Aptitude - Area - Discussion

Simple go for the formulas of Area of Square. There are two formulas:

1) Area of Square = (Sides)2.
2) Area of Square = 1/2 (Diagonals)2.

Let 'S' is a side of square we know that all the sides of a square is equal. By first Formula Area will be S^2. Now putting this S^2 in 2nd Formula. We got the diagonal.

By second formula:

S^2 = 1/2(Diagonal)^2.

Taking square root on both sides.

Square root of 2*S = Diagonal.

Now we have to find Diagonal.

Diagonal = 1.414S(Square root of 2 = 1.414).

So The difference between Diagonal and Edges is:

2S-1.414S = 0.59S.

Percentage saved not by walk on edges = Extra Walk/Total Walk*100.

= 0.59S/2S*100 = 30%.

If we go diagonal means a square cut in between gives a triangle
so a total 3 sides.
let's say 2 sides and 1 hypotenuse.

Let sides = x.
x + x = hypotenuse^2.
2x = hypotenuse^2.

This power 2 becomes root on the other side.
√ 2x is = hypotenuse.
√ 2x is 1.414x.

So now, in 2x we minus 1.414x gives us 0.59x // we do this because we calculate the total distance here.

The maximum distance was 2x but we traveled only 1.414x so 0.59 x distance was extra so we used that further.

The remaining distance/total distance to the percentage.
which is,
0.59X/2X * 100
= 30%.

Hope you got it.

You (Amit) choose the sides of a rectangle not SQUARE.

EXAMPLE METHOD.

According to Pythagoras FOR SQUARE the sides are 2 & 2 or diagonal 2*sqrt(2) = 2*1.141 = 2.828 {where sqrt(2) = 1.141}.

Edges are 2+2 then 4-2.828 = 1.172 (Difference).

(1.172/4)*100 = 29.3 = 30 approx.

Simple method:

Edge of square let be x.

Then diagonal of square will be 2 sqrt(x).

Edge distance - Diagonal distance (difference).

2x-2sqrt(x) = x (2-sqrt(2)) = (2-1.414)x = 0.586x.

%AGE.

(0.586x/2x)*100 = (0.586/2)*100 = 29.3 (30 approx).

Let we have a square. starting from left top to right A,B,C,D.

After that make a diagonal AC.

As we know that we don't know the side of square then we assume x.

Now we start to run from A to C then path is AB+BC and its value,
is x+x = 2x.

Then equation is AB+BC = 2x.

Now the diagonal of the square formula is,
d = square root of 2 x.
d = 1.41x.

Saving per cent is ((2x-1.41x)/2x)*100 .

29.5. approx 30.

We have to find saving distance from A to C.
First, from edges the distance we have to calculate.
So let's take side of a square as X
So for A to C.
We have to cross A to B then B to C.
So, X + X = 2X.
If we travel diagonally means the distance took is root of 2 multiplied by X.
So, saving will b 2X - root 2 X.
So (2.00 - 1.41) X = 0.59 X.
Saving % is 0.59X/2X * 100 = 30%.

let us assume the side of square is 10 cm.
In a square all sides are equal.

To find the diagonal of a square we can use the following formula= Squareroot of (2*(side^2)).

In this case= Square root of (2*(10^2)) = 14.142.
Distance Traveled=14.142.
Distance if used the edge of the square= 10*2= 20.
Saving=20-14.142=5.858.
Saving in %=5.858/20 =30%.

@Riya.

Name the sqaure as ABCD. The person is moving from A to C (diagonally) instead of AB + BC (along the edges). As we can see the souce is point A and the destination is point C.

We actually need to find how much distance (edges) he saved by chosing to move from displacement (daigonally) aprt.

@All.

Here is the solution.

Let side of squares be "5" on alll sides.
so area=5*5 = 25,
so: AB2 = 5*5 =25,
AD2 = 5*5 =25.

pythogorus: AC2 = AB2+BC2.
AC2 = 25+25.
AC = root(50).
AC = 7.07.

Now,
= AC/areax100.
= 7.07/25 x100 = 28.48~ 30 ans.

Suppose ab = 2, ac = 2 then ab+ac=4.

Diagonal bc = ?

Pythagoras Proof bc^2 = ac^2+ab^2 (here ^2 means square).
bc^2= 4+4.
bc=(square root of 8).
bc= 2.83.

Gain = 4-2.83 = 1.17.

Gain(%) = (1.17/4)*100 = 29.28%.

Lets say sides of square is 10 and let diagonal be x therefore by Pythagoras theorem.

X2 = (10) 2 + (10) 2 = 200.

X = 10root2 = 10 x 14.1 (root2 = 1.412).

Therefore 20 - 14.1 = 5.9.

5.9/20 = 0.30 i.e. approx 30%.