please continue in this new thread for all the geometry related discussions.
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semi perimeter = 84+x
use the formula, r=(area of triangle)/(semi perimeter)
we get x = 42.
so length of smaller side = 36+42 = 78cm.
area of the trapezium = x(a+x). Substituting 'a' we get <...
if the perimeter is 40cm. what can be the maximum area of the trapezium
now there can be only one square in the equilateral triangle. Let its side length be 'x'.
clearly tan(60) = x/ = 2x/(a-x)
So, on solving we get x = 3r*(2-sqrt(3)).
One of the diagonals would be of length r. Now, as the diagonals of a rhombus are perpendicular bisectors of each other, the other diagonal has to be of length 2 * sqrt (r^2 - r^2/4) i.e. * r
now, area of a rhombus is...
Options are a. 64 b. 8m c. 32m d. 46m.
How to approach this problem....
=> Radius of in-circle = 8/(3)
Side of an equilateral triangle inscribed in circle of radius r = r3
=> Side of inscribed equilateral triangle = (8/3)*3 = 8
=> Area = (3/4)*8*8 = 163
probably the q is asking fr area of the inner equilateral triangle.then only the answer follows...
because each perpendicular cut +increase in 2 plane circular area
therefore total increase in area would be=6/4*100=150%
hope it's clr....:)
radius of circle=r
height of each of the 6 equilateral triangles formed within the hexagon (joinng cntr of the circle and vertices of hexagon)=root3/2*a=r
thus the conclusion follows..
1.If a circle is inscribed in a regular hexagon hw to arrive to conc dat side of the hexagon is :
2r/root 3 (r is d radius of the circle inscribed)
Ans:(2/3) -(root3/2 pie)
Q2.An Equilateral Triangle side 16 cms has circle inscribed in it.There is anoder equilateral tria...