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My problem is : Given a triangle with base 55in and height 10in, find the dimensions of the largest possible inscribed rectangle, where one of the sides of the rectangle lies on the base of the triangle.

So I know how to solve this if it is equilateral or even if it is a right triangle. It's all about finding similar triangles and using what you know to solve for the missing rectangle dimensions in terms of one variable.

I'm confused for this example because it is not specifying if it is a right triangle or what. It is just a triangle and I can't find any useful information to help me find dimensions of my proportional triangles.

Any tips on how to set this up would be greatly appreciated.

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    $\begingroup$ The answer isn't determined, it can be as small as you like. Picture the triangle with base on the $x-$axis between $0$ and $55$ and third vertex somewhere on the line $y=10$. If the third vertex is far away, at $(N, 10)$ for huge $N$, there is nearly no room to fit a rectangle. $\endgroup$
    – lulu
    Commented Sep 1, 2023 at 22:22
  • $\begingroup$ @lulu sowe can't solve it unless we know what type of triangle it is, right? $\endgroup$
    – k12345
    Commented Sep 1, 2023 at 22:24
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    $\begingroup$ Well, you need to know more than you have provided, in any case. $\endgroup$
    – lulu
    Commented Sep 1, 2023 at 22:25
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    $\begingroup$ Maybe they wanted the smallest area which you can guarantee fitting into such a triangle. In that case, the answer is $0$. $\endgroup$
    – lulu
    Commented Sep 1, 2023 at 22:33
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    $\begingroup$ You could go for the triangle(s) which allow the largest possible largest rectangle subject to the constraints in question $\endgroup$
    – Henry
    Commented Sep 1, 2023 at 23:00

1 Answer 1

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enter image description here

Suppose hight of triangle $CD=h$ and $AB=c$. we draw $A_1B_1$ parallel with AB. $C_1$ and $D_1$ are projections of $B_1$ and $A_1$ respectively. Let $A_1B_1=x$ and $B_1C_1=y$, so the area of rectangle $A_1B_1C_1D_1$ is:

$S=xy$

Triangles $CA_1B_!$ and $CAB$ are similar, so we have:

$\frac xc=\frac{h-y}h\Rightarrow x=\frac{c(h-y)}h$

Therefore:

$S=\frac ch(h-y)y$

we want S maximum; sincs $h-y+y=h$ is constant for a given triangle, then the product $(h-y)y$ is maximum if $h-y=y$ which gives $y=\frac 12 h$.In this way maximum area is:

$S=\frac 14 c h$

As can be seen S maximum is a function of c and h. Now we have $h=10$ and $c=55$, so maximum area is:

$S=\frac 14\times 55\times 10=137.5$

$y=\frac 12 \times 10=5$

so :

$x=\frac{137.5}5=27.5$

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