Question

A wire is bent in the form of a semicircle of radius 21 cm. What will be the area covered by the same
wire (in cm^2) if it is bent into a rectangle of length 10 cm?
(a) 440
(b) 436
(c) 256
(d) 224​

Answer 1

Answer:

¥a)440 is write answer of this question

Answer 2

Answer:

$\bold\red{(a)440{cm}^{2}}$

Step-by-step explanation:

For the semicircular wire,

Radius, r = 21 cm

Now, Perimeter of the semicircle ,

P = πr

=> $P =\frac{ 22×21}{7}\\=>P=66\:cm$

Therefore, total length of wire is

= πr + 2r

= 66 + 42

= 108 cm

Now, it is bent into a rectangle of

length, l = 10cm

Since, the wire is same,

therefore, length will also be same,

Let, the breadth of recatangle is 'b' .

=> 2( l+b ) = 108

=> l + b = 54

=> 10 + b = 54

=> b = 54- 10

=> b = 44 cm

Therefore, the area of the rectangle ,

A' = l × b

=> A' = 10 × 44

=> A' = 440

Hence, correct option is $\bold{(a)440{cm}^{2}}$