Question

A circle placed against a right angle triangle entered in o at radius 14 cm. What is tha raidus of tha smaller circle placed in tha remaining gap​

Answer 1

Answer:

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Step-by-step explanation:

Let the radius of the larger circle be “R” and the smaller circle be “r”.

R = 2 ….. [given]  

It can be seen from the figure attached below that the larger circle is placed against a right angle, therefore, AOCB forms a square with each side = 2 and each angle 90°.

Since the diagonal of the square is given as a√2 where a = side of the square. Here, a = R = 2.

∴ OB = R√2  = 2√2  

Now, from the figure, we can also say,

OB = OD + DB

⇒ OB = R + (r + O’B)

Substituting OB = 2√2, R = 2 & O’B = diagonal of the smaller square = r√2

⇒ 2√2 = 2 + r + (r√2)

⇒ 2√2 – 2 = r (1+√2)

⇒ 2(√2 - 1) = r (1+√2)

⇒ r = 2(√2 - 1) / (1+√2)

On dividing & multiplying by (1 - √2), we get

⇒ r = [2(√2 - 1) / (√2 + 1)] * [(√2 - 1) / (√2 - 1)]

⇒ r = [2(√2 - 1)²] / [(√2)² – (1)²]

⇒ r = [2(√2 - 1)²] / [2 – 1]  

⇒ r = 2 [2 - 2√2 + 1]

⇒ r = 2 [3 - 2√2]

⇒ r = 6 – 4√2

Thus, the radius of the smaller circle is (6 – 4√2) .

Hope this is helpful!!!!

Answer 2

Answer:

the answer will be 14(√2-1)²

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