anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
need help..
anonymous
  • anonymous
still here?
anonymous
  • anonymous
yups

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

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