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Find the ratio of areas of :

1>A triangle inscribed in a square inscribed in a circle.
square base means side of it
Answer is Π:2:1 if base of triangle is side of square.
You're right AcharyaVII.
question should be the MAXIMUM area triangle in a square.


This answer is true ,if you consider that base of triangle lies on side of square. Let the radius of circle be r, then the diagonal=diameter i.e. side of square, s= base length of Δ=height of triangle=√2r. Then ratio of areas of circle, square, triangle = Πr²:2r²:r² Hence, answer is Π:2:1
0 0 0
No problem. I am way happier than troubled.
jackie, it helps t o draw diagrams. for you as well as for other readers.
Sir, but is going kill my time for answering other questions
There are 46 points for this!Come on

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See diagram.

Let the radius of the circle be R.   Area = π R²

The maximum square ABCD that can be inscribed in the circle is when the diameter of the circle is the diagonal (AC or BD) of the square. 

Hence,  AC = BD = √2 AB = 2 R   =>   AB = √2 R.

Area of square = 2 R²

Regarding the triangle inscribed inside the square, there are possibilities.  Let us examine the triangles right angle ΔDCB,  isosceles Δ DCE and the isosceles Δ AFG.

Area of ΔDCB = 1/2 DC*BC = 1/2 AB² =  2 R²/2 =

Area of ΔDCE = 1/2 DC * AD = 1/2 AB² =

Area of Δ AFG:
   We see that it is half of the product of base FG and altitude from A on to FG.  Let us move the point F from B towards C and at the same time move G from D towards C.   If the altitude increases by an amount "h", then the base decreases by "2h" due to the linearity of the geometry.

Hence, the area of Δ AFG, will be less than the area of ABD.  

Maximum area of a triangle inscribed in the square = R²

Hence the ratios = π R² : 2 R² : R²   = π : 2 : 1.

1 5 1
thanx n u r welcom
For the area of triangle AFG to be maximum , F and G have to lie on the vertices of the square.Then why have u not taken it ?
thanx Acharya. and u r welcom
There are three types of inscribed triangles possible. AFG, DCB, DCE. so i compared the maximum areas of these. when F and G lie on vertices of square, then AFG becomes like DCB or DCE.
same thing , right?
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