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