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Consider a $3$ x $3$ square grid and consider the following arrangement:-enter image description here

Find the number of triangles of area $3$ $unit^2$ that can be formed using the points marked above as vertices.

PS:- Because of the limitation on the area( $3$ $unit^2$) I do not know how to do this problem. I tried taking the total number of triangle and subtracting the triangle with area not equal to $3$ $unit^2$. But the same problem occurs. The problem I am encountering is that - I do know how to solve the question for $area = 3 unit^2$

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    in this question, length and width of each small square is 1, am i right? – 2017-01-24
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    @Kiran Yes. the length of the sides of each small square is 1. – 2017-01-24

2 Answers 2

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To build off of @Kiran answer, but to correct the mistake (and perhaps give a slightly different approach), the final answer should in fact be 16.

In order for a triangle, with integer length sides, to have an area of $3$ you must have that $b \times h \times \frac{1}{2} = 3$ or that $b \times h = 6$. This is the motivation for looking at rectangles of area $6$. The possible rectangle dimensions (ignoring the chart for a moment) would be $(1,6), (2,3), (3,2), (6,1)$. Since the rectangle must fit into the table drawn, the first and last options are no longer feasible, and so you must be able to take a rectangle of either $(3,2)$ or $(2,3)$.

If the vertices on such a rectangle are labelled (clockwise) $ABCD$ then you can form two identical triangles by connecting either $AC$ or $BD$. Thus, from each rectangle we can obtain $4$ identical triangles!

enter image description here

So, the question becomes how many $2\times3$ or $3\times2$ rectangles exist in the chart? This is a much easier question to answer, and can be enumerated rather easily by simple inspection. By starting with the first obvious rectangle, and then simply transitioning it sideways, then rotating and doing the same we realize that there are exactly $4$ such rectangles.

enter image description here

Given this and the first fact (that there are $4$ triangles per rectangle, we get the answer that $16$ triangles exist!

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First count the number of rectangles that can be formed with area $6$ unit$^2$ such that each rectangle can give $4$ triangles of required type.

i.e, we need to count the possible ways of getting $6$ adjacent squares first. Horizontally and vertically, we can select $2$ blocks of $6$ adjacent squares. i.e., there are $4$ ways in which adjacent $6$ square blocks can be selected.

Each of this selection can give four triangles of ares $3$ unit$^2$.

So, required count of triangles is $4 \times 4 = 16$

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    I disagree. Each rectangle can give 4 triangles of the correct size ! – 2017-01-24
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    @Dylan, corrected the mistake. your answer is right, i believe. – 2017-01-25