Bunuel wrote:

A circle is inscribed inside right triangle ABC shown above. What is the area of the circle, if the leg AB of the right triangle is 4?

A. \(4\pi(\sqrt{2}-1)^2\)

B. \(8\pi(\sqrt{2}-1)^2\)

C. \(2\pi\)

D. \(4\pi\)

E. \(8\pi(1-\sqrt{2})\)

Solution :Given : 1. ABC is a right angled triangle

2. A circle is inscribed in the triangle

3. AB = 4 cm and angle ACB = 45 degrees

Find : Area of the circle.

Approach :

To find the area of the circle, we need to

find the radius of the circle. It is given that ABC is a right angled triangle and angle ACB is 45 degrees.

From the above information, we can infer that -

angle BAC = \(180 - ACB - ABC = 180 - 45 - 90 = 45\)

Therefore we can conclude that

ABC is an isosceles right-angled triangle.

And

AB will be equal to BC.We know that area of the triangle =1/2 x base x height.........(i)

AND

We can also find the area of the triangle by using the formula =

in-radius of the circle x semi-perimeter.......(ii)

Equating the above two formulas, we can find the in-radius of the circle.

Once we have the in-radius, we can find the area of the circle.

Working :

Since AB is equal to BC

We can write,

AB = BC = 4cm

Area of the triangle = 1/2 x AB x BC

Area of the triangle = 1/2 x 4 x 4

Therefore,

Area of the triangle = 8..................(i)

Also

Using Pythagorus Theorem, we can write,

\(AC^2 = AB^2 + BC^2\)

\(AC^2 = 4^2 + 4^2\)

\(AC^2 = 16 + 16\)

AC = 4\(\sqrt{2}\)

Perimeter of the triangle of ABC = AB + BC + CA

= \(4 + 4+ 4\sqrt{2}\)

= \(4 (2 + \sqrt{2})\)

Semi - perimeter =\(\frac{Perimeter of Triangle}{2}\)

= \(4 (2 + \sqrt{2})/2\)

= \(2 (2 + \sqrt{2})\)

As discussed in the analysis,

area of the triangle can also be written as = in-radius x semi-perimeter

Area of the triangle = in-radius x \(2 (2 + \sqrt{2})\) ..........(ii)

Equating (i) and (ii), we get

in-radius x \(2 (2 + \sqrt{2})\) = 8

in - radius = \(4/(2+\sqrt{2})\)

= \(4(2-\sqrt{2})/(2-\sqrt{2})(2+\sqrt{2})\)

= \(4(2-\sqrt{2})/(4 -2)\)

Therefore,

in-radius = \(2(2- \sqrt{2})\)

= \(2\sqrt{2}(\sqrt{2} - 1)\)

Therefore, Area of the circle = pi x \((in-radius)^2\)

Area of the circle = pi x \([2\sqrt{2}(\sqrt{2} - 1)]^2\)

Area of the circle = pi x \(8(\sqrt{2}- 1)]^2\)

Hence, the correct answer is

Option BThanks,

Saquib

_________________

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