square root of the area of Trapezoid- Pls help

This topic has expert replies
Senior | Next Rank: 100 Posts
Posts: 62
Joined: Wed May 04, 2011 9:50 pm
Thanked: 2 times
Followed by:2 members
In a rectangular co-ordinate system, what is the square root of the area of a trapezoid whose vertices have the co-ordinates (2,2), (2,3),(20,2),(20,-2)?

Answer choices are .: a)7 b) 9 c) 10.22, d) 12.25 e) 14

correct answer choice is B-9

can anyone explain me how to solve this problem in some easy steps?
Source: — Problem Solving |

User avatar
Senior | Next Rank: 100 Posts
Posts: 35
Joined: Sun Jul 31, 2011 9:39 pm
Thanked: 2 times
Followed by:1 members

by Elena89 » Mon Dec 12, 2011 7:20 am
kishokbabu wrote:In a rectangular co-ordinate system, what is the square root of the area of a trapezoid whose vertices have the co-ordinates (2,2), (2,3),(20,2),(20,-2)?

Answer choices are .: a)7 b) 9 c) 10.22, d) 12.25 e) 14

correct answer choice is B-9

can anyone explain me how to solve this problem in some easy steps?
Here's what you do:

First draw the trapezoid in the xy-plane according to the four ordered pairs.

You'll get a figure like the one attached. Take the two parallel sides as the two bases. And draw the perpendicular from one parallel side to the other. This is the height. Now just plug in values in the formula for area of a trapezoid [(Base 1 + Base 2)/2]*Height:

[(4+5)/2]*18 = 81

And root of 81 = 9

And so the answer is option B) 9
Attachments
trapezoid.JPG
Last edited by Elena89 on Mon Dec 12, 2011 9:39 am, edited 1 time in total.

Newbie | Next Rank: 10 Posts
Posts: 3
Joined: Mon Dec 05, 2011 11:11 am

by cpschott » Mon Dec 12, 2011 9:11 am
image is misleading, line h should intersect two vertices- (2,2) and (20,2)

then i have one triangle with area =1X18/2
and the second with area =4X18/2

18/2 + 72/2= 45
sqrt(45) = <7

what gives-- would need points at (2,2); (2,6); (20,2); (20,-3) to get 9

Master | Next Rank: 500 Posts
Posts: 385
Joined: Fri Sep 23, 2011 9:02 pm
Thanked: 62 times
Followed by:6 members

by user123321 » Mon Dec 12, 2011 9:13 am
i think area should be 45 with those coordinates.
i am not sure where i did mistake. can you check the coordinates once again?

user123321
Just started my preparation :D
Want to do it right the first time.

User avatar
Senior | Next Rank: 100 Posts
Posts: 35
Joined: Sun Jul 31, 2011 9:39 pm
Thanked: 2 times
Followed by:1 members

by Elena89 » Mon Dec 12, 2011 9:44 am
my bad! with these coordinates, the area is indeed 45. :?

User avatar
GMAT Instructor
Posts: 15539
Joined: Tue May 25, 2010 12:04 pm
Location: New York, NY
Thanked: 13060 times
Followed by:1906 members
GMAT Score:790

by GMATGuruNY » Mon Dec 12, 2011 10:44 am
kishokbabu wrote:In a rectangular co-ordinate system, what is the square root of the area of a trapezoid whose vertices have the co-ordinates (2,2), (2,3),(20,2),(20,-2)?

Answer choices are .: a)7 b) 9 c) 10.22, d) 12.25 e) 14

correct answer choice is B-9

can anyone explain me how to solve this problem in some easy steps?
The problem has been transcribed incorrectly.
The intended coordinates are (2,-2), (2,3), (20,2) and (20,-2).

Thus:
b1 = 3-(-2) = 5.
b2 = 2-(-2) = 4.
h = 20-2 = 18.
Area = (b1 + b2)/2 * h = (5+4)/2 * 18 = 81.
√81 = 9.

The correct answer is B.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3