[clsuresh]

Here comes our next series of questions

1. A SQUARE IS ROTATED ABOUT AT IT'S CENTRE THROUGH AN ANGLE OF 45 degrees. FIND THE AREA COMMON TO THE TWO POSITIONS OF THE SQUARE.

2. ABCD IS A PARALLELOGRAM. THE ANGULAR BISECTOR OF ANGLE ABC INTERSECTS AD AT P. IF DP =5 cms, CP = 6cms AND BP = 6 cms FIND AB?

regards

[andy_ravi]

[ QUOTE=clsuresh]Hi, Thakarsagar

I think that's not the right answer but I think u got into the right way of thinking. Try on the same lines. All the best and I appreciate u for answering my previous question asking the highest power of 3 in 1! x 2! x 3! x 4! x .......... x 98! x 99! x 100![/QUOTE]

[thakarsagar]

Originally Posted by clsuresh

Here comes our next series of questions

1. A SQUARE IS ROTATED ABOUT AT IT'S CENTRE THROUGH AN ANGLE OF 45 degrees. FIND THE AREA COMMON TO THE TWO POSITIONS OF THE SQUARE.

2. ABCD IS A PARALLELOGRAM. THE ANGULAR BISECTOR OF ANGLE ABC INTERSECTS AD AT P. IF DP =5 cms, CP = 6cms AND BP = 6 cms FIND AB?

regards

1. Already solved in the previous thread.... sqrt(2) - 1 if side is of unit length....
-sagar

[Anurag83]

2.AB = 10 cms.

Am i correct

Originally Posted by thakarsagar

1. Already solved in the previous thread.... sqrt(2) - 1 if side is of unit length....
-sagar

[ankit.3k]

[
2) AB=4 cm?????

plz confirm.


QUOTE=clsuresh]Here comes our next series of questions

1. A SQUARE IS ROTATED ABOUT AT IT'S CENTRE THROUGH AN ANGLE OF 45 degrees. FIND THE AREA COMMON TO THE TWO POSITIONS OF THE SQUARE.

2. ABCD IS A PARALLELOGRAM. THE ANGULAR BISECTOR OF ANGLE ABC INTERSECTS AD AT P. IF DP =5 cms, CP = 6cms AND BP = 6 cms FIND AB?

regards[/QUOTE]

[thakarsagar]

Originally Posted by ankit.3k

[
2) AB=4 cm?????

plz confirm.


QUOTE=clsuresh]Here comes our next series of questions

1. A SQUARE IS ROTATED ABOUT AT IT'S CENTRE THROUGH AN ANGLE OF 45 degrees. FIND THE AREA COMMON TO THE TWO POSITIONS OF THE SQUARE.

2. ABCD IS A PARALLELOGRAM. THE ANGULAR BISECTOR OF ANGLE ABC INTERSECTS AD AT P. IF DP =5 cms, CP = 6cms AND BP = 6 cms FIND AB?

regards

[/QUOTE]


Correct....
solution:

Angle ABP = Angle PBC = Angle BCP = Angle CPD = Angle BPA = Theta (say)
hence, AB = AP = x ....
using cosine rule in Triangle PDC we have,
Cos(theta) = (36 + 25 - x^2)/60....
using cosine rule in Triangle PBC we have,
Cos(theta) = (36 + (x+5)^2 - 36) / 2 * 6 * (x+5)
Soving both of them we get x = 4...
-sagar

[con-CAT-nation]

Hi all!
I am probably the newest member. Been going through the thread lately.


sagar, i dont think that the last square rotation problem and this one are identical. According to my calculation this is coming out to be 2(a^2)/(sqrt(2)+1) where a is the side of the initial square.... i may be wrong

Suresh ... wonderful job..

thanks
concat.

[Devil13]

Hi suresh

I think the answer to the first one will be:-
assume side = a
area 0f sqaure =a^2
after rotation 4 trainges will be uncomon to both the squares
each side will be divided into 3 ..
thus area of all four triangles = 1/2 * a/3 * a/3 * 4
common area = a^2 - 2a^2/9
= 7a^2/9

I hope this is right....my first post on PG

[thakarsagar]

Originally Posted by con-CAT-nation

Hi all!
I am probably the newest member. Been going through the thread lately.


sagar, i dont think that the last square rotation problem and this one are identical. According to my calculation this is coming out to be 2(a^2)/(sqrt(2)+1) where a is the side of the initial square.... i may be wrong

Suresh ... wonderful job..

thanks
concat.

yes dude,
ur right...i thought it was the same questions as before...Haste is bad....
If the side of the sqaure is 1,
Area common = 1 - 4 * 1/2 * 1/3 * 1/3
= 7/9....
-sagar

[con-CAT-nation]

doesnt this solution give rise to right angled equilateral triangles?