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# pls help Math question

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

Note by Rajdeep Dhingra
2 years, 4 months ago

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I used some simple analitical geometry and get the side of the square is $$\frac{8}{3}$$ and thus the area is $$\left(\frac{8}{3}\right)^2=\frac{64}{9}$$ · 2 years, 4 months ago

yes

how did you do it · 2 years, 4 months ago

Just put the triangle in the Cartesian diagram with the coordinates are (0,0),(4,0) and(0,8) · 2 years, 4 months ago

But how to maximize · 2 years, 4 months 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 . · 2 years, 4 months ago

so · 2 years, 4 months ago

Please could you give the full solution · 2 years, 4 months 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 $$\dfrac{8}{3}$$ · 2 years, 4 months ago

Thanks! a lot · 2 years, 4 months ago

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 $$\frac{64}{9}$$ · 2 years, 4 months ago

See this problem. Staff · 2 years, 4 months ago

Could you elaborate his solution @Calvin Lin · 2 years, 4 months ago

You should comment on his solution and ask him to clarify. Staff · 2 years, 4 months ago

Comment deleted Jan 08, 2015

Comment deleted Jan 08, 2015

Sorry i misread the question · 2 years, 4 months ago