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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
use the formula, r=(area of triangle)/(semi perimeter)
we get x = 42.
so length of smaller side = 36+42 = 78cm.

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)

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.

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

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

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

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

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)).

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?

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)

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....

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

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

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...

Q3.A solid Sphere is cut into 8 identical pieces by three mutually perpendicular cuts .By what percentage is the sum of total surface areas of 8 pieces more than total surface area of original sphere?

Detailed explanationsss plzzz in dis question?

Ans:150%

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....:)

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

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 triangle which is inscribed in circle.Find Area of inscribed circle?

Ans:16 root3


Q3.A solid Sphere is cut into 8 identical pieces by three mutually perpendicular cuts .By what percentage is the sum of total surface areas of 8 pieces more than total surface area of original sphere?

Detailed explanationsss plzzz in dis question?

Ans:150%

Hi...
with respect to figure(attached)

AE=DR

3^2-x^2=2^2-y^2
similerly

EB= 4^2-x^2
EB=RC
So

PC^2=y^2+RC^2
=y^2+4^2-x^2

by using 1st one

PC=squareroot(11)

So D is correct answer.
Let me know if you get it...

thankss a lot saar

ABCD is a rectangle where P is a point lying inside it. PA = 3 cm, PB = 4 cm , PD = 2 cm, What is the value of PC = ?
(a) 10 cm (b) 5 cm (c) 8/3 (d) 11


Please help

originally posted by evensense in Official Quant Thread for CAT 2011

Messy_19 Says

Is the answer (d) 11??

bhai please post your approach, I dont know the OA.

Hi...
with respect to figure(attached)

AE=DR

3^2-x^2=2^2-y^2
similerly

EB= 4^2-x^2
EB=RC
So

PC^2=y^2+RC^2
=y^2+4^2-x^2

by using 1st one

PC=squareroot(11)

So D is correct answer.
Let me know if you get it...

ABCD is a rectangle where P is a point lying inside it. PA = 3 cm, PB = 4 cm , PD = 2 cm, What is the value of PC = ?
(a) 10 cm (b) 5 cm (c) 8/3 (d) 11


Please help

originally posted by evensense in Official Quant Thread for CAT 2011

Is the answer (d) 11??

ABCD is a rectangle where P is a point lying inside it. PA = 3 cm, PB = 4 cm , PD = 2 cm, What is the value of PC = ?
(a) 10 cm (b) 5 cm (c) 8/3 (d) 11


Please help

originally posted by evensense in Official Quant Thread for CAT 2011

Cyclic quadrilateral thry plzz.............And how to calculate area of a Pentagon any specific formula for area of a polygon?

Moreover how to calculate the side of a hexagon in which circle is inscribed

New Question::
In triangle ABC,D is on side BC such that BD: DC is 2:3 and E is on side AC such that AE:EC is 1:2.Lines AD and BE intersect at F. Find the ratio of area of triangle AFE and BFD.


Can any body provide the solution for this question??

Quadrilateral ADEC is a cyclic quadrilateral because sum of opp angles is 180,so angle EDC =angle EAC (ANGLES IN THE SAME SEGMENT ARE EQUAL)
hence angle EDC= 20
correct me if i m wrong.

Yes, you got it right ! Its 20.
:)

I did not actually read the question properly , I though they are squares. :splat:

Okay .
See The outer surface area for discreet smaller piecex is 1/4*4*pi*r^2+3*1/4pi*r^2
It has one curved surface (the first part) and three plane surface each same as 1/4th of a circle .
So,

For r=4,
16pi+12pi=26 pi
for the bigger one it's 36pi+27pi=63 pi
89 pi
But one plane side of smaller piece is attached to the bigger one so 1/4*pi*r^2=4pi wasted from each one so 89-8 =81 pi

I could reach at 81 pi so I think may be I did not get the picture correctly

I figured out the solution.
See, if you cut the sphere, then visualise 1/4th of the sphere. One part would be curved with the surface area of 1/4 times of a sphere, and the other two parts would be semicircles.
So, for a piece like that, total surface area = 1/4 (4*pi*r^2) + 2 * 1/2 (pi*r^2) = 2*pi*r^2.

So, as u see in the above solution, we are subtracting the areas of one of the semicircles of the smaller sphere. This is so, because, smaller sphere is joined with the bigger sphere, i.e., one of the semicircles of smaller sphere is joined to that of the bigger sphere.

Hope u get it. If not, then let me know.

See the first equation and try to visualize (may be with an apple )
It has 3 surfaces two plane and one curved.
So here the curved area is 1/4*4*pi*r^2 and the plane areas are 1/2*pi*r^2
So total is 2pi*r^2
Total is 2*pi*(16+36)=104*pi

Now when we fuse them the one surface of the smaller one disappeared which is 1/2*pi*4^2 and same amount of surface from the bigger one also disappears as it got attached to the smaller part.
So 8pi+8pi out of 104 pi
Answer=88 pi

I did a blunder first taking as circle and then taking 1/8th for the planed surface

I got the solution, but was unable to comprehend it. Hope u wud be able to help me. Below is the solution :

@pkaman- we were busy to explain it to each other while we both got that

subhakimi Says

I think it's 1/4*pi(r1^2+r2^2)=1/4*pi*(16+36)=13*pi

I got the solution, but was unable to comprehend it. Hope u wud be able to help me. Below is the solution :

subhakimi Says

I think it's 1/4*pi(r1^2+r2^2)=1/4*pi*(16+36)=13*pi

Just as I mentioned above, OA is 88 sq. units

dreamsohini Says

total surface area=74 pi

I think it's 1/4*pi(r1^2+r2^2)=1/4*pi*(16+36)=13*pi

dreamsohini Says

total surface area=74 pi

Sorry, the OA is 88 sq. units

One-fourth portion of each of the two solid spheres having radii 4 units and 6 units are joined together such that the centres of the two spheres coincide as shown in the figure given below. What is the total surface area of the solid thus formed?

total surface area=74 pi

Thanks a lot .. Gotcha !!!

A circle is divided into 12 equal parts by the points A1, A2, A3, A4, ..... A10, A11 and A12 as shown in the figure. Find the ratio of the measure of angles A1PA4 to A3QA4.

Attached is the image.

SOL:
To find A1PA4:
join A1 and A4
let the centre is O
NOW, angle A4OA9=(30*5)=150
Therefore angle A4A1A9 = 75(angle subtended by arc at the centre is double the angle subtended by the arc in the remaining part of circle)
SIMILARLY, angle A1A4A12=1/2 OF Angle A1OA12 =15
Thus from triangle A1PA4,
angle A1PA4=(180-75-15)=90
To find A3QA4
join A3 and A4
Angle A8A3A4=1/2 of angle A8OA4=1/2*120=60
AND angle QA4A3=45
THUS from triangle QA4A3,A4QA3=(180-60-45)=75
THEREFORE required ratio=90:75=6:5
hope it's clear

dreamsohini Says

Ratio is 18:15 ?

Yep.. u are correct.. answer is 6 : 5.
Could you please share the approach ... ?

One-fourth portion of each of the two solid spheres having radii 4 units and 6 units are joined together such that the centres of the two spheres coincide as shown in the figure given below. What is the total surface area of the solid thus formed?

A circle is divided into 12 equal parts by the points A1, A2, A3, A4, ..... A10, A11 and A12 as shown in the figure. Find the ratio of the measure of angles A1PA4 to A3QA4.

Attached is the image.

Ratio is 18:15 ?

A circle is divided into 12 equal parts by the points A1, A2, A3, A4, ..... A10, A11 and A12 as shown in the figure. Find the ratio of the measure of angles A1PA4 to A3QA4.

Attached is the image.