Question

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.

Answer 1

need help..

Answer 2

still here?

Answer 3

yups

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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.

Answer 9

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.

Answer 10

ok..

Answer 11

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

Answer 12

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

Answer 13

hmm..ya ok

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

yes please

Answer 21

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

Answer 22

yes

Answer 23

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

Answer 24

mensuration

Answer 25

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

Answer 26

no

Answer 27

man...

Answer 28

oh yes..
i understood

Answer 29

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

Answer 30

understanding has no restrictions ;) ,

Answer 31

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

Answer 32

yeahh

Answer 33

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

Answer 34

ok

Answer 35

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

Answer 36

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

Answer 37

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.

Answer 38

hmm

Answer 39

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.

Answer 40

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

Answer 41

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

Answer 42

Congruent triangles and Pythagoras' Theorem were used.

Answer 43

yes

Answer 44

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