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  • Created by @bhars18
closing this old thread. please continue in this new thread for all the geometry related discussions. http://www.pagalguy.com/forum/quantitative-questions-and-answers/69681-geometry-for-cat-2011-a.html
closing this old thread.
please continue in this new thread for all the geometry related discussions.
http://www.pagalguy.com/forum/quantitative-questions-and-answers/69681-geometry-for-cat-2011-a.html
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> speedthestar Says > > 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. the other two sides are 36+x and 48+x. semi perimeter = 84+x us...
speedthestar Says
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.


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.
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> 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 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 n...
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


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
area = 20*x-. Differentiate area w.r.t 'x' to get x= 5*(sqrt(2))
so area = 50*sqrt(2)
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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.
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.
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> 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 draw the line OB. We get 2 equilateral triangle OAB and OCB NOW supose the radiu...
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

draw the line OB. We get 2 equilateral triangle OAB and OCB
NOW supose the radius is r
area of rmbs = 2(rt3/4*rsqr) =32*sqrt3
r=4
ans
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> DJ Harry Says > > 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? ans is (63-36root3)rsquire
DJ Harry Says
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?


ans is (63-36root3)rsquire
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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
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
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> Junta help solving these* 1.* ABC is a triangle, D lies on the side BC and E lies on the side AC. AE = 3, EC = 1, CD = 2, DB = 5, AB = 8. AD and BE meet at P. The line parallel to AC through P meets AB at Q, and the line parallel to BC through P meets AB at R. Find area PQR/area ABC. [i...
Junta help solving these

1.
ABC is a triangle, D lies on the side BC and E lies on the side AC. AE = 3, EC = 1, CD = 2, DB = 5, AB = 8. AD and BE meet at P. The line parallel to AC through P meets AB at Q, and the line parallel to BC through P meets AB at R. Find area PQR/area ABC.


2.
Triangle APM has A = 90o and perimeter 152. A circle center O (on AP) has radius 19 and touches AM at A and PM at T. Find OP.


please tell me solution
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> DJ Harry Says > > 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 triangle = a = r*sqrt(3) now there can be only one square in the equilateral triangle. Let its side length be 'x'. ...
DJ Harry Says
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 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)).
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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?
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?
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> 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.... ...
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....


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 1/2 * (product of the lengths of the diagonals)

32 * sqrt (3) = 1/2 * r * * r

r^2 = 64

So, radius would be 8 m.

Option (B)
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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....
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....
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> 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 How to reach "Side of ...
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



How to reach "Side of inscribed equilateral triangle = (8/3)*3 = 8"

means side = root3*r
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> Q2.An Equilateral Triangle side 16 cms has circle inscribed in it.There is anoder equilateral triangle which is inscribed in circle.Find Area of inscribed circle? Ans:16 root3 > I think there is a typo error in the q probably the q is asking fr area of the inner equilateral triangl...
Q2.An Equilateral Triangle side 16 cms has circle inscribed in it.There is anoder equilateral triangle which is inscribed in circle.Find Area of inscribed circle?

Ans:16 root3

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...


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
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