Find the ratio of areas of :
1>A triangle inscribed in a square inscribed in a circle.
raoatchut191:
so ur brain is the bk how did u get that ans
diagonal=diameter and side=base . So now,
take the radius as r cm
so u considered it as base i took the square base=triangle base
square base ?? how can a square have a base?
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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.
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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
THEN BY 1/2BH IT SHOULD BE 1/2*ROOT2^2
OK now what's the area of the triangle?
You there?
KKK UNDERSTOOD THNQ VERY VERY VERY MUCH FOR SPENDING THIS MUCH TIME TO CLEAR MY DOUBT
You sure did put me into a lot of trouble
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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 = R²
Area of ΔDCE = 1/2 DC * AD = 1/2 AB² = R²
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.
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 = R²
Area of ΔDCE = 1/2 DC * AD = 1/2 AB² = R²
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.
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hope this is easily understood
Whoa! My answer fades away in comparison, please delete mine
this can b done in a very easy manner , far more easier than the above un
finding of maximum area triangle in a square takes some steps. i wonder if there are simpler answers. that will be good to know
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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