Solve the following mathematical problems:

A circle is inscribed inside a right angled triangle with equal sides. What is the area of the circle if the length of one side of the right angled triangle is 4cm?
(Geometry (Triangle))

Created: 1 year ago | Updated: 1 year ago

Updated: 1 year ago

প্রশ্নে বলা হচ্ছে যে, একটি সমকোণী সমদ্বিবাহু ত্রিভুজে একটি বৃত্ত অন্তর্লিখিত আছে। ত্রিভুজটির একবাহুর দৈর্ঘ্য (সমান সমান বাহুর একটি) 4cm হলে বৃত্তটির ক্ষেত্রফল কত?

Depict the following figure according to question

Area of ABC triangle $= \frac{1}{2} \times b a s e \times h e i g h t$

$= \frac{1}{2} \times 4 \times 4 = 8 c m^{2}$

Now, $A C^{2} = {(A B)}^{2} + {(B C)}^{2} = 4^{2} + 4^{2} = 16 + 16 = 32$

$∴ A C = \sqrt{32}$

The radius of a circle inscribed in a triangle $= \frac{2 \times A r e a o f t r i a n g l e}{P e r i m e t e r o f t r i a n g l e} = \frac{2 \times 8}{(4 + 4) + \sqrt{32}} = \frac{16}{8 + (\sqrt{2 \times 4^{2}})}$

$= \frac{16}{8 + 4 \sqrt{2}} = \frac{16}{4 (2 + \sqrt{2})} = \frac{4}{2 + \sqrt{2}} = \frac{2 \times \sqrt{2} \times \sqrt{2}}{\sqrt{2} (\sqrt{2} + 1)} = \frac{2 \sqrt{2}}{\sqrt{2} + 1} = \frac{2 \sqrt{2} (\sqrt{2} - 1)}{(\sqrt{2} + 1) (\sqrt{2} - 1)} = \frac{2 \sqrt{2} (\sqrt{2} - 1)}{2 - 1} = 2 \sqrt{2} (\sqrt{2} - 1) ∴ R a d i u s = 2 \sqrt{2} (\sqrt{2} - 1) c m ∴ A r e a o f c i r c l e = {πR}^{2} = π {\{2 \sqrt{2} (\sqrt{2} - 1)\}}^{2} = π 4 \times 2 {(\sqrt{2} - 1)}^{2} {cm}^{2} = 8 π 2 \sqrt{2} {(\sqrt{2} - 1)}^{2} {cm}^{2} (ans .)$