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### Chittagong-Higher secondary 2016

Posted: Mon Dec 24, 2018 12:56 pm
The lengths of the sides of the rectangle ABCD are 10 and 11. An equilateral triangle is drawn in such a way that no point is situated outside ABCD. The maximum area of the triangle can be expressed as (pāq)/r where p, q and r are positive integers and q is not divisible by any square number. p+q+r=?

I have determined the maximum area of the triangle (221ā3 - 400).But the matter of irony is that the area isn't in the form as mentioned in the question.

### Re: Chittagong-Higher secondary 2016

Posted: Mon Feb 18, 2019 4:28 pm
mizan_24 wrote: ā
Mon Dec 24, 2018 12:56 pm
The lengths of the sides of the rectangle $ABCD$ are $10$ and $11$. An equilateral triangle is drawn in such a way that no point is situated outside ABCD. The maximum area of the triangle can be expressed as $\dfrac{p\sqrt q}{r}$ [where $p, q$ and $r$ are positive integers and $q$ is not divisible by any square number. $p+q+r=?$

I have determined the maximum area of the triangle $(221\sqrt 3 - 400)$.But the matter of irony is that the area isn't in the form as mentioned in the question.
What is the question number?

### Re: Chittagong-Higher secondary 2016

Posted: Thu Oct 31, 2019 7:26 pm
$(221\sqrt 3 - 400)$ equals to a negative value. So your solution is wrong.

Go back to the drawing boards!

### Re: Chittagong-Higher secondary 2016

Posted: Thu Oct 31, 2019 7:56 pm
The solution is as such=>

Let $ABCD$ be a rectangle where $AB$ is $10$ and $BC$ is $11$

Take the base of an isosceles triangle to be $BC$. Let the longer sides be $x$.

So,

$x^2=(5\frac{1}{2})^2 + 100$
$x^2= 130\frac {1}{4}$

Therefore, $x > 11$

Now, it is confirmed that $BC$ can be used as the base of our required equilateral triangle.

Therefore, area of the equilateral triangle=>

$\frac{1}{2} \times 11 \times 11 \times sin (60)= \frac {121\sqrt3}{4}$

Hence, $p= 121, q=3, r=4$ which gives the final answer $p+q+r= 121+3+4= 128.$