For T, the end points are (0,0,1),(0,1,0) and (1,0,0). Hence side of equilateral triangle lying on axes is rt(2).

Therefore area of Equi triangle = rt(3)/4 * (side)^2 = rt(3)/4 * (rt2)^2 = rt(3).

Similarly for S the end points are (1/2,0,0), (0,1/3,0), (0,0,1/6).

Now find the three sides by distance formula and find the area by Heron's formula.

Ans. 56.25 days or 56 & 1/4th days

Actually, after 15 days, 200 men had provisions for 45 days.

Since 40 men left then, (200 - 40) = 160 men had provisions for 45*200/160 days = 56.25 days.

See the figure

Answer:

According to Pythagoras theorem,the area of a  square drawn on the hypotenuse of a right angled triangle equals the sum of the areas of the squares drawn on the other two sides.The sides of  right triangled triangles  bear an integral ratio and are known as Pythagorean triples.Therfore any square ,the length of whose  side(element of N) is the hypotenuse of a right angled triangle  can be dissected into smaller squares (no two of which are of same size).

The Pythagorean triples can be generated by the relation (m^2 -n^2), 2mn, (m^2+n^2),where m&n are elements of N.  Thus any square whose side length can be expressed as the sum of two square  numbers  or simple multiples of such side  lengths  can be dissected into smaller squares, of which no two squares are of same size.

Starting with 5 we can see that 5 is the sum of 2 square numbers 1&4(5= 1+4) and so 5^2 =  3^2+4^2. The next possible number is 10 which is 1+9 and so we can express 10^2 = 6^2+8^2. The next number is 13 which is 4+9 and so 13^2 =5^2+12^2.The next number in sequence is 15.15 cannot be expressed as the sum of two square numbers but it is a multiple of 5,which is sum of two square numbers.Therefore 15^2 = 9^2 +12^2. Thus you can generate side lengths of all squares which can be dissected into smaller squares of which no two squares are of  the same size. Following are the side lengths of squares between N= 1 to 60 that can be dissected into smaller squares (of which no two of them are of the same size) :  5, 10,13,15,17,20,25,26,29,30,34,35,37,40,41,45,50,51,52,53,55,58,60

A cube can also be dissected into a finite number of smaller cubes out of which no two cubes are of the same size.For example 

6^3 = 3^3+4^3 + 5^3.

9^3 =  1^3+6^3 +8^3

13^3 = 1^3 + 5^3 + 7^3 +12^3

14^3 =  2^3 +3^3 + 8^3 + 13^3

However, unlike squares, no formula is available to generate all such cubes. They have to be found  by trial & error only.

Further all squares having a Square number (N) as their lengths can be dissected into N number of smaller identical sized squares.For example a square of side length 4 units can be dissected  into 4 squares of side length of 2 units. That is 4^2  =   4*(2^2) = 16 . Similarly a square of side length 9 units can be dissected into 9 squares of side  length 3 units. That is  9^2  =  9*(3^2) =81 and so on

Similarly all cubes having a  Cube number(N) as their side length can be dissected into N smaller identical sized cubes. For example a cube of size 8 units  can be dissected into 8 smaller cubes of side length 4 units. That is 8^3  = 8*(4^3) = 512.Similarly a cube of side length 27 units can be dissected into 27 cubes of side length 9 units. That is 27^3  =  27*(9^3) = 19683 and so on.

Answer:  It is given that 30x^2 + 517  =  3 x^2*y^2 + y^2 ,that is  

y^2( 3x^2+1)  =   30x^2 +517 .Therefore   y^2 = ( 30x^2+517)/ (3x^2 +1)

=  (30x^2+10+507)/ (3x^2 +1)  =  { (30x^2+10)/ (3x^2+1) + 507/(3x^2+1)}=

10(3x^2+1)/(3x^2+1) + 507/(3x^2+1)   = 10+{507/(3x^2+1)}

Therefore y^2 =  10+ {507/(3x^2+1)}   or

y^2 -  10  =  507/(3x^2+1).   —-----------------------A        

As x and y both are integers,both LHS and RHS are also integers

 Hence it follows that 3x^2+1 should be a factor of 507 so as to make

 RHS integers. The only factors of 507 are 3,13,39,169&507.

Hence  the value of 3x^2+1 should be any one of them.

Putting 3x^2+1 = 3 gives x^2 =2 which makes x irrational and cannot be a

valid  solution. Putting 3x^2+1 = 13 gives 3X^2  =12 or x^2 =4 and so 

X =   +/- 2 which is a valid solution .

Other factors like 39,169,  507 will also gives irrational values to x 

and so not valid solutions.

Hence x = +/- 2 is the only valid solution for the given  equation.

Now putting x =2 in equation A ,we get  y^2 -  10  =  507/13 = 39 or 

Y^2 = 49 and therefore  y  = +/- 7

Hence X^2+y^2 = 4+49 =   53

In terms of radio buttons we may select the button for "5" in first row and '3" in second row so as to

chose the answer 53.

In case of Square we have  Length =Breadth.   We  assume Length or breadth =a =Side of the Square. Then the perimeter of square =2(length+Breadth)= 2*2*Length/Breadth =4* side of the square=4a.

So 1/4 th of perimeter= 1/4* 4a=a = Side of square.

Total selling price,SP= 504+504= Rs 1008 Gain after selling one toy= 12% Selling Price, SP of the toy=(1+.12) Cost Price,CP or, CP=SP/1.12=504/1.12=450 Rs Now loss after selling another toy=4% So,SP=(1-.04)*CP or, CP=SP/0.96=504/0.96=525Rs. So,total cost price,CP= 450+525= 975 Rs So, total profit %= (1008-975)/975*100=33/975 *100=44/13= 3 5/13 (ans)

S is second to the right of Q in English alphabet.

( 2 + 2 ) / ( 2 * 2  - 2 ) = 4 / ( 4 - 2 ) = 4 / 2 = 2

Case 1 : Exactly one digit repeated twice --

eg AABCD     Words= 10C1 X 9C3 X {5!/2!} =50400

Case 2 : Exactly one digit repeated thrice --

eg AAABC     Words= 10C1 X 9C2 X {5!/3!}= 7200

Case 3 : Exactly one digit repeated four times --

eg AAAAB     Words= 10C1 X 9C1 X {5!/4!}= 450

Case 4 : Exactly one digit repeated five times --

eg AAAAA     Words= 10C1 X {5!/5!}= 10

Case 5 : Exactly two digits repeated two times each --

eg AABBC     Words= 10C2 X 8C1 X {5!/(2! 2!)}= 10800

Case 6 : Exactly two digits repeated one of them twice and the other thrice --

eg AAABB     Words= 10C1 X 9C1 X {5!/(2! 3!)}= 900

FINAL ANSWER= Sum of all cases= 69760