Area of the triangle

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gmattesttaker2: Legendary Member; Posts: 641; Joined: Tue Feb 14, 2012 3:52 pmFollowed by: 8 members; Thanked: 11 times

by gmattesttaker2 » Sun Oct 13, 2013 6:08 pm

Hello,

I was wondering if you can please assist with the following question:

A triangle in the xy-coordinate plane has vertices with coordinates (7, 0), (0, 8), and (20, 10). What is the area of this triangle?

OA: 87

Sorry if my question is trivial but I was just wondering why I cannot calculate the area of the triangle here as 1/2 x sq. root 269 x sq. root 113 ? Thanks for your help - Sri

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Source: Beat The GMAT — Problem Solving | Sun Oct 13, 2013 6:08 pm

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16206; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

by Brent@GMATPrepNow » Sun Oct 13, 2013 6:14 pm

Hi Sri,

You are assuming that this is a right triangle. Are you sure it is?

There's another approach you haven't considered.
Hint: Try drawing a rectangle around the triangle.
Another hint here: https://www.beatthegmat.com/coordinate-p ... 63268.html

Cheers,
Brent

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theCodeToGMAT: Legendary Member; Posts: 1555; Joined: Tue Aug 14, 2012 11:18 pmFollowed by: 34 members; Thanked: 448 times; GMAT Score:650

by theCodeToGMAT » Sun Oct 13, 2013 9:03 pm

Area of Trapezium - Area of Triangle 1 - Area of Triangle 2
= 1/2 (20) (8+10)) - 1/2 (7)(8) - 1/2 (13)(10)
= 180 - 28 - 65
= 87

R A H U L

Faheem.Mustafa: Newbie | Next Rank: 10 Posts; Posts: 2; Joined: Sun Oct 13, 2013 8:38 pm

by Faheem.Mustafa » Sun Oct 13, 2013 9:29 pm

gmattesttaker2 wrote:Hello,

I was wondering if you can please assist with the following question:

A triangle in the xy-coordinate plane has vertices with coordinates (7, 0), (0, 8), and (20, 10). What is the area of this triangle?

OA: 87

Sorry if my question is trivial but I was just wondering why I cannot calculate the area of the triangle here as 1/2 x sq. root 269 x sq. root 113 ? Thanks for your help - Sri

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lunarpower: GMAT Instructor; Posts: 3380; Joined: Mon Mar 03, 2008 1:20 amFollowed by: 1535 members; Thanked: 2256 times; GMAT Score:800

by lunarpower » Mon Oct 14, 2013 3:02 am

Hey, I wrote this problem!

If I were 11 years old, I would just draw a 20 x 10 rectangle around everything. Then I could subtract out the three areas I don't want, which, conveniently enough, are all right triangles.

Ron has been teaching various standardized tests for 20 years.

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lunarpower: GMAT Instructor; Posts: 3380; Joined: Mon Mar 03, 2008 1:20 amFollowed by: 1535 members; Thanked: 2256 times; GMAT Score:800

by lunarpower » Mon Oct 14, 2013 3:05 am

To the original poster:

gmattesttaker2 wrote:Sorry if my question is trivial but I was just wondering why I cannot calculate the area of the triangle here as 1/2 x sq. root 269 x sq. root 113 ? Thanks for your help - Sri

Ironically, the answer to this question is in the diagram that you provided yourself.

Take a look at the "rise" and "run" values that you put in your own diagram. The slope of the left-hand leg is -8/7; the slope of the right-hand leg is 10/13.
These aren't perpendicular. (The slope perpendicular to -8/7 is 7/8; the slope perpendicular to 10/13 is -13/10.)

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