## anonymous one year ago Geometry Help! Please!

1. welshfella

The 4 triangles right angled formed are congruent Let the smallest side of these = x and let the area of shaded part equal y area of each triangle = 2* x/2 = x consider one of the large rectangles y + 2x = 6 consider the shaded part which is a paralelogram the area = base * perpendicular height

2. welshfella

area= BE * DE - x y = 2* (3 - x) y = 6 - 2x

3. welshfella

which is the same equation as we had before! how embarrassing!! I must be missing something here.....

4. welshfella

the area is a rhombus - so is there a formula for this in terms of the sides - or diagonals?

5. ganeshie8

Yes, here is an outline of solution step1 : show that all the four triangles outside the shaded region are congruent |dw:1440259209437:dw|

6. ganeshie8

step2 : By CPCTC, it follows that BG = BH = DH = DG, making BGDH a rhombus

7. amistre64

|dw:1440259500328:dw|

8. amistre64

solve for x x^2 + 4 = (3-x)^2

9. ganeshie8

step3 : $$\triangle OBG\sim\triangle ADB$$ by $$AA$$ similarity, so $$\dfrac{OG}{OB}=\dfrac{2}{3}$$ |dw:1440259468027:dw|

10. ganeshie8

you can find OB trivially using the fact that OB = BD/2 (diagonals bisect each other in rhombus)

11. welshfella

yes

12. welshfella

and the area of the rhombus = pq/2 where p and q are the diagonals right?

13. ganeshie8

yes, otherwise it is easy to see that all the four internal triangles are congruent : |dw:1440260040970:dw|

14. ganeshie8

$$A_1 = A_2 = A_3 = A_4$$

15. Loser66

Is it not that it is 6 -2x And xis calculated by pythagorean, x=5÷6

16. Loser66

Or I missed something?

17. ganeshie8

you might be correct, but i like similar triangles more :)

18. welshfella

amistre's method is correct also simple really

19. Loser66

Of course we need prove the 2 triangle are equal first

20. ganeshie8

why do you want to solve a quadratic when you can setup a simple proportion ?

21. amistre64

lol, its not a quadratic, its linear

22. anonymous

Picking up from @amistre64 $2^2 + x^2 = (3-x)^2$$x^2 + 4 = x^2-6x +9$$6x = 9-4$$x = \frac{ 5 }{ 6 }$The figure is also a parallelogram with height 2 and length 3-5/6 = 13/6.$A = bh = (3-\frac{5}{6})2 = \frac{13}{3}$

23. welshfella

once you know the area of the triangles thats it Yes the congruency proff is necessary

24. ganeshie8

it looked like a quadratic to me haha

25. welshfella

yea - but the x^2 cancel out

26. anonymous

Oh i get it now!

27. ganeshie8

i mean my brain felt it is a quadratic, so rejected initially ;p

28. anonymous

you all are so helpful thanks so much!

29. welshfella

yw

30. amistre64

one thing i aint, is a geometrer ;)

31. ganeshie8

looking at it again, i find it is simpler and kinda cute !

32. welshfella

interesting question...