The sides of a quadrilateral taken in order are 16,16,14 and : PS Archive
Check GMAT Club App Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 06 Dec 2016, 10:55

### 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

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# The sides of a quadrilateral taken in order are 16,16,14 and

Author Message
Senior Manager
Joined: 30 Aug 2003
Posts: 329
Location: BACARDIVILLE
Followers: 1

Kudos [?]: 18 [0], given: 0

The sides of a quadrilateral taken in order are 16,16,14 and [#permalink]

### Show Tags

07 Mar 2004, 12:25
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions

### HideShow timer Statistics

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

The sides of a quadrilateral taken in order are 16,16,14 and 10. The angle contained by the frist two sides is 60 degrees. What is teh area of the quadrilateral?

12(sqrt252)
104(sqrt3)
16(sqrt3)(4+sqrt15)
14(sqrt3)+5
what the...
_________________

Pls include reasoning along with all answer posts.
****GMAT Loco****
Este examen me conduce jodiendo loco

Last edited by sunniboy007 on 07 Mar 2004, 13:23, edited 1 time in total.
SVP
Joined: 30 Oct 2003
Posts: 1793
Location: NewJersey USA
Followers: 5

Kudos [?]: 97 [0], given: 0

### Show Tags

07 Mar 2004, 13:04
I dont know how to solve this without knowing the other angles.
Senior Manager
Joined: 30 Aug 2003
Posts: 329
Location: BACARDIVILLE
Followers: 1

Kudos [?]: 18 [0], given: 0

### Show Tags

07 Mar 2004, 13:07
yes, it's a tough bugger!
I will give others a chance to answer it before I post the the official answer and reasoning.
_________________

Pls include reasoning along with all answer posts.
****GMAT Loco****
Este examen me conduce jodiendo loco

Senior Manager
Joined: 05 Feb 2004
Posts: 290
Location: USA
Followers: 1

Kudos [?]: 5 [0], given: 0

### Show Tags

07 Mar 2004, 13:24
B it is.......... !!
here's the soln:

1st 2 sides are 16 each and they form angle = 60 degrees
So If u draw a diagonal the triangle that contains diagonal and these 2 sides will be equilateral, with sides 16.
Therefore area of Triangle = (sqrt3/4)*16*16 =64*sqrt3
For the other triangles sides are=16,14,10
Hence its area = 40sqrt3
Area of quad = sum of 2 triangles = 104 sqrt3!!
Senior Manager
Joined: 30 Aug 2003
Posts: 329
Location: BACARDIVILLE
Followers: 1

Kudos [?]: 18 [0], given: 0

### Show Tags

07 Mar 2004, 13:30
cbrf3
WOuld you like to take the GMAT for me?
_________________

Pls include reasoning along with all answer posts.
****GMAT Loco****
Este examen me conduce jodiendo loco

Director
Joined: 03 Jul 2003
Posts: 652
Followers: 3

Kudos [?]: 91 [0], given: 0

### Show Tags

07 Mar 2004, 19:57
cbrf3 wrote:
For the other triangles sides are=16,14,10
Hence its area = 40sqrt3

What is the formula to find the area from the length of sides?
GMAT Club Legend
Joined: 15 Dec 2003
Posts: 4302
Followers: 38

Kudos [?]: 418 [0], given: 0

### Show Tags

07 Mar 2004, 20:47
cbrf3 wrote:
For the other triangles sides are=16,14,10
Hence its area = 40sqrt3

What is the formula to find the area from the length of sides?

This requires 2 simple pythagorean formulas with 2 unknown. The workout is a bit longer than the simple answer posted by cbrf3 though
We can let base (b) = 16
The height of the second triangle will separate that triangle into 2 right triangles. Let X be the length of the shorter side of the base and (16-X) be the length of the longer side of the base.
h^2 = 10^2 - X^2
h^2 = 14^2 - (16-X)^2
100 - X^2 = 196 - (256-32X+X^2)
100 - X^2 = -60 + 32X - X^2
32X = 160 --> X = 5
Height would be equal to:
h^2 = 10^2 - 5^2
= 100 - 25 = 75
h = 5sqrt3
Area of second triangle is b*h/2 = 16*5sqrt3 / 2 = 40sqrt3
_________________

Best Regards,

Paul

Senior Manager
Joined: 06 Dec 2003
Posts: 366
Location: India
Followers: 1

Kudos [?]: 11 [0], given: 0

### Show Tags

07 Mar 2004, 21:32
Quote:
For the other triangles sides are=16,14,10
Hence its area = 40sqrt3

What is the formula to find the area from the length of sides?

Paul and KPADMA, the formula to find area of a triangle from measures of its sides is:

A = SQRT of (s*(s-a)*(s-b)*(s-c)) ; where s=1/2 (perimeter of triangle)

Hope, this will help

Dharmin
GMAT Club Legend
Joined: 15 Dec 2003
Posts: 4302
Followers: 38

Kudos [?]: 418 [0], given: 0

### Show Tags

07 Mar 2004, 21:45
Dharmin wrote:
Quote:
For the other triangles sides are=16,14,10
Hence its area = 40sqrt3

What is the formula to find the area from the length of sides?

Paul and KPADMA, the formula to find area of a triangle from measures of its sides is:

A = SQRT of (s*(s-a)*(s-b)*(s-c)) ; where s=1/2 (perimeter of triangle)

Hope, this will help

Dharmin

Awesome Dharmin. It sure simplifies things.
_________________

Best Regards,

Paul

Display posts from previous: Sort by