In isosceles trapezium ✷ABCD, ABkCD. Circle with BCand AD as diameters touch each other at E. If 2BE = AD,BC = 10 and area of ✷ABCD = P then find √P3?

Answer 1

Let's solve the problem step by step.

Given:

In isosceles trapezium ✷ABCD, AB || CD.

Circle with BC and AD as diameters touch each other at E.

2BE = AD

BC = 10

Area of ✷ABCD = P

First, let's find the length of AD.

Since 2BE = AD, we can substitute 2BE for AD in the isosceles trapezium.

In an isosceles trapezium, the diagonals are equal, so we have:

BC = AD

Given that BC = 10, we can conclude that AD = 10.

Now, let's find the length of BE:

2BE = AD

2BE = 10

BE = 5

Now, let's find the length of CD:

In an isosceles trapezium, the bases are parallel and equal in length. Therefore, AB = CD.

Let's denote AB = CD = x.

Now, we can use the Pythagorean theorem to find the height (h) of the trapezium.

Since the circle with BC and AD as diameters touches each other at E, the height of the trapezium is equal to the radius of the circle.

The radius of the circle is equal to BE, so r = BE = 5.

Using the Pythagorean theorem:

h^2 = r^2 + (CD/2)^2

h^2 = 5^2 + (x/2)^2

h^2 = 25 + x^2/4

Next, we can find the area (A) of the trapezium:

A = (AB + CD) * h / 2

A = (x + x) * h / 2

A = 2xh/2

A = xh

Now, substitute the value of h^2 from the previous equation:

A = x * √(25 + x^2/4)

Given that the area of the trapezium is P, we have:

P = x * √(25 + x^2/4)

To find √P3, we can simplify it as follows:

√P3 = √(P^(3/2))

= (P^(3/2))^(1/2)

= P^(3/4)

Therefore, √P3 = P^(3/4).

So, the value of √P3 is P raised to the power of 3/4.

Answer 2

√P3 = √(15 * (5 + c)).