Question

5. From each corner of a square of side 4 cm a quadrant of a circle of radius 1 cm is cut and
also a circle of diameter 2 cm is cut as shown in Fig. 12.23. Find the area of the remainin
portion of the square.​

Answer 1

Step-by-step explanation:

we have square side 4cm

area. equals side*side i.e 4×4.

16cm square

and there are four quadrants in the circle all radius1cm.

area of quadrant is 4πr×r

then area of four quadrant

4( πr×r)÷4

we get πr×r. then putting value

3.14×1×1

Answer 2

Solution:

Note: Diagram of this answer attach in attachment file.

Given:

=> Side of square = 4 cm.

=> Diameter of middle circle = 2 cm.

=>Radius of middle circle = 1 cm.

To find:

=> Area of shaded region.

Formula used:

$\sf{\implies Area\;of\;square = (side)^{2}}$

$\sf{\implies Area\;of\;quadrant=\dfrac{1}{4} \pi r^{2}}$

$\sf{\implies Area\;of\;circle = \pi r^{2}}$

So,

Area of square ABCD = 4 × 4

                                    = 16 cm²

∴ Each corner has a quadrant circle of radius = 1 cm.

$\sf{\implies Area\;of\;4\;quadrant\;circle = 4\times \dfrac{1}{4} \pi r^{2} =\pi r^{2}}$

$\sf{\implies \dfrac{22}{7}\times 1\times 1}$

$\sf{\implies \dfrac{22}{7}\;cm^{2}}$

∴ Diameter of middle circle = 2 cm.

∴ Radius of middle circle = 1 cm.

So, Area of middle circle = πr²

$\sf{\implies \dfrac{22}{7}\times 1\times 1}$

$\sf{\implies \dfrac{22}{7}\;cm^{2}}$

Now, Area of shaded region = [Area of square] - [(Area of 4 quadrant + Area of the middle circle)]

$\sf{\implies [16]-\bigg[\dfrac{22}{7} +\dfrac{22}{7} \bigg]}$

$\sf{\implies 16-2\times \dfrac{22}{7}}$

$\sf{\implies 16 - \dfrac{44}{7}}$

$\sf{\implies \dfrac{112-44}{7}}$

${\boxed{\boxed{\sf{\implies \dfrac{68}{7} \; cm^{2}}}}}$