Question 1171116
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Andrew was given a rectangular cardboard 28cm by 16cm. How many right angle triangles 4cm high base 3cm could cut out.
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<pre>
Two right angled triangles with the legs of 3 cm and 4 cm, placed hypotenuse to hypotenuse, form a 3x4-rectangle.


Let's calculate how many such rectangles can be placed onto the 28x16 cm cardboard.


We have two basic placements.


One placement is to direct the 4-cm side of the rectangle along its 28 cm dimension.

By doing this way, we have 28/4 = 7 rectanles in this direction and, OBVIOSLY, 5 rectangles in the perpendicular direction,
                                                     (because 16/3 = 5.33 = 5 when rounded to the closest smaller integer).


In all, we have 7*5 = 35 rectangles and, hence, 35*2 = 70 right angled triangles at such placement.




Next, consider another placement, directing the 4-cm side side along the 16-cm side of the cardboard.

We have then the 3-cm side along the 28-cm side of the cardboard.

By doing this way, we have 4 rectangles along the 16-cm side of the cardboard and, OBVIOUSLY, 9 rectangles along its 28 cm side 
                                                               (because 28/3 = 9.33 = 9 rounded to the closest smaller integer).


In all, we have 4*9 = 36 rectangles and, hence, 36*2 = 72 right angled triangles at such placement.



Of these two opportunities, we chose the placement, which gives maximum numer of rectangles (36) and maximum number of triangles (72).


<U>ANSWER</U>.  Maximum number of triangles is 72.
</pre>

Solved.


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Such problems teachers give to young &nbsp;(advanced ?) &nbsp;students of &nbsp;5th - 6th &nbsp;grades to check if they are able 
to think on self-standing basis and to teach them to solve such problems accurately &nbsp;(and to think accurately) 
on the self-standing manner.


The solution by &nbsp;@MathLover1, &nbsp;based on consideration the areas &nbsp;ONLY, &nbsp;gives an &nbsp;ESTIMATION &nbsp;ONLY 
for the maximum number of triangles from the top, &nbsp;but does not give the exact number of triangles.


The solution by &nbsp;@MathLover1 &nbsp;IS &nbsp;NOT &nbsp;what is expected.


Learn on how to solve such problems &nbsp;(and how to teach your students, &nbsp;if you are a teacher) - from my post.