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.

- anonymous

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- anonymous

need help..

- anonymous

still here?

- anonymous

yups

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## More answers

- anonymous

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- anonymous

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

- anonymous

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

- anonymous

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

- 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:\[A=lw \rightarrow 320=20w \rightarrow w = 16\]

- 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.

- 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.

- anonymous

ok..

- anonymous

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

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- anonymous

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

- anonymous

hmm..ya ok

- 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\[\sin(45^o)=\frac{z}{k}\rightarrow \frac{1}{\sqrt{2}}=\frac{z}{k}\]

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

yes please

- anonymous

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

- anonymous

yes

- anonymous

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

- anonymous

mensuration

- anonymous

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

- anonymous

no

- anonymous

man...

- anonymous

oh yes..
i understood

- anonymous

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

- anonymous

understanding has no restrictions ;) ,

- anonymous

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

- anonymous

yeahh

- anonymous

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

- anonymous

ok

- anonymous

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

- anonymous

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

- 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.

- anonymous

hmm

- 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.

- anonymous

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

- anonymous

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

- anonymous

Congruent triangles and Pythagoras' Theorem were used.

- anonymous

yes

- anonymous

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

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