eucledian geometry

Question

from facebook A square. Calculate the ratio of shaded area to the area of square

but one solution appear in the second fiqure used ratios could any one explained it geometrical thank you i try melanous and similarity without results any hints

Answer 1

One way to solve this problem is:-

(1) Apply co-ordinate geometry technique to find the co-ordinates of Z, which the intersection point of the lines OP and MN.

(2) Once we got that, we can determine the ratio of MZ : ZN (and OZ : ZP if necessary).

Here is another approach.

For simplicity, (1) we can assume it is a $60 \times 60$ square; (2) let the area of the smallest triangle = [OMZ] = x square units.

The areas of those triangles, that can easily be found, are directly marked in red.

[a] = 600 – (x); [b] = 300 – (x)

Further, [c] = 1800 – 300 – 600 – [b] = 900 – (300 – (x)) = 600 + (x)

Note that, $\dfrac {(x)}{[b]} = \dfrac {[a]}{[c]} $. That is, $\dfrac {(x)}{300 – (x)} = \dfrac {600 – (x)}{600 + (x)}$. Then, x = 120. Result follows

Answer 2

???Are you given the equations of the lines or the coordinates of their endpoints? How did you get the values in the second picture?

If we take the square to have corners at (0,0), (1,0), (1,1), and (0,1) it looks to me like the steeper line passes through (0,0) and (1/2,1) so has equation y= 2x. I think the other line passess through (0, 1/3) and (1, 2/3) so has equation y= 1/3(x+ 1).

The two lines cross where 2x= x/3+ 1/3. 5x/3= 1/3. x= 1/5 and then y= 2/5. Dropping a perpendicular from the point of intersection to the base, on the left we have a right triangle with base 1/5 and height 2/5 so area 1/2(1/5)(2/5)= 1/25. On the right we have a trapezoid with bases 2/3 and 2/5 and height 1- 1/5= 4/5. Its area is 1/2(2/3+ 2/5)(4/5)= 1/2(16/15)(4/5)= 32/75.

The total area is 1/25+ 32/75= 35/75= 7/15. Since we took the square to have area 1, the area of the shaded region is 7/15 of the area of the square.