Rectangle PQST, with dimensions w*h, is inscribed in a circl

I don't get your equation.
My equations says that the area of the rectangle wh must equal the area of the triangle.
The height of the triangle is the radius minus the height of the rectangle-->(1-h)
Hence the area of the triangle is w*(1-h)
I don't see why we have to divide h by 2!

Thanks for help

Chuck it, I misread it and over looked the same area part and hence was confused that what's the relation between the rectangle and the triangle.

Thanks anyway..!

Cheers,

Because, The rectangle & Circle share the same centre.

Damn, no I get. I always looked at it and thought the rectangle stands on the center!
THX!

So why did they give us that the areas where the same?

Would love to hear some reactions

Cheers!
J

The equation in red above equates the areas of the rectangle and the triangle.

How do we get height of triangle=(1-h/2)?

In other words how do we know that the centre of the circle and the rectangle are the same?What am I missing here?

A right triangle inscribed in a circle must have its hypotenuse as the diameter of the circle. A rectangle is the sum of two right triangles, thus the diagonals of a rectangle must lie on the diameter of the circle. Therefore the intersection of the diagonals must be the center of the circle.

Hope it's clear.

Got it!
A very nice observation!Thanks,Bunuel!

I am still confused at how we know the height of the triangle is 1-(h/2). Could someone please explain?

Catalysis..letme try

As the triangle is a isosceles triangle, a perpendicular drawn from point R to QS will pass through the centre if extended further.
So, the the length of the line connecting R to centre is 1 (radius)
To get the height of triangle, we need to remove the height of the area covered by rectangle.
Given it is a rectangle, QS = PT (properties of rectangle)
Now QS can only be equal to PT if they are equidistant from the centre of circle (only equidistant chords can be equal)
So, in other words center lies in the middle of the rectangle and hence height from center to QS is h/2
Hence if we remove this from 1 (radius), we get the height of the triangle = 1 - h/2

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I think I got it. So given that QRS is an isosceles triangle we have that the diameter = 2, is equal to 2a + h

We also know that wh= aw/ 2

Therefore replacing we have that h=2/5

Answer is thus B

Or also:

Let's call the height of the isosceles triangle 'A'. So, we have wh = aw/2. 2h=a/. Now, we also know that 2h+h+2h=2 which is the diameter of the circle. Therefore, h=2/5. B is the correct answer

Hope this clarifies
Gimme some freaking Kudos if it helps

Cheers!
J

How do we know that w = 2 ?

r= 1, d= 2

but neither the width of the rectangle nor that of the triangle equal the diameter of the circle! ?

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