Set Alpha – Formulas

Set A

Rectangle

\(\text{Perimeter} = 2(l + b)\)

\(\text{Area} = l \times b\)

Trapezium

\(\text{Area} = \frac{1}{2} \times (a + b) \times h\)

Parallelogram

\(\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}\)

where \( s = \frac{a + b + c}{2} \)

Circles

\(\text{Circumference} = 2 \pi r\)

\(\text{Area} = \pi r^2\)

Ratio of Areas

\(\left( \frac{r_1}{r_2} \right)^2\)

Set B

Right Triangle

\(\text{Area of the largest inscribed square} = \frac{ab}{a+b}\)

Square

\(\text{Perimeter after cutting} = \text{Original perimeter} + 4 \times \text{side of the cut square}\)

Trapezium

\(\text{Area} = \frac{1}{2} \times (a + b) \times h\)

Circle

\(\text{Area} = \pi r^2\)

Set C

Rhombus

\(\text{Area} = \frac{1}{2} \times d_1 \times d_2\)

\(a = \frac{1}{2} \sqrt{d_1^2 + d_2^2}\)

Right Triangle

\(\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}\)

Quadrilateral

\(\text{Area} = \text{Area of right triangle} + \text{Area using Heron’s formula}\)

\(\text{Area using Heron’s formula} = \sqrt{s(s-a)(s-b)(s-c)}\)

where \( s = \frac{a + b + c}{2} \)

Trapezium

\(\text{Area} = \frac{1}{2} \times (a + b) \times h\)

Increase in Area

\(\text{Increase Percentage} = \left( \frac{\text{New Area} – \text{Original Area}}{\text{Original Area}} \right) \times 100\)

Set D

Heron’s Formula

\(\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}\)

where \( s = \frac{a + b + c}{2} \)

Semicircle

\(\text{Area} = \frac{1}{2} \pi r^2\)

Equilateral Triangle

\(\text{Area} = \frac{\sqrt{3}}{4} s^2\)

Circle Inscribed in Triangle

\(r = \frac{s \sqrt{3}}{6}\)

Sector

\(L = \frac{\theta}{360} \times 2\pi r\)

\(\text{Perimeter} = 2r + L\)

Rhombus

\(a = \frac{1}{2} \sqrt{d_1^2 + d_2^2}\)

Set E

Circle and Semicircle

\(\text{Area of semicircle} = \frac{1}{2} \pi R^2\)

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

Right Triangle

\(\text{Area} = \frac{1}{2} \times \text{hypotenuse} \times \text{altitude}\)

Equilateral Triangle

\(\text{Area} = \frac{\sqrt{3}}{4} s^2\)

Sector

\(L = \frac{\theta}{360} \times 2\pi r\)

\(\text{Perimeter} = 2r + L\)

Right Triangle

\(\text{Area} = \frac{1}{2} \times \text{hypotenuse} \times \text{altitude}\)

Triangle formed by Circles

\(\text{Area} = \frac{\sqrt{3}}{4} s^2\)

Wire bent into Circle

\(\text{Circumference} = 2\pi r\)

\(\text{Area} = \pi r^2\)

Distance Travelled by Clock Hands

\(\text{Distance} = 2 \pi \times \text{length of hand}\)

Triangle Inscribed Circle

\(r = \frac{A}{s}\)

\(A = \sqrt{s(s-a)(s-b)(s-c)}\)

where \( s = \frac{a + b + c}{2} \)

Isosceles Right Triangle Inscribed Circle

\(r = \frac{a + b – \text{hypotenuse}}{2}\)

\(\text{Area} = \pi r^2\)

Intersecting Circles

\(\text{Area of sector} = \frac{1}{4} \pi r^2\)

\(\text{Area of triangle} = \frac{1}{2} \times \text{base} \times \text{height}\)

Touching Circles

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

Set A

Problem 4

The difference between the length and breadth of a rectangle is 23 m. If the perimeter is 206 m, then calculate the area.

Let the length be \( l \) and the breadth be \( b \).

\[ \begin{aligned} l – b &= 23 \quad \text{…(i)} \\ 2l + 2b &= 206 \quad \text{…(ii)} \end{aligned} \]

From equation (ii), we have:

\[l + b = 103 \quad \text{…(iii)}\]

Adding equations (i) and (iii):

\[ \begin{aligned} (l – b) + (l + b) &= 23 + 103 \\ 2l &= 126 \\ l &= 63 \, \text{m} \end{aligned} \]

Using equation (iii):

\[ \begin{aligned} 63 + b &= 103 \\ b &= 40 \, \text{m} \end{aligned} \]

Area \( A = l \times b = 63 \times 40 = 2520 \, \text{sq.m} \)

Problem 5

The perimeter of a trapezium is 52 cm and its non-parallel sides are each equal to 10 cm, and its altitude is 8 cm. Find its area.

Let the lengths of the parallel sides be \( a \) and \( b \).

Given:

\[ a + b + 10 + 10 = 52 \\ a + b + 20 = 52 \\ a + b = 32 \]

Using the area formula for a trapezium:

\(\text{Area} = \frac{1}{2} \times (a + b) \times \text{height} \\ \text{Area} = \frac{1}{2} \times 32 \times 8 = 128 \, \text{sq.cm}\)

Problem 6

The adjacent sides of a parallelogram are 8 cm and 9 cm. The diagonal joining the ends of these sides is 13 cm. Find its area.

Using the formula for the area of a parallelogram with sides \( a \) and \( b \) and diagonal \( d \):

\(\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}\)

where \( s = \frac{a + b + c}{2} \) and \( c \) is the diagonal.

Given:

\[ a = 8 \, \text{cm}, \, b = 9 \, \text{cm}, \, c = 13 \, \text{cm} \\ s = \frac{8 + 9 + 13}{2} = 15 \]

\[ \begin{aligned} \text{Area} &= \sqrt{15(15-8)(15-9)(15-13)} \\ &= \sqrt{15 \times 7 \times 6 \times 2} \\ &= \sqrt{1260} \approx 35.5 \, \text{sq.cm} \end{aligned} \]

Problem 7

The ratio between the length and the perimeter of a rectangular plot is 1:3 and the ratio between the breadth and perimeter of that plot is 1:6. What is the ratio between the length and area of that plot?

Let the length be \( l \) and the breadth be \( b \).

Given:

\[ \frac{l}{P} = \frac{1}{3} \\ \frac{b}{P} = \frac{1}{6} \]

where \( P \) is the perimeter.

\[P = 2l + 2b\]

Using the given ratios:

\[l = \frac{P}{3} \\ b = \frac{P}{6}\]

\[ \begin{aligned} P &= 2 \left( \frac{P}{3} \right) + 2 \left( \frac{P}{6} \right) \\ P &= \frac{2P}{3} + \frac{P}{3} \\ P &= P \end{aligned} \]

Area \( A = l \times b = \frac{P}{3} \times \frac{P}{6} = \frac{P^2}{18} \)

Ratio between length and area:

\(\frac{l}{A} = \frac{\frac{P}{3}}{\frac{P^2}{18}} = \frac{18}{3P} = \frac{6}{P} \)

Problem 8

If the ratio of circumference of two circles is 4:9, find the ratio of their areas.

Let the radii be \( r_1 \) and \( r_2 \).

Given:

\[ \frac{C_1}{C_2} = \frac{4}{9} \\ \frac{2 \pi r_1}{2 \pi r_2} = \frac{4}{9} \\ \frac{r_1}{r_2} = \frac{4}{9} \]

Area ratio:

\(\frac{A_1}{A_2} = \frac{\pi r_1^2}{\pi r_2^2} = \left( \frac{r_1}{r_2} \right)^2 = \left( \frac{4}{9} \right)^2 = \frac{16}{81} \)

Problem 9

A wire bent in the form of a circle of radius 42 cm is cut and again bent in the form of a square. Find the ratio of the regions enclosed by the circle and the square in the two cases.

Circumference of the circle:

\[ C = 2 \pi r = 2 \pi \times 42 = 84 \pi \]

Side of the square:

\[ 4s = 84 \pi \\ s = \frac{84 \pi}{4} = 21 \pi \]

Area of the circle:

\[ A_{\text{circle}} = \pi r^2 = \pi \times 42^2 = 1764 \pi \]

Area of the square:

\[ A_{\text{square}} = s^2 = (21 \pi)^2 = 441 \pi^2 \]

Ratio of areas:

\(\frac{A_{\text{circle}}}{A_{\text{square}}} = \frac{1764 \pi}{441 \pi^2} = \frac{1764}{441 \pi} = \frac{4}{\pi} \)

Set B

Problem 10

In the adjoining figure, find the radius of the inner circle if other circles are of radii 1 m.

Since the outer circles each have a radius of 1 m, and they are touching each other, the radius of the inner circle is the same, 1 m.

Problem 11

The sides of a right triangle other than the hypotenuse are 4 cm and 8 cm. Find the area (in cm²) of the largest square inscribed in the triangle.

The largest square inscribed in a right triangle with legs \(a\) and \(b\) has a side \(s\) given by:

\(\frac{ab}{a+b}\)

Given:

\[ a = 4 \, \text{cm}, \, b = 8 \, \text{cm} \]

\(s = \frac{4 \times 8}{4 + 8} = \frac{32}{12} = \frac{8}{3} \, \text{cm}\)

Area of the square:

\( \text{Area} = s^2 = \left(\frac{8}{3}\right)^2 = \frac{64}{9} \approx 7.11 \, \text{cm}^2\)

Problem 12

From one corner of a square of side 8 cm, a small square of side 1 cm is cut off. What is the perimeter of the remaining figure?

The perimeter of the original square:

\[ P = 4 \times 8 = 32 \, \text{cm} \]

After cutting off a small square, the perimeter increases by the perimeter of the small square:

\[ P_{\text{new}} = 32 + 4 \times 1 = 32 + 4 = 36 \, \text{cm} \]

Problem 13

Area of a trapezium is 91 cm² and its height is 7 cm. If one of the parallel sides is longer than the other side by 8 cm, find the length (in cm) of the smaller side.

Let the lengths of the parallel sides be \( a \) and \( b \).

Given:

\(\text{Area} = \frac{1}{2} \times (a + b) \times h \\ 91 = \frac{1}{2} \times (a + b) \times 7 \)

Simplifying, we get:

\[ 91 = \frac{7}{2} \times (a + b) \\ 182 = 7(a + b) \\ a + b = 26 \]

Let \( a = b + 8 \).

Substitute \( a \) in the equation:

\[ b + 8 + b = 26 \\ 2b + 8 = 26 \\ 2b = 18 \\ b = 9 \, \text{cm} \]

Thus, \( a = 9 + 8 = 17 \, \text{cm} \).

Set C

Problem 29

Find the area of the shaded region.

Given:

\[ AB = 16 \, \text{cm}, \, BC = 48 \, \text{cm}, \, AC = 52 \, \text{cm} \]

We need to find the area of the triangle \( ABC \).

Using Heron’s formula:

\(s = \frac{a + b + c}{2} = 58 \, \text{cm}\)

\[ \text{Area} = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{58(58-16)(58-48)(58-52)} = \sqrt{58 \times 42 \times 10 \times 6} \]

Calculating the area:

\(\text{Area} = \sqrt{146160} \approx 382 \, \text{sq.cm}\)

The area of the shaded region is \( 382 \, \text{sq.cm} \).

Problem 30

The perimeter of a triangle is 540 cm and its sides are in the ratio 13:12:5. Find the length of altitude to the smallest side.

Given:

\(a + b + c = 540 \, \text{cm}\)

\(a : b : c = 13 : 12 : 5\)

Let the sides be:

\(a = 13k, \, b = 12k, \, c = 5k\)

Given the perimeter:

\[ 13k + 12k + 5k = 540 \\ 30k = 540 \\ k = 18 \]

Length of the sides:

\[ a = 13 \times 18 = 234 \, \text{cm} \\ b = 12 \times 18 = 216 \, \text{cm} \\ c = 5 \times 18 = 90 \, \text{cm} \]

The smallest side is \( c = 90 \, \text{cm} \).

Area of the triangle:

\[ s = \frac{a + b + c}{2} = 270 \, \text{cm} \]

Using Heron’s formula:

\[ \text{Area} = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{270(270-234)(270-216)(270-90)} = \sqrt{270 \times 36 \times 54 \times 180} \]

Calculating the area:

\[ \text{Area} = \sqrt{2624400} \approx 1620 \, \text{sq.cm} \]

Altitude \( h \) to the smallest side \( c \):

\[ \text{Area} = \frac{1}{2} \times c \times h \\ 1620 = \frac{1}{2} \times 90 \times h \\ h = \frac{1620 \times 2}{90} = 36 \, \text{cm} \]

Problem 31

Find the area of a rhombus whose perimeter is 80 cm and one diagonal is 24 cm.

Given:

\[ \text{Perimeter} = 4a = 80 \\ a = 20 \, \text{cm} \]

Let the diagonals be \( d_1 \) and \( d_2 \). Given:

\(d_1 = 24 \, \text{cm}\)

Using the relation between sides and diagonals of a rhombus:

\[ a = \frac{1}{2} \sqrt{d_1^2 + d_2^2} \\ 20 = \frac{1}{2} \sqrt{24^2 + d_2^2} \\ 40 = \sqrt{576 + d_2^2} \\ 1600 = 576 + d_2^2 \\ d_2^2 = 1024 \\ d_2 = 32 \, \text{cm} \]

Area of the rhombus:

\[ \text{Area} = \frac{1}{2} \times d_1 \times d_2 = \frac{1}{2} \times 24 \times 32 = 384 \, \text{sq.cm} \]

Problem 32

In the given figure \( AC = CD = DB \) and semicircles are drawn with diameters \( AB, BC \), and \( AC \). If the radius of the largest semicircle is 6 cm, then find the area and perimeter of the shaded region in terms of \(\pi\).

Given:

\[ \text{Radius of the largest semicircle} = 6 \, \text{cm} \\ \text{Diameter} = 2 \times 6 = 12 \, \text{cm} \]

Radius of the smaller semicircles:

\(\text{Radius} = \frac{12}{3} = 4 \, \text{cm}\)

Area of the shaded region:

\[ \text{Area} = \frac{1}{2} \pi (6^2) – 2 \times \frac{1}{2} \pi (4^2) \\ = \frac{1}{2} \pi \times 36 – \pi \times 16 \\ = 18\pi – 16\pi = 2\pi \, \text{sq.cm} \]

Perimeter of the shaded region:

\[ \text{Perimeter} = \frac{1}{2} \times 2\pi \times 6 + 2 \times \frac{1}{2} \pi \times 4 \\ = 6\pi + 4\pi = 10\pi \, \text{cm} \]

Problem 33

The perimeter of a semicircular protractor is 36 cm, find its area.

Given:

\[ \text{Perimeter} = \pi r + 2r = 36 \, \text{cm} \\ r(\pi + 2) = 36 \\ r = \frac{36}{\pi + 2} \]

Area of the semicircle:

\[ \text{Area} = \frac{1}{2} \pi r^2 \\ = \frac{1}{2} \pi \left( \frac{36}{\pi + 2} \right)^2 \\ = \frac{1}{2} \pi \left( \frac{1296}{(\pi + 2)^2} \right) \\ = \frac{648 \pi}{(\pi + 2)^2} \, \text{sq.cm} \]

Problem 34

The length of a minute hand of a clock is 14 cm. Find the area swept by it in 5 minutes.

Given:

\(\text{Length of minute hand} = 14 \, \text{cm}\)

Area swept in 60 minutes (full circle):

\[ A_{\text{full}} = \pi r^2 = \pi \times 14^2 = 196\pi \, \text{sq.cm} \]

Area swept in 5 minutes:

\[ A_{\text{5 min}} = \frac{5}{60} \times 196\pi = \frac{1}{12} \times 196\pi = \frac{196\pi}{12} = \frac{49\pi}{3} \, \text{sq.cm} \]

Problem 35

A chord of a circle of radius 12 cm subtends an angle 120° at the centre. Find the area of the minor segment.

Given:

\(r = 12 \, \text{cm}, \, \theta = 120^\circ = \frac{120 \pi}{180} = \frac{2\pi}{3} \, \text{radians}\)

Area of the sector:

\[ A_{\text{sector}} = \frac{1}{2} r^2 \theta = \frac{1}{2} \times 12^2 \times \frac{2\pi}{3} = 48\pi \, \text{sq.cm} \]

Area of the triangle:

\[ A_{\text{triangle}} = \frac{1}{2} r^2 \sin \theta = \frac{1}{2} \times 12^2 \times \sin 120^\circ = \frac{1}{2} \times 144 \times \frac{\sqrt{3}}{2} = 36\sqrt{3} \, \text{sq.cm} \]

Area of the minor segment:

\[ A_{\text{minor segment}} = A_{\text{sector}} – A_{\text{triangle}} = 48\pi – 36\sqrt{3} \, \text{sq.cm} \]

Problem 36

In the given figure, if the radius of the circle is 28 cm, find the shaded area.

Given:

\(\text{Radius of the circle} = 28 \, \text{cm}\)

Area of the circle:

\[ A_{\text{circle}} = \pi r^2 = \pi \times 28^2 = 784\pi \, \text{sq.cm} \]

Area of the hexagon:

\[ \text{Side length of hexagon} = r = 28 \, \text{cm} \]

\[ A_{\text{hexagon}} = \frac{3\sqrt{3}}{2} s^2 = \frac{3\sqrt{3}}{2} \times 28^2 = \frac{3\sqrt{3}}{2} \times 784 = 1176\sqrt{3} \, \text{sq.cm} \]

Area of the shaded region:

\[ A_{\text{shaded}} = A_{\text{circle}} – A_{\text{hexagon}} = 784\pi – 1176\sqrt{3} \, \text{sq.cm} \]

Problem 37

ABCD is a rectangle inscribed in a circle. If \( AB = 12 \, \text{cm} \) and \( BC = 5 \, \text{cm} \), find the area of the shaded region. (Use \( \pi = 3.14 \))

Given:

\[ AB = 12 \, \text{cm}, \, BC = 5 \, \text{cm} \]

Diagonal of the rectangle (which is the diameter of the circle):

\[ d = \sqrt{AB^2 + BC^2} = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \, \text{cm} \]

Radius of the circle:

\[ r = \frac{d}{2} = \frac{13}{2} = 6.5 \, \text{cm} \]

Area of the circle:

\[ A_{\text{circle}} = \pi r^2 = 3.14 \times 6.5^2 = 3.14 \times 42.25 = 132.665 \, \text{sq.cm} \]

Area of the rectangle:

\[ A_{\text{rectangle}} = AB \times BC = 12 \times 5 = 60 \, \text{sq.cm} \]

Area of the shaded region:

\[ A_{\text{shaded}} = A_{\text{circle}} – A_{\text{rectangle}} = 132.665 – 60 = 72.665 \, \text{sq.cm} \]

Set E

Problem 38

In the given figure, the diameter of the biggest semicircle is 108 cm, and the smallest circle touches the three semicircles. Find the area of the shaded region.

Given:

\[ \text{Diameter of the biggest semicircle} = 108 \, \text{cm} \\ \text{Radius of the biggest semicircle} = \frac{108}{2} = 54 \, \text{cm} \]

The radius of the smallest circle is one-third of the radius of the biggest semicircle:

\[ \text{Radius of the smallest circle} = \frac{54}{3} = 18 \, \text{cm} \]

Area of the biggest semicircle:

\[ A_{\text{big semicircle}} = \frac{1}{2} \pi R^2 = \frac{1}{2} \pi (54)^2 = 1458 \pi \, \text{sq.cm} \]

Area of three smaller semicircles:

\[ A_{\text{3 small semicircles}} = 3 \times \frac{1}{2} \pi r^2 = \frac{3}{2} \pi (18)^2 = 486 \pi \, \text{sq.cm} \]

Total area of the shaded region:

\[ A_{\text{shaded}} = A_{\text{big semicircle}} – A_{\text{3 small semicircles}} = 1458 \pi – 486 \pi = 972 \pi \, \text{sq.cm} \]

Problem 39

In the given figure, PR is the diameter, PQ = 24 cm, QR = 7 cm. Find the area of the shaded region.

Given:

\[ PQ = 24 \, \text{cm}, \, QR = 7 \, \text{cm}, \, PR = \text{diameter} = 25 \, \text{cm} \]

Radius of the semicircle:

\[ r = \frac{25}{2} = 12.5 \, \text{cm} \]

Area of the semicircle:

\[ A_{\text{semicircle}} = \frac{1}{2} \pi r^2 = \frac{1}{2} \pi (12.5)^2 = 78.125 \pi \, \text{sq.cm} \]

Area of the triangle \( \Delta PQR \):

\[ \text{Area} = \frac{1}{2} \times PQ \times QR = \frac{1}{2} \times 24 \times 7 = 84 \, \text{sq.cm} \]

Area of the shaded region:

\[ A_{\text{shaded}} = A_{\text{semicircle}} – A_{\text{triangle}} = 78.125 \pi – 84 \, \text{sq.cm} \]

Problem 40

In an equilateral triangle of side 24 cm, a circle is inscribed touching its sides. Find the area of the remaining portion of the triangle. (Use \( \pi = \frac{22}{7} \) and \( \sqrt{3} = 1.732 \))

Given:

\[ \text{Side of the equilateral triangle} = 24 \, \text{cm} \]

Area of the equilateral triangle:

\[ A_{\text{triangle}} = \frac{\sqrt{3}}{4} s^2 = \frac{\sqrt{3}}{4} \times 24^2 = \frac{\sqrt{3}}{4} \times 576 = 144 \sqrt{3} \, \text{sq.cm} \approx 144 \times 1.732 = 249.408 \, \text{sq.cm} \]

Radius of the inscribed circle:

\[ r = \frac{s \sqrt{3}}{6} = \frac{24 \sqrt{3}}{6} = 4 \sqrt{3} \, \text{cm} \]

Area of the inscribed circle:

\[ A_{\text{circle}} = \pi r^2 = \pi (4 \sqrt{3})^2 = 48 \pi \, \text{sq.cm} \approx 48 \times \frac{22}{7} = 150.857 \, \text{sq.cm} \]

Area of the remaining portion:

\[ A_{\text{remaining}} = A_{\text{triangle}} – A_{\text{circle}} = 249.408 – 150.857 \approx 98.551 \, \text{sq.cm} \]

Problem 41

The angle of a sector is 60° and its radius is 10.5 cm. Find the perimeter of the sector.

Given:

\[ \theta = 60^\circ, \, r = 10.5 \, \text{cm} \]

Length of the arc:

\[ L = \frac{\theta}{360} \times 2\pi r = \frac{60}{360} \times 2\pi \times 10.5 = \frac{1}{6} \times 21\pi = 3.5\pi \, \text{cm} \]

Perimeter of the sector:

\[ P = 2r + L = 21 + 3.5\pi \, \text{cm} \approx 21 + 3.5 \times 3.14 = 21 + 10.99 = 31.99 \, \text{cm} \]

Problem 42

Find the area of the right triangle if the radius of its circumcircle is 5 cm and altitude drawn to the hypotenuse is 4 cm.

Given:

\[ \text{Radius of circumcircle} = 5 \, \text{cm}, \, \text{Altitude} = 4 \, \text{cm} \]

Diameter of the circumcircle (which is the hypotenuse of the triangle):

\[ \text{Hypotenuse} = 2 \times 5 = 10 \, \text{cm} \]

Area of the triangle:

\[ \text{Area} = \frac{1}{2} \times \text{Hypotenuse} \times \text{Altitude} = \frac{1}{2} \times 10 \times 4 = 20 \, \text{sq.cm} \]

Problem 43

Three equal circles, each of radius 7 cm, touch each other. Find

i) area between the three circles

ii) perimeter of the region between the three circles.

Given:

\[ r = 7 \, \text{cm} \]

The distance between the centers of any two circles is equal to \(2r = 14 \, \text{cm}\). The area of the equilateral triangle formed by joining the centers:

\[ A_{\text{triangle}} = \frac{\sqrt{3}}{4} s^2 = \frac{\sqrt{3}}{4} \times 14^2 = \frac{\sqrt{3}}{4} \times 196 = 49 \sqrt{3} \, \text{sq.cm} \]

Area of one circle sector within the triangle:

\[ A_{\text{sector}} = \frac{1}{6} \pi r^2 = \frac{1}{6} \pi (7)^2 = \frac{49 \pi}{6} \, \text{sq.cm} \]

Total area of the three circle sectors:

\[ A_{\text{3 sectors}} = 3 \times \frac{49 \pi}{6} = \frac{49 \pi}{2} \, \text{sq.cm} \]

Area between the three circles:

\[ A_{\text{between}} = A_{\text{triangle}} – A_{\text{3 sectors}} = 49 \sqrt{3} – \frac{49 \pi}{2} \, \text{sq.cm} \]

Perimeter of the region between the three circles:

\[ \text{Perimeter} = 3 \times \left(\frac{2\pi r}{6}\right) = \pi r = \pi \times 7 = 7 \pi \, \text{cm} \]

Problem 44

A copper wire, when bent in the form of a square, encloses an area 484 cm². If the same wire is bent in the form of a circle, find the area of the circle.

Given:

\[ A_{\text{square}} = 484 \, \text{sq.cm} \]

Side of the square:

\[ s = \sqrt{484} = 22 \, \text{cm} \]

Perimeter of the square (length of the wire):

\[ P = 4s = 4 \times 22 = 88 \, \text{cm} \]

If the wire is bent in the form of a circle:

\[ C = 88 \\ 2\pi r = 88 \\ r = \frac{88}{2\pi} = \frac{44}{\pi} \]

Area of the circle:

\[ A_{\text{circle}} = \pi r^2 = \pi \left(\frac{44}{\pi}\right)^2 = \frac{1936}{\pi} \, \text{sq.cm} \]

Problem 45

Bicycle wheel makes 5000 revolutions in moving 11 km. Find the diameter of the wheel.

Given:

\[ \text{Distance} = 11 \, \text{km} = 11000 \, \text{m} \]

Each revolution covers a distance equal to the circumference of the wheel:

\[ C = \frac{11000}{5000} = 2.2 \, \text{m} \]

Diameter of the wheel:

\[ d = \frac{C}{\pi} = \frac{2.2}{\pi} \approx \frac{2.2}{3.14} \approx 0.7 \, \text{m} = 70 \, \text{cm} \]

Problem 46

The short and long hand of a clock are 4 cm and 6 cm respectively. Find the sum of the distances travelled by their tips in 2 days (\( \pi = \frac{22}{7} \)).

Given:

\[ \text{Short hand} = 4 \, \text{cm}, \, \text{Long hand} = 6 \, \text{cm} \]

Distance travelled by the short hand in 1 day:

\[ \text{Distance}_{\text{short}} = 2 \pi \times 4 = 8\pi \, \text{cm} \]

Distance travelled by the long hand in 1 day:

\[ \text{Distance}_{\text{long}} = 2 \pi \times 6 = 12\pi \, \text{cm} \]

Total distance in 2 days:

\[ \text{Total distance} = 2 \times (\text{Distance}_{\text{short}} + \text{Distance}_{\text{long}}) = 2 \times 20\pi = 40\pi \, \text{cm} \]

Using \( \pi = \frac{22}{7} \):

\[ 40\pi = 40 \times \frac{22}{7} = \frac{880}{7} \approx 125.71 \, \text{cm} \]

Problem 47

A triangle whose sides are respectively 5 cm, 5 cm, and 8 cm. Find the area of the circle inscribed in it. (\( \pi = 3.14 \))

Given:

\[ a = 5 \, \text{cm}, \, b = 5 \, \text{cm}, \, c = 8 \, \text{cm} \]

Using Heron’s formula:

\[ s = \frac{a + b + c}{2} = 9 \, \text{cm} \]

Area of the triangle:

\[ A = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{9(9-5)(9-5)(9-8)} = \sqrt{9 \times 4 \times 4 \times 1} = \sqrt{144} = 12 \, \text{sq.cm} \]

Radius of the inscribed circle:

\[ r = \frac{A}{s} = \frac{12}{9} = \frac{4}{3} \, \text{cm} \]

Area of the inscribed circle:

\[ A_{\text{circle}} = \pi r^2 = \pi \left(\frac{4}{3}\right)^2 = \pi \times \frac{16}{9} = \frac{16\pi}{9} \approx \frac{16 \times 3.14}{9} \approx 5.58 \, \text{sq.cm} \]

Problem 48

If the hypotenuse of an isosceles right triangle is \( 14\sqrt{2} \) cm, find the area of the circle inscribed in it (\( \pi = \frac{22}{7} \)).

Given:

\[ \text{Hypotenuse} = 14\sqrt{2} \, \text{cm} \]

The legs of the isosceles right triangle are:

\[ a = b = 14 \, \text{cm} \]

Area of the triangle:

\[ A = \frac{1}{2} a^2 = \frac{1}{2} \times 14^2 = 98 \, \text{sq.cm} \]

Radius of the inscribed circle:

\[ r = \frac{a + b – \text{hypotenuse}}{2} = \frac{14 + 14 – 14\sqrt{2}}{2} = \frac{28 – 14\sqrt{2}}{2} = 14 – 7\sqrt{2} \]

Area of the inscribed circle:

\[ A_{\text{circle}} = \pi r^2 = \pi (14 – 7\sqrt{2})^2 = \pi (196 – 196\sqrt{2} + 98) = \pi (294 – 196\sqrt{2}) \approx 3.14 \times 294 \approx 923.16 \, \text{sq.cm} \]

Problem 49

In the given figure, two circles with centres P and Q intersect at A and B. If AP = 4 cm, \( \angle APB = 90^\circ \), and \( \angle AQP = 60^\circ \), find the area of the shaded region.

Given:

\[ AP = 4 \, \text{cm}, \, \angle APB = 90^\circ, \, \angle AQP = 60^\circ \]

Area of sector \( APB \):

\[ A_{\text{sector APB}} = \frac{1}{4} \pi (4)^2 = 4\pi \, \text{sq.cm} \]

Area of sector \( AQP \):

\[ A_{\text{sector AQP}} = \frac{1}{6} \pi (4\sqrt{2})^2 = \frac{1}{6} \pi (32) = \frac{32\pi}{6} = \frac{16\pi}{3} \, \text{sq.cm} \]

Area of the triangle \( APB \):

\[ A_{\text{triangle APB}} = \frac{1}{2} AP \times PB = \frac{1}{2} \times 4 \times 4 = 8 \, \text{sq.cm} \]

Area of the triangle \( AQP \):

\[ A_{\text{triangle AQP}} = \frac{1}{2} AQ \times PQ \sin(60^\circ) = \frac{1}{2} \times 4\sqrt{2} \times 4\sqrt{2} \times \frac{\sqrt{3}}{2} = \frac{1}{2} \times 32 \times \frac{\sqrt{3}}{2} = 8\sqrt{3} \, \text{sq.cm} \]

Total area of the shaded region:

\[ A_{\text{shaded}} = A_{\text{sector APB}} + A_{\text{sector AQP}} – A_{\text{triangle APB}} – A_{\text{triangle AQP}} = 4\pi + \frac{16\pi}{3} – 8 – 8\sqrt{3} \, \text{sq.cm} \]

Problem 50

Two circles touch internally. The sum of their areas is \( 116 \pi \) sq.cm, and the distance between their centres is 6 cm. Find their radii.

Given:

\[ A_1 + A_2 = 116 \pi \, \text{sq.cm} \]

Let the radii be \( r_1 \) and \( r_2 \).

Area of the first circle:

\[ A_1 = \pi r_1^2 \]

Area of the second circle:

\[ A_2 = \pi r_2^2 \]

Given:

\[ \pi r_1^2 + \pi r_2^2 = 116 \pi \\ r_1^2 + r_2^2 = 116 \]

Distance between their centres is:

\[ r_1 – r_2 = 6 \]

Solving for \( r_1 \) and \( r_2 \):

\[ r_1^2 – 2r_1r_2 + r_2^2 = 36 \\ r_1^2 + r_2^2 – 2r_1r_2 = 36 \\ 116 – 2r_1r_2 = 36 \\ 2r_1r_2 = 80 \\ r_1r_2 = 40 \]

Solving the quadratic equation:

\[ r_1^2 + r_2^2 = 116 \\ r_1^2 – 6r_1 – 76 = 0 \]

Using the quadratic formula:

\[ r_1 = \frac{6 \pm \sqrt{6^2 + 4 \times 76}}{2} = \frac{6 \pm \sqrt{484}}{2} = \frac{6 \pm 22}{2} \]

So:

\[ r_1 = 14 \, \text{cm}, \, r_2 = 4 \, \text{cm} \]