Coordinate Geometry

Expert replies

kuiper: Junior | Next Rank: 30 Posts; Posts: 24; Joined: Fri Aug 27, 2010 3:27 pm; Thanked: 1 times

by kuiper » Fri Dec 03, 2010 5:06 pm

https://www.postimage.org/image.php?v=PqPnlgJ

Question on this image along with answer choices. I actually answered [spoiler]C - 10sqrt(2)[/spoiler]

I am using the approach - Area of Triangle - 1/2 * b * h

The base of the triangle can be figured out as 5sqrt(2) given the two coordinates of the base and using the distance formula.

The height of the triangle is - 4? (Drawn from point R to x-axis)

And therefore - [spoiler]10sqrt(2) [/spoiler]- which is incorrect

Seems like I might be miscalculating the height but I am not sure why that is incorrect?

I appreciate any help.

Like this post

Source: Beat The GMAT — Problem Solving | Fri Dec 03, 2010 5:06 pm

Night reader: Legendary Member; Posts: 1337; Joined: Sat Dec 27, 2008 6:29 pmFollowed by: 10 members; Thanked: 127 times

by Night reader » Fri Dec 03, 2010 5:45 pm

kuiper wrote:https://www.postimage.org/image.php?v=PqPnlgJ

Question on this image along with answer choices. I actually answered [spoiler]C - 10sqrt(2)[/spoiler]

I am using the approach - Area of Triangle - 1/2 * b * h

The base of the triangle can be figured out as 5sqrt(2) given the two coordinates of the base and using the distance formula.

The height of the triangle is - 4? (Drawn from point R to x-axis)

And therefore - [spoiler]10sqrt(2) [/spoiler]- which is incorrect

Seems like I might be miscalculating the height but I am not sure why that is incorrect?

I appreciate any help.

Unfortunately your answer is not correct; even if the figures are not drawn to the scale, the answer is not correct.

Look you don't need tedious calculations here besides finding the area of three triangles and one quadrilateral.

Area triangle EFB = (4*3)/2 = 6
Area triangle DFC = (3*4)/2 = 6
Area triangle ADE = (7*1)/2 = 3.5

Area qdrl. ABCD = 7*4 = 28

Area of triangle PQR = Area ABCD - Areas (EFB, DFC, ADE) or 28-6-6-3.5 = 12.5

Answer A.

Attachments

Last edited by Night reader on Fri Dec 03, 2010 5:57 pm, edited 1 time in total.

Like this post

kuiper: Junior | Next Rank: 30 Posts; Posts: 24; Joined: Fri Aug 27, 2010 3:27 pm; Thanked: 1 times

by kuiper » Fri Dec 03, 2010 5:53 pm

No, you are absolutely correct. The answer in the sample image was not correct either.

All I am wondering is - why is the height of the triangle in the image not 4? - (per my post above)

Like this post

Night reader: Legendary Member; Posts: 1337; Joined: Sat Dec 27, 2008 6:29 pmFollowed by: 10 members; Thanked: 127 times

by Night reader » Fri Dec 03, 2010 5:56 pm

kuiper wrote:No, you are absolutely correct. The answer in the sample image was not correct either.

All I am wondering is - why is the height of the triangle in the image not 4? - (per my post above)

Why you are supplying us with the incorrect answer selected on the GMAT Prep then?!

you mentioned different answer of yours...

please quit your question, for you were demonstrated the easiest way how to solve this problem

Like this post

Night reader: Legendary Member; Posts: 1337; Joined: Sat Dec 27, 2008 6:29 pmFollowed by: 10 members; Thanked: 127 times

by Night reader » Fri Dec 03, 2010 6:04 pm

kuiper wrote:https://www.postimage.org/image.php?v=PqPnlgJ

Question on this image along with answer choices. I actually answered [spoiler]C - 10sqrt(2)[/spoiler]

I am using the approach - Area of Triangle - 1/2 * b * h

The base of the triangle can be figured out as 5sqrt(2) given the two coordinates of the base and using the distance formula.

The height of the triangle is - 4? (Drawn from point R to x-axis)

And therefore - [spoiler]10sqrt(2) [/spoiler]- which is incorrect

Seems like I might be miscalculating the height but I am not sure why that is incorrect?

I appreciate any help.

I Don't understand 5sqrt(2) ?

we may not assume the PQR is a right triangle, also we are not given any angle at the base (but we can calculate these).

sqrt(2) coefficient is usual per the right triangles of which the non-hypotenuse sides are equal; this speeds up calculations.

Like this post

beat_gmat_09: Master | Next Rank: 500 Posts; Posts: 437; Joined: Sat Nov 22, 2008 5:06 amFollowed by: 1 member; Location: India; Thanked: 50 times; GMAT Score:580

by beat_gmat_09 » Fri Dec 03, 2010 7:44 pm

OPQ is right angled at origin with sides 3,4 hypotenuse PQ=5 from pythagorean triplets.
PR is a hypotenuse formed if you drop a perpendicular from R to x-axis
This forms another pythagorean triplet, with PR=5
Similarly RQ is one side of triangle formed when perpendicular from R is dropped on Q (y-axis)
RQ=7 and in this triangle another length is 1
Hypotenuse RQ^2 = 7^2 + 1^2 = 50
RQ = Sqrt(50)
PQ= 5, RP=5 from RQ it can be found out that RPQ is right angled at P.
Area of PQR = 1/2(5)(5) = 12.5

Hope is the dream of a man awake

Like this post

Night reader: Legendary Member; Posts: 1337; Joined: Sat Dec 27, 2008 6:29 pmFollowed by: 10 members; Thanked: 127 times

by Night reader » Fri Dec 03, 2010 7:58 pm

beat_gmat_09 wrote: RQ = Sqrt(50)
PQ= 5, RP=5 from RQ it can be found out that RPQ is right angled at P.

the coordinates of the point R could be skewed to new coordinates (4<x<7; 0<y<4) this will invalidate your solution;
that 5^2 and 5^2 should make Sqrt(50) coincidence as long as your assumption of a right triangle - the values could not make up hypotenuse...

Like this post

beat_gmat_09: Master | Next Rank: 500 Posts; Posts: 437; Joined: Sat Nov 22, 2008 5:06 amFollowed by: 1 member; Location: India; Thanked: 50 times; GMAT Score:580

by beat_gmat_09 » Fri Dec 03, 2010 8:10 pm

x+y=90 in triangle OPE 3-4-5 triplet
x+y=90 in triangle PCR 3-4-5 triplet
Angle F = 180 - (x+y) = 180 - 90 = 90
Triangle PQR is right angled at P.

Hope is the dream of a man awake

Like this post

Night reader: Legendary Member; Posts: 1337; Joined: Sat Dec 27, 2008 6:29 pmFollowed by: 10 members; Thanked: 127 times

by Night reader » Fri Dec 03, 2010 8:25 pm

You may have skewed R, different sides (not 3 and 4) x and x are not equal, y and y are not equal, in your example what makes them relative is 3 and 4 in two triangles {3;4}, {4;3}

this solution is unique and works only for this problem...

Attachments

Like this post

beat_gmat_09: Master | Next Rank: 500 Posts; Posts: 437; Joined: Sat Nov 22, 2008 5:06 amFollowed by: 1 member; Location: India; Thanked: 50 times; GMAT Score:580

by beat_gmat_09 » Fri Dec 03, 2010 8:32 pm

If you cut triangle RPC rotate it 90 degrees clockwise and place it on triangle POQ you'll
see the triangles are exactly same, their sides and their angles (common angle 90 degree and angles x and y).
Do you still disagree ?

Hope is the dream of a man awake

Like this post

Night reader: Legendary Member; Posts: 1337; Joined: Sat Dec 27, 2008 6:29 pmFollowed by: 10 members; Thanked: 127 times

by Night reader » Fri Dec 03, 2010 8:36 pm

beat_gmat_09 wrote:If you cut triangle RPC rotate it 90 degrees clockwise and place it on triangle POQ you'll
see the triangles are exactly same, their sides and their angles (common angle 90 degree and angles x and y).
Do you still disagree ?

I don't disagree, your solution is unique to this problem with the given set of conditions. If an aim is to solve the problem - you nailed it; while this solution would not be useful if R is skewed.

By principle, the first solution is more handy; not because it's mine - but for its universality

Like this post

beat_gmat_09: Master | Next Rank: 500 Posts; Posts: 437; Joined: Sat Nov 22, 2008 5:06 amFollowed by: 1 member; Location: India; Thanked: 50 times; GMAT Score:580

by beat_gmat_09 » Fri Dec 03, 2010 8:41 pm

Either way.
Whichever works in 2 minutes

Hope is the dream of a man awake

Like this post

Post new topic Post Reply

• Page 1 of 1

GMAT Course Reviews

563 reviews

1168 reviews

431 reviews

359 reviews

147 reviews

5/5

5 Star (563 Reviews)

"Target Test Prep is the closest to the official version of the GMAT exam, about 99% accuracy in terms of the quality and quantity of information. The course has excellently created singular sets of focused lessons and tests for every possible topic that one could come across in the official GMAT exam."

"The TTP course maximizes the efficiency of the time you spend studying. It will take time and effort but I could almost guarantee that if you complete the course exactly as it is laid out you will get an amazing score. They also have a very responsive team willing to help with any questions you might have."

"TTP has two things that I think no other test prep company offers: A teaching approach that reinforces understanding and an attitude that will give you the mental preparedness needed to succeed on the test. TTP gives you a deep understanding of the concept you need to know while teaching you how to think."

Write a Review