Question

UVWX is a square inscribed in a circle , which in turn is circumscribed by another square PQRS. If the area of the circle is 64 Sq cm ,find the area of the shaded region. ​

Answer 1

Step-by-step explanation:

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Answer 2

Answer:

Area of shaded region is 160 $cm^2$.

Step-by-step explanation:

Since area of circle = 64$\pi$ $cm^2$

⇒ $\pi r^2 = 64 \pi$

⇒ $r^2 = 64$

⇒ $r = 8$ cm

So, radius of circle = 8 cm

Now,

From figure, AO is a radius of circle.

⇒ AO + OB is a diameter.

⇒ diameter, AB = 8 + 8

                          = 16 cm

Also, AB = PQ and PQRS is a square.

⇒ PQ = 16 cm

Thus, sides of square is 16 cm.

Area of PQRS = Side × Side

                        = 16 × 16

                        = 256 $cm^2$

Hence, area outside the circle and inside the square PQRS,

= Area of PQRS - area of circle

= 256 - 64

= 192 $cm^2$

Notice that this area is sum of four equal regions.

Thus, area of 1 region = 192/4

                                    = 48 $cm^2$ ...... (1)

Similarly,

From figure, XO is a radius of circle.

⇒ XO + OV is a diameter.

⇒ diameter, XV = 8 + 8

                          = 16 cm

Also, XV = diagonal of square UVWX.

⇒ $XV=\sqrt{VW^2+XW^2}$, (Using Pythagoras)

⇒ $16=\sqrt{VW^2+VW^2}$ (Sides of square are equal, i.e., VW = XW)

⇒ $16=\sqrt{2VW^2}$

⇒ $16=VW\sqrt{2}$

⇒ $VW = \frac{16}{\sqrt{2} }$

Thus, sides of square UVWX is 16/$\sqrt{2}$ cm.

Area of UVWX = Side × Side

                         = 16/$\sqrt{2}$ × 16/$\sqrt{2}$

                         = 512 $cm^2$

Hence, area inside the circle and outside the square UVWX,

= Area of UVWX - area of circle

= 512 - 64

= 448 $cm^2$

Notice that this area is sum of four equal regions.

Thus, area of 1 region = 448/4

                                    = 112 $cm^2$ ...... (2)

So, area of shaded region = area (1) + area (2)

                                            = 48 + 112

                                            = 160 $cm^2$

Thus, area of shaded region is 160 $cm^2$.

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