Question

100 point,
[tex}Maths question[/tex]

The number of points having both co-ordinates as a integers, that lies interior of a triangle with vertices (0,0) , (0,41) and (41,0) ;

a) 901

b) 861

c) 820

d) 780
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Answer it with solution.

Answer 1

D 780 is u answer  

4

down vote

The problem can be approached by finding the area of the triangle in two ways:

First, as A=12(412)=16812

Second, by an application of Pick's theorem A=i+b2−1

Now boundary points b can be calculated by considering that there is a single origin point (0,0) common to both cathetuses, 40 points unique to each of the two cathetuses, and 42 points along the hypotenuse (inclusive of the end points where it intersects with the cathetuses).

So b=1+2(40)+42=123, giving A=i+1212

Equating the two expressions for A, we get the number of interior points i=780.

Answer 2

Method - 1:

From figure:

⇒ y = -x + 41

⇒ x + y = 41 {x > 0 , y > 0}

⇒ (40,1).

Integral points on line x = 1 is 39.

Integral points on line x = 2 is 38.

⇒ 39,38,37,.....1

We know that sum of n terms = n(n + 1)/2

                                                 = 39(39 + 1)/2

                                                 = 1560/2

                                                 = 780.

Therefore, answer is 780 - option(D).

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Method - 2:

Let us assume that the points on the side of triangle are not inside it., we have x = 1, the point goes inside from y = 1 and y = 39.  We should not include y = 40 because it is on the line itself, not inside.

The next possible value is x = 38, the possible values are y = 1 and y = 37.

So, we can add up the numbers.

39 + 38 + 37+....1

We know that sum of n natural terms = n(n + 1)/2

⇒ 39(39 + 1)/2

⇒ 1560/2

⇒ 780.

Therefore, answer is 780.

Hope this helps!