|
Author |
Message |
|
TAGS:
|
|
|
Senior Manager
Joined: 30 Aug 2003
Posts: 335
Location: BACARDIVILLE
Followers: 1
Kudos [?]:
0
[0], given: 0
|
The sides of a quadrilateral taken in order are 16,16,14 and [#permalink]
07 Mar 2004, 13:25
Question Stats:
0% (00:00) correct
0% (00:00) wrong based on 0 sessions
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, 14:23, edited 1 time in total.
|
|
|
|
|
|
|
SVP
Joined: 30 Oct 2003
Posts: 1963
Location: NewJersey USA
Followers: 3
Kudos [?]:
25
[0], given: 0
|
I dont know how to solve this without knowing the other angles.
|
|
|
|
|
|
Senior Manager
Joined: 30 Aug 2003
Posts: 335
Location: BACARDIVILLE
Followers: 1
Kudos [?]:
0
[0], given: 0
|
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: 417
Location: USA
Followers: 1
Kudos [?]:
1
[0], given: 0
|
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: 335
Location: BACARDIVILLE
Followers: 1
Kudos [?]:
0
[0], given: 0
|
 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: 716
Followers: 2
Kudos [?]:
4
[0], given: 0
|
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: 4441
Followers: 10
Kudos [?]:
81
[0], given: 0
|
kpadma wrote: 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: 376
Location: India
Followers: 1
Kudos [?]:
0
[0], given: 0
|
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: 4441
Followers: 10
Kudos [?]:
81
[0], given: 0
|
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
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|