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

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The sides of a quadrilateral taken in order are 16,16,14 and [#permalink] New post 07 Mar 2004, 12:25
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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...
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Last edited by sunniboy007 on 07 Mar 2004, 13:23, edited 1 time in total.
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 [#permalink] New post 07 Mar 2004, 13:04
I dont know how to solve this without knowing the other angles.
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 [#permalink] New post 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.
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 [#permalink] New post 07 Mar 2004, 13:24
B it is.......... :-D !!
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!! :punk
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 [#permalink] New post 07 Mar 2004, 13:30
:yikes :pray cbrf3
WOuld you like to take the GMAT for me? :lol:
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 [#permalink] New post 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?
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 [#permalink] New post 07 Mar 2004, 20:47
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
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 [#permalink] New post 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

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 [#permalink] New post 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.
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  [#permalink] 07 Mar 2004, 21:45
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