Question

in fig . A is the centre of the circle ∠ABC =45° and AC = 7√2 cm . find area of seg BXC

Answer 1

Answer:

28 cm^2

Step-by-step explanation:

In Tr ABC

AB = AC ( Radii of the circle )

=> Ang(ABC) = Ang(ACB)

Since equal sides have equal angles

WKT

Sum of angles in a triangle = 180

=> Ang(ABC) + Ang(ACB) + Ang(BAC) = 180

=> 2*Ang(ABC) + Ang(BAC) = 180

=> 90 +Ang(BAC) = 180

=> Ang(BAC) = 90

Therefore Tr ABC is a Right angle Triangle

Now

Area of segment BXC = Area of Sector ABXCA - Area of Tr ABC

Ar of Tr ABC = 0.5 * AB * AC

Ar of Tr ABC = 0.5 * 7√2 *7√2

Ar of Tr ABC = 49 cm^2

Area of Sector ABXCA = pi * AB * AC * Ang(BAC) / 360

Area of Sector ABXCA = (22/7) * 7√2 * 7√2 * 90 / 360

Area of Sector ABXCA = 77 cm^2

=> Area of segment BXC = 77 - 49 cm^2

=> Area of segment BXC = 28 cm^2

$Formulae\\\\Area\; \;of\;\;a\;\;Triangle\;\;=\;\;\frac{1}{2}\times b\times h\\\\b=base\;\;h=height\\\\Area\;\;of\;\;a\;\;sector\;\;=\;\;\pi \times r^2\times \frac{\theta}{360}\\\\ r=radius \;\;of \;\;sector\;\;\theta=\;angle\;\;made\;\;by\;\;the\;\;sector$

Hope it Helps