Question

ABCD is a square of Side 2 CM if each vertex as Centre and one centimetre as radius four circles are drawn then the radius of the circle which touches which touches these four circles externally is

Answer 1

The radius of the circle touching the four smaller circles is (√2 + 1) cm

Step-by-step explanation:

Each side of the square is 2 cm long.

[ Refer to the attachment added. ]

If circles of radius 1 cm are drawn at the vertices and then draw an outer circle touching all the four circles, we get the attached figure.

We have to find the radius of the outer circle. This is given by

= (1/2 * diagonal of the square) + radius of a circle

= (1/2 * side * √2) + radius of a circle

= {(1/2 * 2 * √2) + 1} cm

= (√2 + 1) cm

Some important questions:

1. In Figure 4, ABCD is a square of side 4 cm. A quadrant of a circle of radius 1 cm is drawn at each vertex of the square and a circle of diameter 2 cm is also drawn. Find the area of the shaded region. (Use π = 3.14)

- https://brainly.in/question/7457872

2. In the below figure,find the area of the shaded region [use pi = 3.14]

- https://brainly.in/question/2388155

Answer 2

Answer:

$(\sqrt{2}+1)$ cm

Step-by-step explanation:

In the attach figure, ABCD is a square of side 2 cm.

There are four circle each vertices of square.

There are two possible circle which touches four circles externally.

In ΔOAB, ∠OBA = 90°

$OA^2=OB^2+AB^2$

$OA^2=1^2+1^2$

$OA=\sqrt{2}$

The radius of inner required circle,

r = OA - R

$r=\sqrt{2}-1$

The radius of outer required circle,

r' = OA + R

$r'=\sqrt{2}+1$

Hence, the radius of the circle which touches four circles externally is $\sqrt{2}+1$ cm