Introduction To Three Dimensional Geometry

The correct option is (A):

The equation of the OD line is

equation of diagonal BE is

S.D =
 

Inradius Calculation

Given: We are tasked with calculating the inradius using the provided formula.

Solution:

The formula for the inradius is:

Where is the area and is the semi-perimeter. Given the expression for the inradius:

Thus, the value of the inradius is 0.25.

The value of r² is 2

Radical Axis and Circle Equation

Radical Axis and Circle Equation

Given: Points and are the midpoints of line segments and respectively. The line serves as the radical axis connecting two circles. The equation of the line is:

Solution:

We are given the equation of the radical axis as: 

On solving for , we find:

Thus, the correct answer is .

We start by applying the law of sines to the sides and angles of the triangle:

From this, we deduce that .

Using the identity for sine, we have:

Now, using the relationship between the sides:

Expanding and simplifying, we get:

From this, we get Equation (1):

Next, we use the cosine law to relate angle :

Simplifying further, we get:

Now, we can find the area of the triangle using the formula:

Substitute the values of and :

Finally, we find that:

Thus, the value of is 1008.

The vertices form a regular octagon inscribed in a circle with a radius of 2. We need to evaluate , which is the result of:

Multiplying the terms:

Thus, the expression simplifies to:

Finally, the value of is:

Therefore, the correct answer is .

Consider points M and N as the midpoints of line segments PQ and ST respectively.
As a result, the line MN serves as the radical axis connecting two circles.
Equation are as follows :
C₁ : x² + y² = 1 ……(i)
C₂ : (x - 4)² + (y - 1)² = r²
⇒ x² + y² - 8x - 2y + 17 - r² = 0 …….(ii)
Now, from the equation (i) and (ii), we get :
Equation of MN :
8x + 2y - 18 + r² = 0
As, B is on x-axis
⇒
So, AB =
…… (By distance formed a)
⇒ On solving, we get :
r² = 2

According to the question, Ai are 8^th root of 2⁸ & let P be 2e^iα.
Now, z⁸ – 2⁸ = (z – A₁) (z – A₂) ……… (z – A₈)
So, Put z = 2e^iα
⇒ 2⁸e^i8α – 2⁸ = (2e^iα – A₁) (2e^iα – A₂) (2e^iα – A₃) ….. (2e^iα – A₈)
⇒ 2⁸ |e^i8α – 1| = |(2e^iα – A₁) (2e^iα – A₂) (2e^iα – A₃) ….. (2e^iα – A₈)|
⇒ 2⁸ |e^i4α – e^–i4α | = |(2e^iα – A₁) (2e^iα – A₂) (2e^iα – A₃) ….. (2e^iα – A₈)|
⇒ 2⁹ |sin4α| = |(2e^iα – A₁) (2e^iα – A₂) (2e^iα – A₃) ….. (2e^iα – A₈)|
So, from this we get that :
Maximum value of the PA₁.PA₂ …… PA₈ = 2⁹
Therefore, the correct answer is 2⁹ = 512.