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.
Apr 2512:30 AM EDT
-01:30 AM EDT
At one point, she believed GMAT wasn’t for her. After scoring 595, self-doubt crept in and she questioned her potential. But instead of quitting, she made the right strategic changes. The result? A remarkable comeback to 695. Check out how Saakshi did it.
Apr 2601:30 AM EDT
-02:30 AM EDT
Learn how Keshav, a Chartered Accountant, scored an impressive 705 on GMAT in just 30 days with GMATWhiz's expert guidance. In this video, he shares preparation tips and strategies that worked for him, including the mock, time management, and more.
Apr 2610:00 AM EDT
-11:00 AM EDT
The Target Test Prep course represents a quantum leap forward in GMAT preparation, a radical reinterpretation of the way that students should study. Try before you buy with a 5-day, full-access trial of the course for FREE!
Apr 2711:00 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
Apr 2808:00 AM PDT
-11:00 AM PDT
Whether you are just beginning your MBA journey or fine-tuning your path to a 700+ score, GMAT Day is designed to give you a competitive edge.
May 0111:00 AM PDT
-12:00 PM PDT
Free- full-length test + 15 concept videos + 150+ short videos + study plan!
25 Jun 2020, 23:49
Kudos
Bookmarks
D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
63% (02:59) correct
38%
(02:45)
wrong
based on 40
sessions
History
Date
Time
Result
Not Attempted Yet
Attachment:
2.jpg [ 14.82 KiB | Viewed 2187 times ]
On the xy-plane above, if the equation of \(l_{1}\) is \(y=\frac{1}{2}x\) and if point B is defined by the xy-coordinate pair (5,0), what is the area of OAB ?
(A) 4
(B) 3\(\sqrt{2}\)
(C) 2\(\sqrt{5}\)
(D) 5
(E) 7
26 Jun 2020, 00:36
Kudos
Bookmarks
Explanation:
Slope of line Ol1 is 1/2. Since angle at A is 90 degree, So slope of AB must be -2.
Draw a line from Pt.A towards OB , It will divide it by OB into 4 to 1. Height of Triangle OAB is 2.
Area = 1/2 x 2 x 5
=5
IMO-D
Slope of line Ol1 is 1/2. Since angle at A is 90 degree, So slope of AB must be -2.
Draw a line from Pt.A towards OB , It will divide it by OB into 4 to 1. Height of Triangle OAB is 2.
Area = 1/2 x 2 x 5
=5
IMO-D
28 Jun 2020, 06:00
Kudos
Bookmarks
Let us follow a more concept oriented approach for this sum. It may seem long, but it really isn't.
Area of the Triangle = ½ * Base * Height
The first aim is to get the coordinates of A, and then the lengths of OA and AB
If the base is AB, then the height is the perpendicular OA
We are given that O is (0,0), B is (5,0) and OA is the line y = 1/2x ----- (1) .
The slope of OA = m1 = 1/2
Since OA is perpendicular to AB, therefore the product of the slopes = -1 i.e m1 x m2 = -1
Therefore m2 = -2
In line AB, we use the slope point form, y - y1 = m(x - x1) to find the equation of the line
Slope = -2, and the line passes through (5,0). Therefore y – 0 = -2(x – 5)
Or y = -2x + 10 ----- (2)
Since OA and AB intersect, solving equations (1) and (2), we get the co-ordinates of A
y = ½ x = - 2x + 10
2.5x = 10
x = 4
y = 1/2x = 4/2 = 2
Therefore the co ordinates of A are (4,2)
Using the distance formula between 2 points A (x1,y1) and B (x2,y2) = Sqrt(x1 – x2)^2 + (y1 – y2)^2:
OA = Sqrt[(4 – 0)^2 + (2 – 0)^2] = Sqrt(20)
AB = Sqrt[(5 – 4)^2 + (0 – 2)^2] = Sqrt(5)
Area of Triangle AOB = ½ * sqrt(20) * sqrt(5) = ½ * sqrt(100) = ½ * 10 = 5
Area of the Triangle = ½ * Base * Height
The first aim is to get the coordinates of A, and then the lengths of OA and AB
If the base is AB, then the height is the perpendicular OA
We are given that O is (0,0), B is (5,0) and OA is the line y = 1/2x ----- (1) .
The slope of OA = m1 = 1/2
Since OA is perpendicular to AB, therefore the product of the slopes = -1 i.e m1 x m2 = -1
Therefore m2 = -2
In line AB, we use the slope point form, y - y1 = m(x - x1) to find the equation of the line
Slope = -2, and the line passes through (5,0). Therefore y – 0 = -2(x – 5)
Or y = -2x + 10 ----- (2)
Since OA and AB intersect, solving equations (1) and (2), we get the co-ordinates of A
y = ½ x = - 2x + 10
2.5x = 10
x = 4
y = 1/2x = 4/2 = 2
Therefore the co ordinates of A are (4,2)
Using the distance formula between 2 points A (x1,y1) and B (x2,y2) = Sqrt(x1 – x2)^2 + (y1 – y2)^2:
OA = Sqrt[(4 – 0)^2 + (2 – 0)^2] = Sqrt(20)
AB = Sqrt[(5 – 4)^2 + (0 – 2)^2] = Sqrt(5)
Area of Triangle AOB = ½ * sqrt(20) * sqrt(5) = ½ * sqrt(100) = ½ * 10 = 5
