Question

Class-
IIT Foundation Explorer
scribed inside in it
9. The figure shows the rectangle ABCD with a semicircle and a circle inscribed
as shown. What is the ratio of the area of the circle to that of the semicircle
It is of 20 point I think so

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Answer 1

Some constructions are required, refer the attachment.

→  Mark the midpoints of AB and CD as M and N respectively such that AD = MN = BC since ABCD is a rectangle.

→  Join BN.

→  Draw EF which is a tangent to both the semicircle and the small circle.

Since a semicircle is inscribed in the rectangle, CD is the diameter of the semicircle and N will be the center, so that CN = DN. And also CN = MN and MN = BC gives CN = DN = BC, thereby implying CD = 2BC. Hence we get that the length of the rectangle is twice its breadth.

Let  BC = CN = x.  So from ΔBCN, we get  BN = x√2  by using Pythagoras' theorem.

Since  PN = CN = x,  we also get  BP = BN - PN = x√2 - x = (√2 - 1) x.  

ΔBCN is an isosceles right triangle because of BC = CN. Hence ∠CBN = 45°. (Angles of an isosceles right triangle is always 45°, 45° and 90°.)

Since EF is the tangent of the semicircle, ∠EPN = ∠FPN = ∠BPF = ∠BPE = 90°.

So, from  ΔBPF,  ∠FBP = 45°  and  ∠BPF = 90°  imply  ∠BFP = 45°.

Hence, from ΔBFE,  ∠FBE = 90°  and  ∠BFE = 45°  imply  ∠BEF = 45°.

So we get ΔBFE is an isosceles right triangle. Now we have to consider this triangle.

BP is the altitude from B to EF, so it bisects EF, since of the fact that the altitude from the vertex joining congruent sides of an isosceles triangle bisects the opposite side. So  PE = PF.

And another feature for the isosceles right triangle BFE is that PE = PF = PB  too. Hence PE = PF = (√2 - 1) x  and  EF = 2 (√2 - 1) x.  

And also we get  BE = BF = √2(√2 - 1) x.

Now we have to find the area and semiperimeter of this triangle.

Area = (1/2) · BP · EF = (1/2)((√2 - 1) x)(2(√2 - 1) x) = ((√2 - 1) x)²

Perimeter = BE + BF + EF = √2(√2 - 1) x + √2(√2 - 1) x + 2 (√2 - 1) x = (2√2 + 2)(√2 - 1) x = 2(√2 + 1)(√2 - 1) x = 2x

Hence semiperimeter = x

Now we're going to find the length of radius of the small circle.

We can see that the circle is inscribed in the triangle BFE. So the radius of this circle will be equal to the inradius of ΔBFE, won't it be?

Here we use  "A = rs",  where A is the area of the triangle, r is its inradius and s is its semiperimeter.

So,

    A = rs

⇒  r = A / s

⇒  r = (((√2 - 1) x)²) / x

⇒  r = (√2 - 1)² x

⇒  r = (3 - 2√2) x

Now, area of the small circle,

π · ((3 - 2√2) x)² = (17 - 12√2) x²π

And, area of the semicircle will be  x²π / 2.

Now let's find out the ratio!

    (17 - 12√2) x²π  :  x²π / 2

⇒  17 - 12√2  :  1/2

⇒  34 - 24√2  :  1

Hence this is the ratio!