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.

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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.

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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You need to find z in this set up given the info. you have.
You can find z by finding x, and you find x using the information you have about the rectangle.
The perimeter of the big square is equal to the perimeter of the rectangle.
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:\[A=lw \rightarrow 320=20w \rightarrow w = 16\]
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.
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.
ok..
The diagonal of the square bisects the angle it cuts through, so we have the following (see angles):
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I've called one side of one triangle h and the other k, so that x=h+k.
hmm..ya ok
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\[\sin(45^o)=\frac{z}{k}\rightarrow \frac{1}{\sqrt{2}}=\frac{z}{k}\]
Since sin(45)=z/h=z/k, we must have h=k, and so h+k=2h=18, which means h=9.
So from any one of those two sine equations, you have\[\frac{z}{h}=\frac{1}{\sqrt{2}}\rightarrow z=\frac{h}{\sqrt{2}}=\frac{9}{\sqrt{2}}\]
The area of the square is\[z^2=\frac{81}{2}\]
but lokisan i dont know trignometry...i am in standard 9..
Hmmmm...let me see if there's another way.
yes please
Are you taught Pythagoras' Theorem? I don't know how the Indian system works.
yes
What are you currently learning? That might help me understand what your teacher's thinking.
mensuration
Are you 'allowed' to know that the diagonals of a square bisect the right angles?
no
man...
oh yes.. i understood
I know you understood...I'm just trying to stick to what you're allowed to know.
understanding has no restrictions ;) ,
yeah...have you covered congruence tests for triangles?
yeahh
Okay, I'll try and put something together with that.
ok
I might have something, but I want to double check.
hey i think i'hv got the answer ..it's coming to 648..is it correct.. ??
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.
hmm
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.
okay.. thanks a lot .need to go and practice some more problems.will bug u if in need.thanx again
Oh...okay...anyway, I got \[z= \frac{\sqrt{648}}{3}\]so that the area is \[z^2=\frac{648}{9}=72\]
Congruent triangles and Pythagoras' Theorem were used.
yes
Cool...well, if you're happy, I'm happy :)

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