please help me with the math problem \[The\quad area\quad of\quad the\quad largest\quad square\quad which\quad can\quad be\\ inscribed\quad in\quad a\quad right\quad angled\quad triangle\quad with\quad legs\\ 4\quad units\quad and\quad 8\quad units\quad is\]

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TopNewestI used some simple analitical geometry and get the side of the square is \(\frac{8}{3}\) and thus the area is \(\left(\frac{8}{3}\right)^2=\frac{64}{9} \) – Poetri Sonya · 2 years ago

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how did you do it – Rajdeep Dhingra · 2 years ago

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– Poetri Sonya · 2 years ago

Just put the triangle in the Cartesian diagram with the coordinates are (0,0),(4,0) and(0,8)Log in to reply

– Rajdeep Dhingra · 2 years ago

But how to maximizeLog in to reply

– Poetri Sonya · 2 years ago

The largest square in the triangle is when the square's coordinates are \((0,0),(0,x),(x,0)\)and\((x,x)\) with \((x,x)\) lies in the hypotenuse .Log in to reply

– Rajdeep Dhingra · 2 years ago

soLog in to reply

– Rajdeep Dhingra · 2 years ago

Please could you give the full solutionLog in to reply

– Poetri Sonya · 2 years ago

Since \((x,x)\) lies in the hypotenuse, it lies in the line \(y=-2x+8\). Then \[x=-2x+8\] \[3x=8\] \[x=\frac{8}{3}\] So the side is \(\dfrac{8}{3}\)Log in to reply

– Rajdeep Dhingra · 2 years ago

Thanks! a lotLog in to reply

This the procedure to do this kind of a sum:

Let the side of the required square be \(a\).

Length of the remaining parts of the two legs are respectively \(8-a\) and \(4-a\).

So the total area which is \(\frac{1}{2}.a.(8-a)+a^{2}+\frac{1}{2}.a.(4-a)\) will be equal to \(16\) (the area of the whole triangle)

Solving the equation you get \(x=\frac{8}{3}\)

and hence the area is \(\frac{64}{9}\) – Anik Mandal · 2 years ago

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See this problem. – Calvin Lin Staff · 2 years ago

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@Calvin Lin – Rajdeep Dhingra · 2 years ago

Could you elaborate his solutionLog in to reply

– Calvin Lin Staff · 2 years ago

You should comment on his solution and ask him to clarify.Log in to reply

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– Soutrik Bandyopadhyay · 2 years ago

Sorry i misread the questionLog in to reply

– Rajdeep Dhingra · 2 years ago

could you tell me your approachLog in to reply