**Q.1: The side of a square is 10 cm. Find **

**(i) The area of the inscribed circle, and**

**(ii) The area of the circumscribed circle. [Take \(\pi = 3.14\) ]**

**Sol:**

**(i)** If a circle is inscribed in a square, then the side of the square is equal to the diameter of the circle.

Side of the square = 10 cm

Side = Diameter = 10

\(Therefore,\)

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

**(ii)** If a circle is circumscribed in a square, then the diagonal of the square is equal to the diameter of the circle.

Diagonal of the square = \(\sqrt{2} \times\)

\(= \sqrt{2} \times 10\)

\(= 10 \sqrt{2}\)

Diagonal = Diameter = \(= 10 \sqrt{2}\)

\(Therefore, r = 5 \sqrt{2}\)

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

\(= 3.14 \times (5\sqrt{2})^{2}\)

**Q.2: If a square is inscribed in a circle, find the ratio of the areas of the circle and the square.**

**Sol:**

If a square is inscribed in a circle, then the diagonals of the square are diameters of the circle.

Let the diagonal of the square be d cm.

Thus, we have:

Radius, r = \(\frac{d}{2}\)

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

We know:

d = \(\sqrt{2} \times Side\)

\(\Rightarrow Side = \frac{d}{\sqrt{2}} \; cm\)

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

\(= \left ( \frac{d}{\sqrt{2}} \right ) ^{2}\)

\(= \left ( \frac{d^{2}}{2} \right ) \; cm^{2}\)

Ratio of the area of the circle to that of the square :

\(= \frac{\pi \frac{d^{2}}{4}}{\frac{d^{2}}{2}}\)

Thus, the ratio of the area of the circle to that of the square is \(\pi : 2\)

**Q.3: The area of a circle inscribed in an equilateral triangle is 154 \(cm^{2}\). Find the perimeter of the triangle. [Take \(\sqrt{3} = 1.73\) ]**

**Sol:**

Let the radius of the inscribed circle be r cm.

Given:

Area of the circle = 154 \(cm^{2}\)

We know:

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

\(\Rightarrow 154 = \frac{22}{7} r^{2}\)

\(\Rightarrow r^{2} = \frac{154 \times 7}{22}\)

\(\Rightarrow r^{2} = 49\)

\(\Rightarrow r = 7\)

In a triangle, the centre of the inscribed circle is the point of intersection of the median and the altitudes of the triangle. The centroid divides the median of a triangle in the ratio 2:1.

Here, AO: OD = 2 : 1

Now, Let the altitude be h cm.

We have:

\(\angle ADB = 90^{\circ}\)

OD = \(\frac{1}{3} AD\)

\(OD = \frac{h}{3}\)

\(\Rightarrow h = 3r\)

\(\Rightarrow h = 21\)

Let each side of the triangle be a cm.

In the right- angled triangle ADB, we have:

\(AB^{2} = AD^{2} + DB^{2}\)

\(a^{2} = h^{2} + \left ( \frac{a}{2} \right )^{2}\)

\(3a^{2} = 4h^{2}\)

\(a^{2} = \frac{ 4h^{2}}{3}\)

\(a= \frac{2h}{\sqrt{3}}\)

\(a= \frac{42}{\sqrt{3}}\)

\(Therefore,\)

\(= 3 \times \frac{42}{\sqrt{3}}\)

\(= \sqrt{3} \times 42\)

**Q.4: The radius of the wheels of a vehicle is 42 cm. How many revolutions will it complete in a 19.8 km long journey?**

**Sol:**

Radius of the wheels = 42 cm

Circumference of the wheels = \(2 \pi r\)

\(2 \times \frac{22}{7} \times 42\)

Distance covered by the wheel in 1 revolution = 2.64 m

Total distance = 19.8 km or 19800 m

\(Therefore,\)

**Q.5: The wheels of the locomotive of a train are 2.1 m in radius. They make 75 revolutions in one minute. Find the speed of the train in km per hour.**

**Sol: **

Radius of the wheels = 2.1 m

Circumference of the wheels = \(2 \pi r\)

\(2 \times \frac{22}{7} \times 2.1\)

Distance covered by the wheels in 1 revolution = 13.2 m

Distance covered by the wheels in 75 revolutions \(= \left ( 13.2 \times 75 \right ) = \left ( 990 \times \frac{1}{1000} \right ) km\)

Distance covered by the wheel in 1 minute = Distance covered by the wheels in 75 revolutions \(= \frac{990}{1000}km\)

\(Therefore,\)

**Q.6: The wheels of a car makes 2500 revolutions in covering a distance of 4.95 km. Find the diameter of a wheel.**

**Sol:**

Distance = 4.95 km \(= 4.95 \times 1000 \times 100\)

\(Therefore,\)

\(\frac{4.95 \times 1000 \times 100}{2500} = 198\)

Now,

Circumference of the wheel = 198 cm

\(\Rightarrow 2 \pi r =198\)

\(\Rightarrow 2 \times \frac{22}{7} \times r =198\)

\(\Rightarrow r = \frac{198 \times 7}{44}\)

\(Therefore,\)

**Q.7: A boy is cycling in such a way that the wheels of his bicycle are making 14 revolutions per minute. If the diameter of a wheel is 60 cm, calculate the speed (in km/h) at which the boy is cycling.**

**Sol:**

Diameter of the wheel = 60 cm

\(Therefore,\)

Circumference of the wheels = \(2 \pi r\)

= \(2 \times \frac{22}{7} \times 30\)

Distance covered by the wheel in 1 revolution = \(\frac{1320}{7}\)

\(Therefore, \)

= \(\left ( \frac{1320 \times 140}{7 \times 100} \times \frac{1}{1000} \right ) \; km = \frac{264}{1000} \; km\)

Now,

Distance covered by the wheel in 1 minute = Distance covered by the wheel in 140 revolution \(= \frac{264}{1000} \; km\)

\(Therefore, \)

Hence, the speed at which the boy is cycling is 15.84 km/h

**Q.8: The diameter of the wheels of a bus is 140 cm. How many revolutions per minute do the wheels make when the bus is moving at a speed of 72.6 km/h.**

**Sol:**

Diameter of the wheels = 140 cm

Radius = 70 cm

Circumference = \(2 \pi r\)

\(= 2 \times \frac{22}{7} \times 70\)

Speed of the wheels = 72.6 km /h

Distance covered by the wheels in 1 minute = \(= \frac{76.2 \times 1000 \times 100}{60} = 121000\)

Number of revolutions made by the wheels in 1 minute = \(= \frac{Total \; distance \; covered}{Circumference }\)

\(= \frac{121000}{440}\)

Hence, the wheel makes 275 revolutions per minute.

**Q.9: The diameters of the front and rear wheels of a tractor are 80 cm and 2 m respectively. Find the number of revolutions that a rear wheel makes to cover the distance which the front wheel covers in 800 revolutions.**

**Sol:**

Radius of front wheel = 40 cm = \(\frac{2}{5}\)

Circumference of the front wheel = \(= \left ( 2 \pi \times \frac{2}{5} \right ) m = \frac{4 \pi}{5} \)

Distance covered by the front wheel in 800 revolutions = \(= \left ( \frac{4 \pi}{5}\times 800 \right ) m = \left (640 \pi \right ) m \)

Radius of the rear wheel = 1 m

Circumference of the rear wheel \(= \left (2 \pi \times 1\right ) = 2 \pi\; m\)

\(Therefore,\)

\(=\frac{640 \pi }{2 \pi}\)

**Q.10: Four equal circles are described about the four corners of a square so that each touches woe of the others, as shown in the figure. Find the area of the shaded region, if each side of the square measures 14 cm.**

** **

**Sol:**

Side of the square = 14 cm

Radius of the circle = \(= \frac{14}{2} = 7\)

Area of the quadrant of one circle \(= \frac{1}{4} \times \frac{22}{7} \times 7 \times 7\)

Area of the quadrant of four circles = \(38.5 \times 4 = 154 cm^{2}\)

Now,

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

Area of the shaded region = Area of the square -Area of the quadrants of the four circles = 196 – 154 = 42 \(cm^{2}\)

**Q.11: Four equal circles, each of radius 5 cm , touch each other, as shown in the figure. Find the area included between them. [Take \(\pi = 3.14\) ]**

**Sol:**

Radius = 5 cm

AB= BC= CD= AC = 10 cm

All sides are equal , so it is a square.

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

Area of a square = \(10^{2}= 100 \; cm^{2}\)

Area of the quadrant of one circle = \(= \frac{1}{4} \pi r^{2}\)

= \(= \frac{1}{4} \times \frac{22}{7} \times 5 \times 5\)

Area of the quadrant of four circles = \(19.64 \times 4 = 78.57 \; cm^{2}\)

Area of the shaded portion = Area of the square – Area of the quadrant of four circles

= 100 – 78.57 = 21.43 \(cm^{2}\)

**Q.12: Four circles, each of radius a units, touch each other. Show that the area between them is \(\left (\frac{6}{7} a^{2} \right )\) sq. units .**

**Sol:**

When four circles touches each other, their centres from the vertices of a square. The sides of the square are 2a units.

Area of the square = \((2a)^{2}= 4a^{2}\)

Area occupied by the four sectors = \(4 \times \frac{90}{360} \times \pi \times a^{2}\)

= \(\pi aa^{2}\)

Area between the circles = Area of the square – Area of the four sector

\(= \left ( 4 – \frac{22}{7} \right ) a^{2}\)

\(= \left ( \frac{6}{7} \right ) a^{2}\)

**Q.13: Three equal circles, each of radius 6 cm, touch one another as shown in the figure. Find the area enclosed between them. [\(\pi = 3.14 \; and \; \sqrt{3} =1.732\)]**

**Sol:**

Join ABC, all sides are equal, so it is the equilateral triangle.

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

\(= \frac{1.73}{4} \times 12 \times 12\)

= 62.28 \(cm^{2}\)

Area of the arc of the circle = \(= \frac{60}{360} \pi r^{2}\)

\(= \frac{1}{6} \times \frac{22}{7} \times 6 \times 6\)

Area of the three sectors \(= 3 \times 18.86 = 56.57 \; cm^{2}\)

Area of the shaded portion = Area of the triangle – Area of the three quadrant

= 62.28 – 56.57 = 5.71 \(cm^{2}\)

**Q.14: If three circles of radius a each, are drawn such that each touches the other two, prove that the area included between them is equal to \(\frac{4}{5} a^{2}\). [\(\pi = 3.14 \; and \; \sqrt{3} =1.732\) ]**

**Sol:**

When three circles touch each other, their centres from an equilateral triangle, with each side being 2a.

Area of the triangle = \(\frac{\sqrt{3}}{4} \times 2a \times 2a = \sqrt{3}a^{2}\)

Total area of the three sectors of circles \(=3 \times \frac{60}{360} \times \frac{22}{7} \times a^{2} = \frac{1}{2} \times \frac{22}{7} a^{2} = \frac{11}{7} a^{2}\)

Area of the region between the circles = Area of the triangle – Area of the three sectors

\(= \left ( \sqrt{3} – \frac{11}{7} \right ) a^{2}\)

\(= \left ( 1.73 – 1.57 \right ) a^{2}\)

\(= 0.16 a^{2}\)

\(= \frac{4}{25} a^{2}\)

**Q.15: In the given figure, ABCD is a trapezium of area 24.5 \(cm^{2}\). If \(AD \parallel BC\), \(\angle DAB = 90^{\circ}\), AD = 10 cm and ABE is quadrant of a circle then find the area of the shaded region.**

** **

**Sol:**

Area of trapezium = \(\frac{1}{2} (AD + BC) \times AB\)

\(\Rightarrow 24.5 = \frac{1}{2} (10+4) \times AB\)

\(\Rightarrow AB = 3.5\)

Area of the shaded region = Area of trapezium ABCD – Area of quadrant ABE

= 24.5 – \(\frac{1}{4} \pi (AB)^{2}\)

\(= 24.5 – \frac{1}{4} \times \frac{22}{7} \times (3.5)^{2}\)

= 24.5 – 9.625 = 14.875 \(cm^{2}\)

Hence, the area of shaded region is 14.875 \(cm^{2}\)

**Q.16: ABCD is a field in the shape of trapezium \(AD \parallel BC, \angle ABC = 90^{\circ} \;and \; \angle ADC = 60^{\circ}\). Four sectors are formed with centres A,B,C and D, as shown in the figure. The radius of each sector is 14 m. Find the following **

**(i) Total area of the four sectors,**

**(ii) Area of the remaining portion, given that AD = 55 m , BC = 45 m and AB = 30 m.**

** **

**Sol:**

**(i)** Area of four sectors = Area of sector having central angle \( 60^{\circ} \)

\(\frac{60^{\circ}}{360^{\circ}} .\pi (14)^{2} + \frac{90^{\circ}}{360^{\circ}}. \pi (14)^{2} + \frac{90^{\circ}}{360^{\circ}} .\pi (14)^{2} + \frac{120^{\circ}}{360^{\circ}} .\pi (14)^{2}\)

\(\left ( \frac{60^{\circ}}{360^{\circ}} + \frac{90^{\circ}}{360^{\circ}} + \frac{90^{\circ}}{360^{\circ}} + \frac{120^{\circ}}{360^{\circ}} \right ) \pi (14)^{2}\)

\(\left ( \frac{360^{\circ}}{360^{\circ}} \right ) .\pi (14)^{2}\)

\(= 616 m^{2}\)

**(ii)** Area of the remaining portion = Area of trapezium ABCD – Area of four quadrants

\(= \frac{1}{2}(AD+BC) \times AB – 616\)

\(= \frac{1}{2}(55+45) \times 30 – 616\)

= 1500 – 616 = 884 \(m^{2}\)

**Q.17: Find the area of the shaded region in the given figure, where a circulation arc of radius 6 cm has been drawn with the vertex of an equilateral triangle of side 12 cm as the centre and a sector of the circle of radius 6 cm with centre B is made. [ Use \(\sqrt {3}=1.73 \; and \; \pi = 3.14\)]**

** **

**Sol:**

In equilateral triangle all the angles are of \(60^{\circ}\)

\(Therefore, \angle ABO = \angle AOB = 60^{\circ}\)

Area of the shaded region = ( Area of triangle – Area of sector having central angle \(60^{\circ}\)

= \(\frac{\sqrt{3}}{4}(AB)^{2} – \frac{60}{360} . \pi (6)^{2} + \frac{300}{360}. \pi (6)^{2}\)

\(= \frac{\sqrt{1.73}}{4}(12)^{2} – \frac{1}{6}\times 3.14 (6)^{2} + \frac{5}{6} \times 3.14 (6)^{2}\)

= 62.28 – 18.84 + 94.2

= 137.64 \(cm^{2}\)

Hence, the area of shaded region is 137.64 \(cm^{2}\)

**Q.18: In the given figure, ABCD is a rectangle with AB = 80 cm and BC = 70 cm,\(\angle AED = 90^{\circ}\) and DE = 42 cm. A semicircle is drawn, taking BC as diameter. Find the area of the shaded region.**

** **

**Sol:**

We know that the opposite sides of a rectangle are equal.

AD= BC = 70 cm

In right triangle AED

\(AE^{2}= AD^{2}- DE^{2}\)

\(AE^{2}= 70^{2}- 12^{2}\)

\(= 4900 – 1764\)

\(\Rightarrow AE= 56\)

Area of the shaded region = Area of rectangle – (Area of triangle + Area of semicircle )

= \(AB \times BC – \left [ \frac{1}{2} \times AE \times DE +\frac{1}{2} \pi \left ( \frac{BC}{2} \right )^{2}\right ]\)

\(= 80 \times 70 – \left [ \frac{1}{2} \times 56 \times 42 +\frac{1}{2} \times \frac{22}{7} \times \left ( \frac{70}{2} \right )^{2} \right ]\)

= 5600 – 3101 = 2499 \(cm^{2}\)

Hence, the area of shaded region is 2499 \(cm^{2}\)