## anonymous 5 years ago length of a rectangle is 20 meter and its are is 320 squaremeters .find the area of the square drawn on the diagonal of a square whose perimeter equals that of the given rectangle.

1. anonymous

need help..

2. anonymous

still here?

3. anonymous

yups

4. anonymous

5. anonymous

You need to find z in this set up given the info. you have.

6. anonymous

You can find z by finding x, and you find x using the information you have about the rectangle.

7. anonymous

The perimeter of the big square is equal to the perimeter of the rectangle.

8. anonymous

If the rectangle has area 320m^2 and one side 20m, then the other side of the rectangle must be found from the formula for the area:$A=lw \rightarrow 320=20w \rightarrow w = 16$

9. anonymous

So the perimeter of the rectangle is 16x2 + 20x2 = 72. But we're told this is the perimeter of the big square, so the x-value in the diagram is $x=\frac{72}{4}=18$since each side of a square is equal.

10. anonymous

Now we have to use information about the side length (x) and the angles that have been made by inserting that smaller square alongside the diagonal.

11. anonymous

ok..

12. anonymous

The diagonal of the square bisects the angle it cuts through, so we have the following (see angles):

13. anonymous

I've called one side of one triangle h and the other k, so that x=h+k.

14. anonymous

hmm..ya ok

15. anonymous

In the bottom left triangle, we have$\sin(45^o)=\frac{z}{h}$but$\sin(45^o)=\frac{1}{\sqrt{2}}$so$\frac{z}{h}=\frac{1}{\sqrt{2}}$Also, in the bottom right, you have$\sin(45^o)=\frac{z}{k}\rightarrow \frac{1}{\sqrt{2}}=\frac{z}{k}$

16. anonymous

Since sin(45)=z/h=z/k, we must have h=k, and so h+k=2h=18, which means h=9.

17. anonymous

So from any one of those two sine equations, you have$\frac{z}{h}=\frac{1}{\sqrt{2}}\rightarrow z=\frac{h}{\sqrt{2}}=\frac{9}{\sqrt{2}}$

18. anonymous

The area of the square is$z^2=\frac{81}{2}$

19. anonymous

but lokisan i dont know trignometry...i am in standard 9..

20. anonymous

Hmmmm...let me see if there's another way.

21. anonymous

22. anonymous

Are you taught Pythagoras' Theorem? I don't know how the Indian system works.

23. anonymous

yes

24. anonymous

What are you currently learning? That might help me understand what your teacher's thinking.

25. anonymous

mensuration

26. anonymous

Are you 'allowed' to know that the diagonals of a square bisect the right angles?

27. anonymous

no

28. anonymous

man...

29. anonymous

oh yes.. i understood

30. anonymous

I know you understood...I'm just trying to stick to what you're allowed to know.

31. anonymous

understanding has no restrictions ;) ,

32. anonymous

yeah...have you covered congruence tests for triangles?

33. anonymous

yeahh

34. anonymous

Okay, I'll try and put something together with that.

35. anonymous

ok

36. anonymous

I might have something, but I want to double check.

37. anonymous

hey i think i'hv got the answer ..it's coming to 648..is it correct.. ??

38. anonymous

You have to scrap what I wrote before too - I stuffed one of the sines up. And your 648 is a number I'm getting too.

39. anonymous

hmm

40. anonymous

You can do it using congruent triangles. Only catch is that you need to know that the diagonal of a square bisects the right-angle in to 45 degrees.

41. anonymous

okay.. thanks a lot .need to go and practice some more problems.will bug u if in need.thanx again

42. anonymous

Oh...okay...anyway, I got $z= \frac{\sqrt{648}}{3}$so that the area is $z^2=\frac{648}{9}=72$

43. anonymous

Congruent triangles and Pythagoras' Theorem were used.

44. anonymous

yes

45. anonymous

Cool...well, if you're happy, I'm happy :)