Geometry Quantitative

What is the area of a parallelogram with an angle 45 degrees, height 4 cm and a diagonal 5 cm?
P.S: Please post your answers in the thread only after 6 PM., April 25 2003.
10 9
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the other two sides are 36+x and 48+x.
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.
2 1
Let distance between the two parallel sides be 'x' and let one of the sides be 'a', then the other parallel side is x+a+x = a+2x. The two non-parallel sides will be x*sqrt(2). So (as perimeter = 40) 2x+2a+(2*sqrt(2)*x) = 40
=> x+a=20
area of the trapezium = x(a+x). Substituting 'a' we get <...
The radius of an incircle of a triangle is 24cm and the segments in which one side is divided by the point of contact are 36cm and 48cm. Find the length of the smaller of two sides of the triangle.
ans is (63-36root3)rsquire
ABCD is a trapezium such that abcd. Angle A and B are equal and 45 degree.
if the perimeter is 40cm. what can be the maximum area of the trapezium
please tell me solution
Side of the triangle = a = r*sqrt(3)
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)).
1 1
Q. ABC is a equilateral triangle inscribed in a circle of radius r. What is the area of the largest square that can be inscribed inside it?
Side of the rhombus is equal to the radius of the circle. Let this be r.
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...
2 2
A rhombus OABC is drawn inside a circle, whose centre is at O, in such a way that vertices A,B,C of the rhombus are on the circle. If the area of the rhombus is 32* sqrt(3) cm. Find the radius of the circle.
Options are a. 64 b. 8m c. 32m d. 46m.
How to approach this problem....
How to reach "Side of inscribed equilateral triangle = (8/3)*3 = 8"
means side = root3*r
Radius of in-circle of an equilateral triangle of side a is a/(23)
=> 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
1 1
I think there is a typo error in the q
probably the q is asking fr area of the inner equilateral triangle.then only the answer follows...
after cutting the sphere into 8 pieces the area increases by=(2pi r^2)*3=6pi r^2
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....:)
side of hexagon=a
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..
And about hexagon guyzzz
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)
Solutionss plzz
Q1.2 Circles intersect each oder .The Circumfrence of each passes thro centre of d other.What part of the area of each circle is the area of thier intersecting region?
Ans:(2/3) -(root3/2 pie)
Q2.An Equilateral Triangle side 16 cms has circle inscribed in it.There is anoder equilateral tria...