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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
3 years 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 $$\left(\frac{8}{3}\right)^2=\frac{64}{9}$$

- 3 years ago

yes

how did you do it

- 3 years ago

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

- 3 years ago

But how to maximize

- 3 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 .

- 3 years ago

so

- 3 years ago

Please could you give the full solution

- 3 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 $$\dfrac{8}{3}$$

- 3 years ago

Thanks! a lot

- 3 years 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 $$\frac{64}{9}$$

- 3 years ago

See this problem.

Staff - 3 years ago

Could you elaborate his solution @Calvin Lin

- 3 years ago

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

Staff - 3 years ago

Comment deleted Jan 08, 2015

Comment deleted Jan 08, 2015

could you tell me your approach

- 3 years ago