB.
\(y-7 = 0.5(x-2)\)
\((-2x)-7 = 0.5(x-2)\)
\(-2x-7=0.5x-1\)
\(-6 = 2.5x\)
\(x = -6/2.5 = -12/5 = -2.4\)
We plug in again for y and get that y = 4.8.
Now we need the distance between C and D
--> C lies at (2, 7)
--> D lies at (-2.4, 4.8)
\(\sqrt{(2 - (-2.4))^2 + (7 - 4.8)^2}\)
\(=\sqrt{(4.4)^2 + (2.2)^2}\)
\(=\sqrt{19.36 + 4.84}\)
\(= \frac{1}{10} * \sqrt{1936 + 484}\)
\(= \frac{1}{10} * \sqrt{2420}\)
\(= \frac{{11 * 2} }{ 10} * \sqrt{5}\)
And we need the distance from A to B:
\(\sqrt{2^2 + 4^2}\)
\(=\sqrt{4 + 16}\)
\(=\sqrt{20}\)
\(= 2\sqrt{5}\)
Then we multiply the distance from A to B by the distance from C to D and then multiply that by 0.5
Area =
\(\frac{{11 * 2} }{ 10} * \sqrt{5} * 2\sqrt{5} * \frac{1}{2}\)
Area = 11
On the coordinate plane there are 3 points đŽ(â1, 1), đ”(1, â3), and đ¶(3, 4). What is the area of the triangle connecting the 3 points?
(A) 10
(B) 11
(C) 12
(D) 14
(E) 13
=================================
We want to take the area. Area is 0.5 * base * height -- where height is the altitude of the triangle.
I want AB to be the base of my triangle. This means that we want the altitude dropped from C to be our height.
To simplify things, I'm going to move the triangle such that B sits on the origin; this moves the triangle -1 in the x-direction and +3 in the y direction.
A -> (-2, 4)
B -> (0, 0)
C -> (2, 7)
Let's check if we're lucky and it's a right triangle.
-> Slope of AB is -4 / 2 = -2
-> Slope of BC is 7 / 2 = 3.5
-> Nope!
So we have to tackle the altitude in two steps:
(1) Figure out the point D where the continuation of AB intersects a perpendicular line that runs through point C
(2) Take the distance between C and D
Because it runs through the origin, we can just write the equation of AB as y = -2x --> we know this describes our point D
We then can write the equation of CD.
--> The slope of CD is the negative reciprocal of the slope of AB
--> CD runs through C
So we get y - 7 = 0.5(x -2); we can plug in our equation from AB.
\(y-7 = 0.5(x-2)\)
\((-2x)-7 = 0.5(x-2)\)
\(-2x-7=0.5x-1\)
\(-6 = 2.5x\)
\(x = -6/2.5 = -12/5 = -2.4\)
We plug in again for y and get that y = 4.8.
Now we need the distance between C and D
--> C lies at (2, 7)
--> D lies at (-2.4, 4.8)
\(\sqrt{(2 - (-2.4))^2 + (7 - 4.8)^2}\)
\(=\sqrt{(4.4)^2 + (2.2)^2}\)
\(=\sqrt{19.36 + 4.84}\)
\(= \frac{1}{10} * \sqrt{1936 + 484}\)
\(= \frac{1}{10} * \sqrt{2420}\)
\(= \frac{{11 * 2} }{ 10} * \sqrt{5}\)
And we need the distance from A to B:
\(\sqrt{2^2 + 4^2}\)
\(=\sqrt{4 + 16}\)
\(=\sqrt{20}\)
\(= 2\sqrt{5}\)
Then we multiply the distance from A to B by the distance from C to D and then multiply that by 0.5
Area =
\(\frac{{11 * 2} }{ 10} * \sqrt{5} * 2\sqrt{5} * \frac{1}{2}\)
Area = 11
22 Oct 2022, 23:28
Kudos
Bookmarks
On the examination, use the plastic notebook to write a huge coordinate system and pin point the points mentioned above.
Then, once you find the triangle, find out the area of the rectangle. What is mean is the complete length of the Y axis that the triangle takes up and the length of the X axis.
Now, the area comes up to be 28. 7 x 4.
When our triangle is enclosed within the rectangle, there are 3 other triangles formed. Each of them have easily measurable lenghts. Because they are parallel to the X and Y axis. Find the area of the triangles. 28 - 4 - 7 - 6 = 11.
Took me a little less than 3 minutes but as long as we figure out a solid way to find the correct answer... why not?
Then, once you find the triangle, find out the area of the rectangle. What is mean is the complete length of the Y axis that the triangle takes up and the length of the X axis.
Now, the area comes up to be 28. 7 x 4.
When our triangle is enclosed within the rectangle, there are 3 other triangles formed. Each of them have easily measurable lenghts. Because they are parallel to the X and Y axis. Find the area of the triangles. 28 - 4 - 7 - 6 = 11.
Took me a little less than 3 minutes but as long as we figure out a solid way to find the correct answer... why not?
