A CIRCLE WITH RADIUS 2 IS PLACED AGAINST A RIGHT ANGLE ANOTHER SMALLER CIRCLE IS ALSO PLACED AS SHOWN IN THE FIGURE WHAT IS THE THE RADIUS OF SMALLER CIRCLE
Answers
Hi there,
The figure required for the given question is missing. So, I have attached a figure which satisfies the question given and have solved it accordingly:
Answer:
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:
Step-by-step explanation:
OR=
2
OP (length of diagonal of square OPQR)
OR=2
2
From figure
OR=2+r+
2
r
r be radius of small circle.
2
r is length of diagonal of square in fig. 2
2
2
=2+r+
2
t
2
2
−2=r(1+
2
)
r=
(
2
+1)×(
2
−1)
2(
2
−1)×(
2
−1)
r=
(
2
)
2
−(1)
2
2(
2
−1)
2
r=
1
2(2+1−2
2
)
r=6−4
2
.