**Question 1:** **In the given figure, APB and AQO are semi-circles and AO = OB. If the perimeter of the figure is 40 cm, find the area of the shaded region.**

**Solution:**

Perimeter of shaded region = Length of the arc AQO + Length of the arc APB + Length of OB

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

Area of the shaded portion = Area of semicircle AQO + Area of semicircle APB

= \( \frac{1}{2}\Pi \left ( \frac{7}{2} \right )^{2}+\frac{1}{2}\Pi (7)^{2}\)

= \( \frac{1}{2}\times \frac{22}{7}\times \left ( \frac{7}{2} \right )^{2}+\frac{1}{2}\times \frac{22}{7}\times (7)^{2}\)^{2}

Hence, the area of the shaded portion is 96.25 cm^{2}.

**Question 2:** **Find the area of a quadrant of a circle whose circumference is 44 cm.**

**Solution:**

Let the radius of the circle be r.

Now,

Circumference = 44

\(\Rightarrow\)

\(\Rightarrow\)

Now,

Area of quadrant = \( \frac{1}{4}\Pi r^{2}=\frac{1}{4}\times \frac{22}{7}\times (7)^{2}=38.5cm^{2}\)

Hence, the area of the quadrant of the circle is 38.5 cm^{2}.

**Question 3:** **In the given figure, find the area of the shaded region, where ABCD is a square of side 14 cm and all circles are of the same diameter.**

**Solution:**

Area of the square = Side^{2} = 14^{2} = 196 sq. cm

Area of the circle = 4 x \(\Pi\)

Area of the shaded region = Area of the square â€“ Area of four circles = 196 â€“ 154 = 42 cm^{2}

**Question 4:** **Find the area of the shaded region in the given figure, if ABCD is a rectangle with sides 8 cm and 6 cm and O is the centre of the circle.**

**Solution:**

In right triangle ABC

AC^{2} = AB^{2} + BC^{2} = 8^{2} + 6^{2} = 64 + 36 = 100

Therefore, AC^{2} = 100

\(\Rightarrow\)

Now, Radius of circle (OA) = \(\frac{1}{2}\)

Area of the shaded region = Area of circle â€“ Area of rectangle OABC

= \( \Pi (OA)^{2}-AB\times BC\)^{2}

Hence, the area of the shaded region is 30.57 cm^{2}.

**Question 5:** **A wire is bent to form a square enclosing an area of 484 m ^{2}. Using the same wire, a circle is formed. Find the area of the circle.**

**Solution:**

Area of the circle = 484 cm^{2}

Area of the square = Side^{2}

\(\Rightarrow\)^{2}

\(\Rightarrow\)^{2} = Side^{2}

\(\Rightarrow\)

Perimeter of the square = 4 x Side

Perimeter of the square = 4 x 22 = 88 cm

Length of the wire = 88 cm

Circumference of the circle = Length of the wire = 88 cm

Now, let the radius of the circle be r cm.

Thus, we have:

\(\Rightarrow\)

\(\frac{22}{7}\)

Area of the circle = \(\Pi r^{2}\)^{2}

Thus, the area enclosed by the circle is 616 cm^{2}.

**Question 6:** **A square ABCD is inscribed in a circle of radius r. Find the area of the square.**

**Solution:**

Let the diameter of the square be d and having circumscribed circle of radius r.

We know that if a circle circumscribes a square, then the diameter of the circle is equal to the diagonal of the square.

Therefore, d = 2r

Now,

Area of square = \(\frac{1}{2}d^{2}\)^{2} sq. units

Hence, the area of the square ABCD is 2r^{2} sq. units.

**Question 7:** **The cost of fencing a circular field at the rate of Rs. 25 per metre is Rs. 5500. The field is to be ploughed at the rate of 50 paise per m ^{2}. Find the cost of ploughing the field. [Take \(\Pi =\frac{22}{7}\)**

**Solution:**

Circumference = \( \frac{Total\, cost\, of\, fencing}{Rate\, of\, fencing}=\frac{5500}{25}=220\)

Let the radius of the circle be r.

Now,

Circumference = 220

\(\Rightarrow\)

\(\Rightarrow\)

Now,

Area of field = \(\Pi r^{2}\)

Cost of plugging = Rate x Area of field = 0.5 x 3850 = Rs 1925

Hence, the cost of plugging the field is Rs 1925.

**Question 8:** **A park is in the form of a rectangle 120 m by 90 m. At the centre of the park, there is a circular lawn as shown in the figure. The area of the park excluding the lawn is 2950 m ^{2}. Find the radius of the circular lawn. (Given, \(\Pi =3.14\))**

**Solution:**

Area of the rectangle = l x b = 120 x 90 = 10800 sq. m

Area of the park excluding the lawn = 2950 m^{2}

Area of the circular lawn = Area of the park â€“ Area of the park excluding the lawn = 10800 â€“ 2950 = 7850 m^{2}

Area of the circular lawn = \(\Pi r^{2}\)

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)^{2} = 2497.72

\(\Rightarrow\)

Or,

r \(\approx\)

Thus, the radius of the circular lawn is 50 m.

**Question 9:** **In the given figure, PQSR represents a flower bed. If OP = 21 m and OR = 14 m, find the area of the flower bed****.**

**Solution:**

Area of the flower bed is the difference between the areas of sectors OPQ and ORS.

Area of the flower bed = \( \frac{\Theta }{360}\times \Pi (PO^{2}-OR^{2})\)

= \( \frac{90}{360}\times \frac{22}{7}(21^{2}-14^{2})\)^{2}

**Question 10:** **In the given figure, O is the centre of the bigger circle, and AC is its diameter. Another circle with AB as diameter is drawn. If AC = 54 cm and BC = 10 cm, find the area of the shaded region.**

**Solution:**

We have:

OA = OC = 27 cm

AB = AC â€“ BC = 54 â€“ 10 = 44

AB is the diameter of the smaller circle.

Thus, we have:

Radius of the smaller circle = \(\frac{AB}{2}\)

Area of the smaller circle = \(\Pi r^{2}\)^{2}

Radius of the larger circle = \(\frac{AC}{2}\)

Area of the larger circle = \(\Pi r^{2}\)

= \( \frac{22}{7}\times 27\times 27\)^{2}

Therefore, Area of the shaded region = Area of the larger circle â€“ Area of the smaller circle = 2291.14 â€“ 1521.14 = 770 cm^{2}

**Question 11:** **From a thin metallic piece in the shape of a trapezium ABCD in which AB || CD and \(\angle\)BCD = 90 ^{o}, a quarter circle BFEC is removed. Given, AB = BC = 3.5 cm and DE = 2 cm, calculate the area of remaining (shaded) part of metal sheet.**

**Solution:**

Since, BFEC is a quarter of a circle.

Hence, BC = EC = 3.5 cm

Now, DC = DE + EC = 2 + 3.5 = 5.5 cm

Area of shaded region = Area of the trapezium ABCD â€“ Area of the quadrant BFEC

= \( \frac{1}{2}\times (AB+DC)\times BC-\frac{1}{4}\times \Pi (EC)^{2}\)

= \( \frac{1}{2}\times (3.5+3.5)\times 3.5-\frac{1}{4}\times \frac{22}{7}\times (3.5)^{2}\)^{2}

Hence, the area of the shaded region is 6.125 cm^{2}.

**Question 12:** **Find the area of the major segment APB of a circle of radius 35 cm and \(\angle\)AOB = 90 ^{o}, as shown in the given figure.**

**Solution:**

Area of minor segment = Area of sector AOBC â€“ Area of right triangle AOB

= \( \frac{90}{360}\times \Pi (OA)^{2}-\frac{1}{2}\times OA\times OB\)

= \( \frac{1}{4}\times \frac{22}{7}\times (35)^{2}-\frac{1}{2}\times 35\times 35\)^{2}

Area of major segment APB = Area of circle â€“ Area of minor segment

= \( \Pi (OA)^{2}-350\)^{2}

Hence, the area of major segment is 3500 cm^{2}.

**Exercise – 18B**

**Let \(\Pi = \frac{22} {7}\) unless stated otherwise.**

**Q.1: The circumference of a circle is 39.6 cm. Find its area.**

**Sol:**

Circumference = 39.6 cm

We know,

Circumference of a circle = \(2 \Pi r\)

\(\Rightarrow 39.6 = 2 \times \frac{22}{7}\times r\)

\(\Rightarrow \frac{39.6 \times 7}{2 \times 22} = r\)

\(\Rightarrow r= 6.3\)

Also,

Area of the circle = \(\Pi r^{2}\)

\(= \frac{22}{7} \times 6.3 \times 6.3\)

**Q.2: The area of a circle is \( 98.56 cm^{2}\). Find its circumference.**

**Sol:**

Let the radius be r.

Now,

Area = 98.56

\(\Rightarrow \pi r^{2} = 98.56\)

\(\Rightarrow r = 5.6\)

Now,

Circumference = \(= 2 \pi r = 2 \times \frac{22}{7} \times 5.6 = 35.2\)

Hence the circumference of the circle is 35.2 cm.

**Q.3: The circumference of a circle exceeds its diameter by 45 cm. Find the circumference of the circle. Â **

**Sol: **

Let the radius of the circle be r.

Now,

Circumference = Diameter +45

\(\Rightarrow 2 \pi r = 2 r + 45\)

\(\Rightarrow 2 \pi r – 2 r = 45\)

\(\Rightarrow 2r (\frac{22}{7} – 1) = 45\)

\(\Rightarrow 2r (\frac{15} {7} ) = 45\)

\(\Rightarrow r = \frac{45\times 7}{30} = 10.5\)

Therefore, Circumference = Diameter + 45 = 2(10.5) + 45 = 66 cm

Hence, the circumference of the circle is 66 cm.

**Q.4: A copper wire bent in the form of a square enclosed an area of \(484 \; cm^{2}\). The same wire is now bent in the form of a circle. Find the area enclosed by the circle.**

**Sol:**

Area of the circle = \(484 \; cm^{2}\)

Area of the square = \(Side^{2}\)

\(\Rightarrow Side^{2} = 484\)

\(\Rightarrow Side^{2} = 22^{2}\)

\(\Rightarrow Side = 22\)

Perimeter of the square = \(\Rightarrow 4 \times Side = 4 \times 22 =88\)

Length of the wire = 88cm

Circumference of the circle = Length of the wire = 88cm

Now, let the radius of the circle be r cm.

Thus we have:

\( 2 \pi r = 88\)

\(\Rightarrow 2 \times \frac{22}{7} \times r = 88\)

\(\Rightarrow r = 14\)

Area of the circle = \(\pi r^{2} = \frac{22}{7} \times 14 \times 14\)

= 616 \(cm^{2}\)

Thus the area enclosed by the circle is 616 \(cm^{2}\)

**Q.5: A wire is bent in the form of an equilateral triangle encloses an area of \(121 \sqrt{3} cm^{2}\). The same wire is bent to form a circle. Find the area enclosed by the circle.**

**Sol:**

Area of an equilateral triangle = \(\frac{\sqrt{3}}{4} \times (Side)^{2}\)

\(\Rightarrow 121 \sqrt{3} = \frac{\sqrt{3}}{4} \times (Side)^{2}\)

\(\Rightarrow 121 \times 4 = (Side)^{2}\)

\(\Rightarrow Side = 22 \; cm\)

Perimeter of an equilateral triangle = \(3 \times Side\)

\(3 \times 22\)

= 66 cm

Length of the wire = 66 cm

Now let the radius of the circle be r cm.

We know,

Circumference of the circle = Length of the wire

\(2 \pi r = 66\)

\(2 \times \frac{22}{7} \times r = 66\)

\(r = \frac{66 \times 7}{2 \times 22}\)

\(r = \frac{21}{2 } = 10.5\)

Thus we have:

Area of the circle = \(\pi r^{2}\)

\(= \frac{22}{7} \times 10.5 \times 10.5\)

\(= 346.5 \; sq. cm\)

Area enclosed by the circles = \(= 346.5 \; sq. cm\)

**Q.6: The length of a chain used as the boundary of a semicircular park is 108 m. Find the area of the park.**

**Sol:**

Let the radius of the park be r.

Length of chain = Perimeter of the semicircular park

\(\Rightarrow 108 = \frac{1}{2} \times 2\pi r + 2r\)

\(\Rightarrow 108 = r \left ( \frac{22}{7} + 2 \right )\)

\(\Rightarrow 108 = \frac{36}{7} r\)

\(\Rightarrow r = 21 m\)

Now, area of park = \(\frac{1}{2} \times \pi r^{2} = \frac{1}{2} \times \frac{21}{7} \times (21)^{2}\)

\(= 693 \; m^{2}\)

Hence, the area of the park is \( 693 \; m^{2}\)

**Q.7: The sum of the radii of two circles is 7 cm, and the difference of their circumference is 8 cm. Find the circumference of the circles.**

**Sol:**

Let the radii of the circles be \(r_{1}\; cm \; and \; r_{2} \; cm\)

Now,

Sum of the radii of the two circles = 7 cm

\(r_{1} + r_{2} = 7\)

Difference of the circumference of the two circles = 8 cm

\(\Rightarrow 2 \pi r_{1} – 2 \pi r_{2} = 8\)

\(\Rightarrow 2 \pi (r_{1} – r_{2}) = 8\)

\(\Rightarrow (r_{1} – r_{2}) = \frac{8}{2\pi }\)

\(\Rightarrow (r_{1} – r_{2}) = \frac{8}{2 \times \frac{22}{7} }\)

\(\Rightarrow (r_{1} – r_{2}) = \frac{8 \times 7}{44 }\)

\(\Rightarrow (r_{1} – r_{2}) = \frac{56}{44 }\)

\(\Rightarrow (r_{1} – r_{2}) = \frac{14}{11}\)

Now, adding (i) and (ii)

\(2 r_{1} = \frac{91}{11}\)

\(r_{1} = \frac{91}{22}\)

Therefore, Circumference of the first circle = \(2 \pi r_{1}\)

\(= 2 \times \frac{22}{7} \times \frac{91}{22}\)

Also,

\(r_{1} – r_{2} = \frac{14}{11}\)

\(\frac{91}{22} – r_{2} = \frac{14}{11}\)

\(\frac{91}{22} – \frac{14}{11} = r_{2}\)

\(r_{2} = \frac{63}{22}\)

Therefore, Circumference of the second circle = \(2 \pi r_{2}\)

\(= 2 \times \frac{22}{7} \times \frac{63}{22}\)

Therefore, the circumference of the first and second circles are 18 cm and 26 cm respectively.

**Q.8: Find the area of a ring whose outer and inner radii are respectively 23 cm and 12 cm.**

**Sol:**

Let the radii of the circles be \(r_{1}\; cm \; and \; r_{2} \; cm \)

We have:

\(r_{1} = 23 \; cm \)

\(r_{2} = 12 \; cm \)

Now,

Area of the outer ring = \(\pi r_{1} ^{2}\)

\(= \frac{22}{7} \times 23 \times 23\)

\(= 1662.57 \; cm^{2} \)

Area of the inner ring = \(\pi r_{2} ^{2}\)

\(= \frac{22}{7} \times 12 \times 12 \)

\(= 452.57 \; cm^{2} \)

Area of the ring = Area of the outer ring – Area of the inner ring

= 1662.57 – 452.57\(= 1210\; cm^{2}\)

**Q.9: A path of 8 m width runs around the outside of a circumference park whose radius is 17 m. Find the area of the path.**

**Sol:**

The radius (r) of the inner circle is 17 m .

The radius (R) of the outer circle is 25 m . Â Â ( includes path i.e. 17+8)

Area of the path = \(\pi R^{2} – \pi r^{2})\)

= \(\pi (R^{2} – r^{2})\)

\(= \frac{22}{7} (25^{2} – 17^{2})\)

\(= \frac{22}{7} (25 – 17) (25 + 17)\)

\(= \frac{22}{7} (8) (42)\)

\(= 1056\; m^{2}\)

Therefore, Area of the path = 1056 \(m^{2}\)

**Q.10: A racetrack is in the form of a ring whose inner circumference is 352 m and outer circumference is 396 m. Find the width and the area of the track.**

**Sol:**

Let r and R be the radii of the inner and outer tracks.

Now,

Circumference of the outer track = \(2 \pi R\)

\(\Rightarrow 396 = 2 \times \frac{22}{7} \times R\)

\(\Rightarrow R = \frac{396 \times 7}{44}\)

\(\Rightarrow R = 63\)

Circumference of the inner track = \(2 \pi R\)

\(\Rightarrow 352 = 2 \times \frac{22}{7} \times r \)

\(\Rightarrow r = \frac{352 \times 7}{44}\)

\(\Rightarrow R = 56 \)

Width of the track = Radius of the outer track – Radius of the inner track

= 63 â€“ 56 = 7 m

Area of the outer circle = \(\pi R^{2}\)

= \(\frac{22}{7} \times 63 \times 63\)

\(= 12474 m^{2}\)

Area of the inner circle = \(\pi r^{2}\)

= \(\frac{22}{7} \times 56 \times 56\)

Area of the track = 12474 – 9856 = 2618 \(m^{2}\)