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Ganit Learning Guides

Advanced GeometryLines, Angles, Shapes, Circles, Polygons, Solids

Author: Raghu M.D.

Advanced-Geometry 1 of 41 ©2014, www.learningforknowledge.com/glg

Contents

GEOMETRY ....................................................................................................... 2

Lines and Angles ............................................................................................................................... 2

Shapes .............................................................................................................................................. 10

Loci ................................................................................................................................................... 16

Cyclic Quadrilaterals ...................................................................................................................... 23

Area of Polygons ............................................................................................................................. 28

Quadrilateral ................................................................................................................................... 29

Curved figures ................................................................................................................................. 30

Solids ................................................................................................................................................ 35

Mensuration Table .......................................................................................................................... 36


A

B

GEOMETRY

Lines and Angles

A line segment is the shortest distance of a path between two points. A line segment is a part

of line. Usually a line segment is denoted by the points, for example, A and B.

All the points on the line are collinear.

Pairs of lines can be parallel or intersecting.

Parallel Intersecting

Angles

Angles are a measure of separation of two lines. Consider AB and BC intersecting and

terminating at point B.

Rules of Angles

Angles formed by two or more intersecting lines and polygons follow a set of rules. Polygons

are plane figures formed by a set of intersecting lines.

1. Vertically opposite angles are equal.

2. Angles around a point add up to 360°.

A

B C


3. Angles at a point on a straight line add up to 180°.

4. Angles in a triangle add up to 180°.

5. Exterior angle in a triangle is equal to sum of the opposite interior angles.

6. Base angles in an isosceles triangle are equal.

7. Angles in an equilateral triangle are each equal to 60°.

8. Angles in a quadrilateral add up to 360°.

9. Interior angles in a rectangle or square are each equal to 90°.

10. In a parallelogram opposite angles are equal.


11. In any polygon exterior angles add up to 360°.

12. When a pair of parallel lines are intercepted by a transversal, pairs of alternate and corresponding angles are equal.

Angle Calculations

In a given situation, knowing the value of some of the angles can be used to find the value of

other angles.

Example 1: From the figure given below, write an equation and solve for x.

Working: (3x+30) and (2x) are values of angles on a straight line.

(3x+30) + (2x) = 180°

5x + 30 = 180°

� =��

� =

��

� = 30

Answer: x = 30

A bearing is an angle used to indicate a direction. Bearings are measured from North

direction. They are useful in reading maps.

Example 2: The main road in a town runs north to south. A sketch below shows the path of a

car traveling from A to C through B. Find the bearing from C to A.

Line DE lies in North to South.

Working: Consider ∆ ABC.

AB = BC = 2 km

∴ ∆ ABC is an isosceles triangle.

Hence ∠ BAC = ∠ BCA

But exterior angle = 60° (given)

3x+30 2x

N

S

W E 60°

D

B

C

A

2km


∴ ∠ BAC + ∠ BCA = 60° (sum of opposite interior angles)

∠ BAC + ∠ BAC = 60° (∠ BAC = ∠ BCA, by above)

or 2 ∠ BAC = 60°

∴ ∠ BAC = 30°

Since DE is North to South, ∠ BAC is the bearing of point C from point A.

Answer: Bearing of C is equal to 30°.

Example 3: In the figure of a pentagon below, pairs of exterior and interior angles are

marked as, (a°, b°), (c°, d°), (e°, f°), (g°, h°) and (i°, j°). Show that the sum of interior angles

(b°, c°, d°, f°, h°, j°) is 540°.

Working:

(a°+b°) = 180° (Angles on a straight line)

Similarly,

(c°+d°) = (e°+f°) = (g°+h°) = (i°+j°) = 180°

∴ (a°+b°) + (c°+d°) + (e°+f°) + (g°+h°) + (i°+j°) = 5×180°

or, (a°+c°+ e°+ g°+ i°) + (b°+d°+f°+h°+j°) = 900°

But, (a°+c°+ e°+ g°+ i°) = 360° (Exterior angles of a polygon add up to 360°)

∴ 360° + (b°+d°+f°+h°+j°) = 900°

∴ b°+d°+f°+h°+j° = 900° − 360° = 540°

where, b, d, f, h and j are interior angles of the pentagon.

Answer: The sum of the interior angles of a pentagon = 540°

Example 4: ∆PQR is an equilateral triangle. PS is a line ⊥ to QR. Show that PS divides the

angle QPR into two equal parts.

Construction: Draw an equilateral ∆PQR and a perpendicular to QR

from P intersecting at S.

a°

b° c°

d°

e° f°

g°

h°

i° j°

P

Q R S


Working: In ∆PSQ, ∠PSQ = 90° (Given PS ⊥ QR)

∠PQS = ∠PQR = 60° (∆PQR is equilateral)

∠PSQ + ∠PQS + ∠QPS = 180° (Angles in a triangle)

∴ 90° + 60° + ∠QPS = 180°

150° + ∠QPS = 180°

∠QPS = 180° − 150° = 30°

Similarly, In ∆PSQ, ∠PSR = 90° (Given PS ⊥ QR)

∠PRS = ∠PRQ = 60° (∆PQR is equilateral)

∠PSR + ∠PRS + ∠RPS = 180° (Angles in a triangle)

∴ 90° + 60° + ∠RPS = 180°

150° + ∠RPS = 180°

∠RPS = 180° − 150° = 30°

Hence, ∠QPS = ∠RPS = 30°

But, ∠QPR = 60° (∆PQR is equilateral)

∠QPR = 60° = ∠QPS + ∠RPS

= 30° + 30° = 60°

Answer: The line PS bisects angle APR and divides it into two equal parts.

Example 5: The figure below shows a parallel lines cut by a pair of intersecting lines. Find

the value of the angle marked x°.

Construction: Mark the points of intersection as A, B

and C, and the parallel lines as AD and CG. Draw a line

EF through point B, parallel to AD and CG.

Working: Since EF || CG,

∠EBC = ∠BCG (Alternate angles)

∴ ∠EBC = ∠BCG = 30° (given ∠BCG = 30°)

Since EF || AD, ∠DAB = ∠ABE (Alternate angles)

x°

30°

60°

A D

C

BE

F

G


But, ∠DAB = ∠ABE = x°

∠ABC = ∠ABE + ∠EBC

or, ∠ABE + ∠EBC = 60° (given ∠ABC = 60°)

Hence, ∠DAB + 30° = 60°

or, ∠DAB = 60° − 30° = 30°

∴ ∠DAB = x° = 30°

Answer: The angle x° = 30°

EXERCISE AdvGeoAngles

Angles

1. Find the value of aº and bº in the figure below.

Ans: a = 60°, b = 30°

2. Find the value of xº and yº in the triangle shown below.

Ans: x = 90°

3. PQRS is a quadrilateral. Diagonal PS bisects angles P and S. Given ∠QPS = 65º and

∠RSP = 35º, find the value of ∠P and ∠S.

Ans: ∠P = ∠S = 80°

4. Find the value of the interior angle of a regular octagon.

Ans: 135°

90º

2aº

aº

bº

xº 140º

130º


5. In the figure below, show that line AB is parallel to line CD, given ∠A + ∠C = 125º.

(Hint: ∠A + ∠B + ∠C = 180º)

6. Timothy travels in a car 30 kms North and then turns East and travels a further distance of

30 kms. Find the bearing of his destination with respect to his starting point.

Ans: 45°

7. Find the value of xº in the figure below.

Ans: 60°

8. Find the value of aº in the figure below.

Ans:

9. Calculate the value of angles marked aº and bº.

Ans:

xº 80º

45º

A

B

C

D

E

xº

2xº

aº

2aº

2aº−30º

90º

90º

aº bº

40º 45º


10. In the figure below write the value of xº in terms of aº and bº.

Ans: xº = ��(��)

�

aº

xº

bº

xº


Shapes

Plane figures are broadly classified according to their shapes. A plane figure is defined as an

area bound by intersecting lines and curves. Simple figures are formed by intersecting lines,

for example, polygon or circle. Composite shapes are formed by a combination of polygons

and curves.

Similar Shapes

Although the size of a polygon can vary, its shape is determined by the lengths, the number

of sides and the angles formed at its vertices. Any two polygons which have exactly the same

shape, but not necessarily the same size, are called Similar polygons. The same rule applies

for composite figures also. Larger figure amongst a pair of a similar figure can be considered

as an enlargement of the smaller figure. Also the smaller figure can be considered as a

reduction of the larger one.

Similar shapes can be denoted using the symbol ⦀. For example, ∆ABC ⦀ ∆PQR.

Properties of Similar Shapes

1. The shape is preserved.

2. Ratio of lengths of corresponding sides is the same.

3. Interior angles of a triangle are equal to the corresponding interior angles of another

triangle.

4. Several figures of the same shape can be similar to one another.

5. Areas of similar polygons are proportional to the square of the ratio of their sides.

6. If two figures are identical, having the same lengths of sides, angles and area, they are said

to be congruent figures.

Similar Triangles

Consider two similar triangles ∆ABC and ∆PQR.

Here, ∠A = ∠P , ∠B = ∠Q , ∠C = ∠R

A P

B C Q R


and, ��

��=

��

��=

��

��= �

Also, Area of ∆ABC = k2 Area of ∆PQR

Conditions for similarity of triangles

Each condition is sufficient to conclude that the two triangles are similar.

SSS Criterion

Two triangles are similar if the ratio of their corresponding sides are the same.

SAS Criterion

If two corresponding sides have the same ratio and the two sides are equal, then the two

triangles are similar.

AAA Criterion

Two triangles will be similar if each of the three angles of one triangle is each equal to each

of the three angles of the other.

Similar Polygons

Any pair of polygons can be partitioned into triangles. If each one of these triangles is similar

to the corresponding one, then the two polygons are similar.

Some of the properties of special quadrilateral can help in finding if they are similar. For

example, any two squares will be similar.

Example 1: State true or false.

a) All equilateral triangles are similar. (Hint: Each of the interior angles of any

equilateral triangle is each equal to 60º).

Answer: True

b) All parallelograms are similar. (Hint: The interior angles can be different).

Answer: False

c) All squares are similar. (Hint: Each of the interior angles is equal to 90º. Also the ratio

of the sides are the same).

Answer: True


d) Two triangles have the same area. Hence they are similar. (Hint: Many dissimilar

triangles can have the same area).

Answer: False

Example 2: In the figure below, ∆PQRis an isosceles triangle. PS is perpendicular to QR.

Show that ∆PQS is similar and congruent to ∆PRS.

Working: Consider triangles PQS and PRS.

∠PSQ = ∠PSR = 90º (PS ⊥ QR)

∠PQS = ∠PQR = ∠PRQ = ∠PRS (∆PQR is isosceles)

∴∠PQS = ∠PRS

Also,PQ=PR (∆PQR is isosceles)

∠PQS + ∠PSQ + ∠QPS = ∠PRS + ∠PSR + ∠RPS = 180º (Angles in a ∆)

∴ ∠PQS + ∠PSQ + ∠QPS = ∠PRS + ∠PSR + ∠RPS

∴ ∠QPS = ∠RPS

Hence, ∆PQS ⫴ ∆PRS (AAA Criterion)

Answer: ∆PQS is similar and congruent to ∆PRS .

Example 3: ABCD and PQRS are two rectangles. Given AB=6cms, PQ=3cms, and their

areas are equal to 24cm2 and 6cm2, show that ABCD and PQRS have the same shape.

Construction: Draw two rectangles and

mark them as ABCD and PQRS.

Working: Area of rectangle ABCD = a

∴ a = AB × BC

a = AB × 6 = 6 AB

But a = 24 cm2 (Given)

6 AB = 24

AB = 24 = 4 cm 6

P

Q R

S

P A

B

D

C Q R

S


Area of rectangle PQRS = p

∴ p = PQ × QR

p = PQ × 3

p = 3 PQ

But p = 6 cm2 (Given)

∴ 3 PQ = 6

PQ = 6 = 2 cm 3

Hence BC = 6 = 2 QR 3

Also AB = 4 = 2 PQ 2

Also BC = AB = CD = AD = 2 QR PQ RS PS

In a rectangle opposite sides are equal.

Hence AB = CD and BC = AD

PQ = RS and QR = PS

Also, each of the interior angles is equal to 90º.

Hence ABCD and PQRS have the same shape.

Answer: ABCD ⦀ PQRS

Example 4: Triangles ABC and PQR are similar. Angles A and Q have been marked in the

figure below. Find the remaining angles.

Given ∆ABC ⦀ ∆PQR

∴ ∠A = ∠P = 50° (Given ∠A = 50°)

Hence ∠P = 50°

And, ∠B = ∠Q = 40° (Given ∠Q = 40°)

Hence ∠B = 40°

A B

C

P Q

R

50° 40°


∠A + ∠B + ∠C = 180° (Angles of a triangle)

Or, 50° + 40° + ∠C = 180°

Or, ∠C = 180° − 90°

∴ ∠C = 90°

Also, ∠C = ∠R = 90°

∴ ∠R = 90°

Answer: ∠A = ∠P = 50° ; ∠B = ∠Q = 40° ; ∠C = ∠R = 90° .

Example 5: ABC is a triangle. BC is extended to D and CE is drawn parallel to AB. Also ED

is drawn parallel to AC. Show that triangles ABC and ECD are similar.

Working: Since AB ∥ BC,

∠ABC = ∠ECD (corresponding angles)

Since AB ∥ BC,

∠ACB = ∠EDC (corresponding angles)

Also, ∠ABC + ∠BAC + ∠ACB = ∠ECD + ∠CED + ∠EDC = 180° (Angles in a ∆)

∴ ∠ABC + ∠BAC + ∠ACB = ∠ECD + ∠CED + ∠EDC

Or, ∠BAC = ∠CED

Since all the three angles of ∆ABC are equal to the three angles of ∆ECD, the two triangles

are similar.

Answer: ∆ABC ⦀ ∆ECD

EXERCISE

Similarity

1. State True or False

a) All Isosceles triangles are similar.

Ans: False.

b) Given ABCD is a rectangle, diagonal AC divides the rectangle into two

similar triangles.

A

B D C

E


A

B F

C

D

E

Ans: True.

c) Any two regular hexagons are similar.

Ans: True.

d) Exterior angles of two polygons add up to 360° in each polygon. Hence they

are similar.

Ans: False.

2. ∆ABC and ∆PQR are two right-angle triangles. Given ∠A=55° and ∠A=35°, show

that ∆ABC and ∆PQR are similar.

Hint: (∠A+∠B+∠C = ∠P+∠Q+∠R = 180°)

3. ABCD is a parallelogram, AC is a diagonal. Show that ∆ABC and ∆ADC are

congruent.

Hint: In a parallelogram opposite sides are equal.

4. In a triangle PQR, S and T are mid-points of PQ and PR respectively. If QR is equal

to 8cms, find the length of ST.

Ans: 4 cms.

5. ∆ABC and ∆PQR are similar triangles shown in the figure below. Given the length

AB = 2x units, PQ = x units, QR = 3 units and AC = 4 units, find the lengths of BC

and PR.

Ans: BC = 6, PR = 2

6. ABCDEF is a regular hexagon. BF and CE are two diagonals.

Show that ∆ABF ⦀∆DCE.

Hint: All sides and angles are equal in a regular hexagon.

7. In the figure below, show that ∆PQR is

similar to ∆RST, given ∠Q=∠S = 90°.

A

B C

2x 4

P

Q R

x

3

P

Q R

S

T


Hint: Opposite angles are equal.

8. ABCD is a trapezium with BC ∥ AD. Diagonals intersect

at E such that BE = 2 cms, ED = 4 cms and BC = 2.5

cms. Using similar triangles find the length of AD.

Ans: AD = 5 cms.

9. In ∆ABC, line CD intersects AB at the point D. If ∠BAC = ∠ACD, show that ∆ABC

is similar to ∆BCA.

Hint: Exterior angle is equal to the sum of

interior opposite angles.

10. Line plan of a land and building is shown below. The land and building plans are

rectangles. Drive way is 3m wide and walk way is 1m wide. The land measures 14m x

10m. Show that the shapes of land and building are similar

and find the dimensions of the building.

Ans: 7m x 5m

Loci

Locus (plural Loci) is the path traced by a point-sized particle with reference to another point,

points, line or lines.

For example, the locus of a point equidistant from two other points is the perpendicular

bisector of the line joining the two points.

Proof:

Construction

A

B

D

C

E

A B

C

D

Building

Walk way

Walk way

Wa

lk w

ay

Dri

ve

wa

y

D

B

C

A E


Draw a line segment CD ⊥ AB. JoinCAandCB.

Mark the point of intersection of AB and CD as E.

Working:

Consider triangles AEC and BEC.

AE = EB since the point E is equidistant from A and B.

Similarly, point C is equidistant from A and B.

∴ AC = BC

Line EC is common to both the triangles. Hence all the three sides of both the

triangles are equal.

∴ ∆ AEC is congruent to ∆ BEC.

∴ ∠ AEC = ∠ BEC

But ∠ AEC + ∠ BEC = 180°

Or 2 ⨉ ∠ AEC = 180°

Or∠ AEC = ��°

� = 90°

∴ ∠ AEC = ∠ BEC = 90°

Since ∠ AEC = ∠ BEC and AE = EB, the line segment CE ⏊ AB.

Hence ED ⏊ AB and bisects AB.

Conclusion: The locus of a point which is equidistant from two other points is its

perpendicular bisector.

Another example of a locus is: the path traced by a point which is equidistant from a pair of

intersecting lines is the bisector of the angle formed by the lines.

Loci can be a curve. The best example of a curved loci is a circle. Locus of a point

equidistant from another fixed point, called the centre, is a circle.

This diagram shows a circle. The fixed point or the centre is C.

The locus or the path traced is the circumference and the distance

between any point on the circle and its centre is called the radius.

Properties of Circles

Following are the names given to parts of a circle.

r

C


A line joining two points on a circle and passing through its centre is called the diameter.

Ratio of the length of the circumference to the length of diameter of

any circle is constant called .

Hence C = 4 × D .

Value of 4 = 3.142 (4 s.f.)

A Sector is the part of a circle bound by two radii and a section of the

circumference.

A Segment is a portion of the circle bound by a chord and a section of

the circumference.

Area of a circle is given by the formula 6 = 78� , where r is the

radius.

A line that intersects the circumference at only one point is called the

Tangent.

A line that is perpendicular to the tangent at the point of intersection is

called the Normal. The Normal passes through the centre of the circle.

Angle Properties

Angle subtended by any two radii of a circle is twice the value of the

angle subtended by any two chords originating from the same two

points.

In the figure shown, ∠AOB = 2 ⨉ ∠ACB .

Example 1: A, B and C are three points. Show that the locus of a point equidistant from A, B

and C is a single point.

Construction: Mark 3 points A, B and C. Draw lines AB

and BC. Draw perpendicular bisectors to AB and BC. Mark

the point of intersection of the bisectors as D.

C

BA

C

O

D

C B

A


Working: Locus of a point that is always equidistant from A and B is the perpendicular

bisector of AB. Point D is on the perpendicular bisector of AB. Hence AD = BD.

Also, locus of a point that is always equidistant from B and C is the perpendicular bisector of

BC. Point D is also on the perpendicular bisector of BC. Hence CD = BD.

∴ AD = BD = CD (Lines not shown in the figure)

Since there can be only one point of intersection for any two lines, D is the only point in the

plane such that AD = BD = CD.

Answer: Point D is the locus equidistant from A, B and C.

Example 2: Two circles with centres P and R touch externally at Q. Show that P, Q and R are

collinear.

Construction: Draw two circles with centres P and R and touching

at Q. Join PQ and QR.

Draw a common tangent to both the circles at Q.

Name the tangent as ST.

Working: A tangent is perpendicular to the radius at the point of

contact.

SQ ⏊PQ

Or, ∠ SQP = 90°

Similarly, SQ ⏊QR

∠ SQR = 90°

Hence, ∠ SQP + ∠ SQR = 180°

∴ PQ and QR are segments of the same line.

Answer: P, Q and R are collinear.

Example 3: In the figure shown below, find the values of angles marked x° and y°, given

∠OBC = 32°.

Working: Consider ∆OBC. OB = OC (radii)

∴ ∠OBC = ∠OCB (∆OBC is isosceles)

T

S

R Q P

32° C B

x°

A

y°

O


∠OBC = ∠OCB = 32°

But ∠OBC + ∠OCB + ∠BOC = 180°

Or, 32° + 32° + ∠BOC = 180°

∴ ∠BOC = 180° - 64° = 116°

Or, x = 116°

∠BOC is formed by joining two points B and C to the centre of the circle O, whereas

∠BAC is formed by joining them to a point A on the circumference.

∴ ∠BOC = 2 x ∠BAC

Or, x = 2 y

∴ y = ��

� = 58°

Answer: x = 116°, y = 58°

Example 4: Circumference of a circle is 31.42 cms. Find the area of the circle. Assume 4 =

3.142.

Working: Circumference C = 4 D

D = C / 4

∴ D = 31.42 / 3.142

Hence D = 10 cms.

∴ r = D / 2 = 5 cms.

Area of the circle A = 78�

= 3.142 ⨉ 25 = 78.55

Answer: Area A = 78.55 sq.cms. (2 dp)

Example 5: The figure below shows two concentric circles with a common centre O. Radii

OA and OC have been extended to intersect the outer circle at B and D. Radii of the two

circles are 3cms and 5cms. If AC measures 4cms, find the length of the chord BD and show

that AC ∥ BD.

Working: Consider the triangles OAC and OBD.


Here ∠A = ∠C because OA = OC (radii)

Hence ∆OAC is isosceles.

Similarly, ∠B = ∠D because OB = OD (radii) and

∆OBD is isosceles.

But, ∠A + ∠C + ∠O = ∠B + ∠D +∠O = 180° (angles in triangles)

Or, 2 ∠A + ∠O = 2 ∠B+ ∠O

∴ ∠A = ∠B

Hence AC ∥ BD (corresponding angles are equal)

Since ∠A = ∠B , ∠C = ∠D, ∠O is common,

∆ OAC ⦀ ∆ OBD (AAA criteria)

∴ 9�

9�=

�:

��

�

�=

�:

;

∴ BD = 6.67

Answer: BD = 6.67 cms

EXERCISES

1. Find the total length of the edge of a semi-circular protractor of radius 4 cms. (Use 4

= 3.14).

Ans: 20.56 cms.

2. In the figure shown below ∠AOB = 78°. Find the values of angles OAB and ACB.

Ans: ∠OAB = 51° and ∠ACB = 39°.

3. AB and CD are diameters of a circle with centre at O and radius = 5 cms. Given

∠AOC = 60°, show that AC is parallel to CD and find the length of CD.

A C

B

D

O

O

C

B A


Ans: 5 sms.

4. A cyclist covers 31.4 kms. During the journey the cycle wheel completes 10,000

revolutions. Find the diameter of the wheel. Assume 4 = 3.14.

Ans: 1m.

5. In the sketch shown below, find the value of ∠x°. O is the centre of the circle. (Sketch

is not to scale).

Ans: x = 22°.

6. In the figure shown below, PR is a tangent to the circle with centre O. The point of

contact is Q. If ∠RQT = 35°, find ∠PQS.

Ans: ∠PQS= 55°.

7. O and P are centres of two circles of radii 3.6 cms and 2.4 cms respectively. AB is a

common transverse tangent. AB intersects OP at Q. Show that 3 OQ = 2 QP.

Hint: ∆ OAQ ⦀ ∆ PBQ

8. Two circles with centres O and P touch internally. Radius of larger circle with centre

O is 4 cms. Distance OP is 1 cm. Find the difference in areas of the two circles.

Ans: 22 cm2 .

112°

x° O

O T

S

R P Q

A

B

P Q

O


9. The figure shown below is drawn using arcs of two concentric circles. AC and BD are

lines joining the arcs and intersecting at O.

Find the area and perimeter of the figure

ABDC. O is the common centre and ∠COD =

60°. (Figure not to scale).

Ans: 27.36 cms, 33.90 cm2 (2dp).

10. Find the perimeter and area of the shaded part of the figure shown below. This figure

is drawn fom four arcs of circles whose centres are the vertices of a square of side 8

cms. (Figure not to scale).

Ans: 214.72 cm2 (2dp).

Cyclic Quadrilaterals

If the vertices of a quadrilateral lie on the circumference of the circle, then the same is called

a Cyclic Quadrilateral.

Properties of a Cyclic Quadrilateral

In a cyclic quadrilateral the interior opposite angles add up to 180°.

Proof:

Construction: Draw a circle with centre O. Draw a quadrilateral

ABCD such that all the vertices are on the circle. Join BO and

OD. Extend BC to E.

Working: ∠BOD is the angle subtended at the centre by

chord BD (shown as dashed line in figure).

∠BAD is the angle subtended at the circumference by the same chord BD.

60° . O

D

C

B

A

A

B C

D O

E


∴ ∠BOD = 2 ∠BAD (Angle property)

Similarly, the reflex ∠BOD = 2 ∠BCD (Angle property)

But ∠BOD + reflex ∠BOD = 360° (Angle around a point)

∴ 2 ∠BAD + 2 ∠BCD = 360°

Or ∠BAD + ∠BCD = 180°

Conclusion: Interior opposite angles of a cyclic quadrilateral add up to 180°.

Example 1: Identify cyclic quadrilaterals from the following list.

(a) Rhombus

(b) Rectangle

(c) Square

(d) Parallelogram

Working:

Rhombus: Interior opposite angles are equal, but they do not add up to 180°. Hence

rhombus is not a cyclic quadrilateral.

Rectangle: Interior opposite angles of a rectangle add up to 180°. Hence a rectangle is a

cyclic quadrilateral.

Square: Interior opposite angles of a square add up to 180°. Hence a square is a cyclic

quadrilateral.

Parallelogram: Interior opposite angles of a parallelogram are equal but they do not

add up to 180°. Hence a parallelogram is a not a cyclic quadrilateral.

Answer: (b) and (c).

Example 2: Figure below shows a cyclic quadrilateral ABCD in a circle with centre O.

Chord AB is equal to chord CD and ∠ABC = 58°. Find the value of the remaining angles.

Working: ABCD is an isosceles trapezium, given AB = CD.

∠ABC = ∠DCB = 58°

∠ABC + ∠ADC = 180° (opposite angles of a cyclic

quadrilateral)

58° + ∠ADC = 180° (∠ABC = 58°)

O

B

A D

C


∴ ∠ADC = 180° − 58° = 122°

∠ABC + ∠DCB + ∠ADC + ∠DAB = 360°

Or 58° + 58° + 122° + ∠DAB = 360°

∴ ∠DAB = 122°

Answer: ∠ABC = ∠DCB = 58°, ∠ADC + ∠DAB = 122°

Example 3: Figure below shows a cyclic quadrilateral PQRS. Sides PQ and SR have been

extended to meet at point T. Given ∠PTS = 29° and ∠PQR = 112°, find the values of

remaining angles of the quadrilateral.

Working: In quadrilateral PQRS

∠PSR + ∠PQR = 180° (interior opposite angles)

∠PSR + 112° = 180°

∴ ∠PSR = 68°

∠PSR or ∠PST + ∠SPT+∠PTS= 180° (angles of a ∆)

68° + ∠SPT+29°= 180°

∴ ∠SPT=180° - 97° = 83°

In other words, ∠SPQ = 83°

∠SPQ + ∠QRS = 180° (interior opposite angles)

∴ ∠QRS = 180° - 83° = 97°

Answer: ∠PSR = 68°, ∠SPQ = 83° and ∠QRS = 97°

Example 4: Square ABCD is a cyclic quadrilateral and has been inscribed in a circle with

centre O and diameter 8 cms. Find the area of the square and the length of its side.

Construction: Draw a circle with centre O. Draw a square with its vertices on the

circumference. Mark the vertices as ABCD. Join the diagonals.

Working: Given, diagonal AB is the diagonal.

AC = 8 cms.

P

Q

R

S

T

A B

C D

O


∴ Radius AO = ��

�=

�

� = 4

Also AO = OD = 4 cms.

Area of ∆ AOD = �

�× 6A × AB (half of base times height)

= �

�× 4 × 4 = 8 cm2

Area of the square ABCD is the sum of the areas of the 4 congruent triangles:

∆ AOB, ∆ BOC, ∆ COD, ∆ DOA

∴ Area of square ABCD = 4 ⨉ 8 = 32 cm2

Length AB = √ 32 = 4 √2 cms. (A = l2)

Answer: Area = 32 cm2 , Length = 4 √2 cms.

Example 5: Show that in a cyclic quadrilateral, the exterior angle is equal to the interior

opposite angle.

Construction: Draw a cyclic quadrilateral PQRS in a circle.

Extend SR to T.

Working: ∠QRT + ∠QRS = 180° (supplementary angles)

∠QRS + ∠QPS = 180° (opposite angles)

∴ ∠QRS + ∠QRT = ∠QRS + ∠QPS

∴ ∠QRT = ∠QPS

Answer: The exterior angle of a cyclic quadrilateral is equal to the interior opposite angle.

EXERCISES

Cyclic Quadrilateral

1. Find the value of the interior angles in the cyclic quadrilateral shown below.

Ans: ∠A = 60°, ∠B = 108°, ∠C = 72° and ∠D = 120°

P

Q

R

S

T

A

D

C

B

x°

1.2x° 2x°


30°

A B

C D

A

B

C

D

O

2. A cyclic quadrilateral has the shape of a kite. Vertices of the kite touch the circumference

of the circle of radius 3 cms. Find the length of the major diagonal.

Ans: 7 cms.

3. ABCD is a cyclic quadrilateral with its vertices on the circumference of a circle O. Given

∠ACD = 30° and AB ∥CD. Find angles ∠A, ∠B, ∠C and ∠D of the quadrilateral.

Ans: ∠A=90°, ∠B=90°, ∠C=90°, ∠D=90°.

4. Figure below shows a cyclic quadrilateral. O is the centre of the circle. ∠BOD = 118°. Find the values of ∠A and ∠C.

Ans: ∠A = 59°, ∠C = 121° .

5. Show that the quadrilateral PQRS is a cyclic quadrilateral.

Hint: 2x + x = 1.3x + 1.7x

6. Find the value of angles marked P and Q.

Ans: ∠P = 95°, ∠Q = 93°

P

Q

R S

1.3x° x°

2x° 1.7x°

P

Q

87° 85°


7. Find the values of x and y marked in the cyclic quadrilateral given below.

Ans: x° = y° = 60°.

8. Find the values of the interior angles of the cyclic quadrilateral ABCD in the figure

shown below, given ∠AED = 30°.

Ans:

9. In the figure shown below PS is extended to T and SU ∥ QR. Given ∠TSU = 35° and ∠P

= 80°, find ∠PQR, ∠PSR and ∠QRS.

Ans: ∠PQR = 115°, ∠PSR = 85°, ∠QRS = 100°.

10. A square shaped lawn has been cultivated inside a fenced square land. Corners of the

lawn are on the circumference of a circle. Fences are tangential to the same circle. Find

the ratio of the area of the lawn to the area of the fenced land.

Ans: 1:2

Area of Polygons

Area is the two-dimensional space enclosed in a figure by lines or curves. Lines or curves are

called edges and the points of intersection of the edges are called vertices.

P

Q

R S

2x° 2y°

x° y°

A

B

C

D

E

P

Q

R

S T

U


Area is measured in square units of length, for example, cm2.

It is defined as the product of the length and breadth of a rectangle, or length x length or l2 in

the case of a square.

If l = 1 unit then the area enclosed in the figure shown is 1 square unit.

Similarly, area A of a rectangle is the product of its length l and width w.

A = l ⨉ w

Area of a triangle is equal to half the area of a rectangle enclosing it. Consider a triangle ABC

enclosed in a rectangle ABDE.

CF is drawn perpendicular to AB.

Here, ∆ ACF ≡ ∆ AEC (All 3 sides are equal)

∴ Area of ∆ ACF = Area of ∆ AEC

But, Areas of ∆ ACF + ∆ AEC = Area of AECF

Or, Area of ∆ ACF = �

� Area of AECF

Similarly, Area of ∆ BCF = �

� Area of BFCD

Hence, Area of ∆ ACF + Area of ∆ BCF = �

� (Area of AECF + Area of BFCD)

Or, Area of ∆ ABC = �

� (Area of Rectangle ABDE)

This statement is true for triangles of all shapes. Hence if the length of the base is b and the

height to the vertex is h, then A = �

� b h .

Quadrilateral

Area of a quadrilateral can be calculated by dividing it into two triangles.

Consider a quadrilateral ABCD where AC is a diagonal,

DE and BF are perpendiculars to AC.

Lengths of BF and DE are h1 and h2 respectively.

∆ ABC and ∆ ADC are the two parts of quadrilateral

ABCD.

l

l

A B

C D E

F

A

B

C D

E

F

h1

h2


Area of ∆ ABC = �

�ℎ� × 6F

Area of ∆ ADC = �

�ℎ� × 6F

∴ Area of ABCD = �

�6F(ℎ� + ℎ�)

If AC = d ,

Then Area of ABCD = �

�G(ℎ� + ℎ�)

Special Quadrilaterals

Kite: In a kite where pairs of adjacent sides are equal, the diagonals are

perpendicular to each other.

Area of kite ABCD = �

�(ℎ� + ℎ�)G�

Or = �

�G�G�

Where d1 = h1+h2, length of BD

And d2 = length of AC

Trapezium: Area of a trapezium is given as:

A = (HI�HJ)

�× ℎ

Where l1 and l2 are lengths of parallel sides and h is the distance between the parallel lines.

Parallelogram: Area A = l ⨉ h , where l is the length of parallel sides and h is the

perpendicular distance between them.

Curved figures

A simple method to measure the area of a figure which has no regular shape is to transfer it

on to a graph sheet with grid lines marked on it. The number of squares occupied by the

figure are counted.

Part of the figure which does not occupy a complete square but covers more than half is

considered as 1 square, and that which covers less than half is considered as zero.

h1

h2

A

B

C

D


A curved figure is shown in the diagram. Number of squares covered by it are counted.

In the figure shown (not to scale) the grid lines are 1 cm apart.

In this case the number of squares covered are 12.

Hence Area = 12 ⨉ 1 = 12 cm2

Example 1: PQRS I quadrilateral. Diagonal PR measures 6 cms. QE is perpendicular to PR

and measures 3 cms and SE ia 2 cms. Find the area of PQRS.

Working:

Area of quadrilateral, A = �

�G�(ℎ� + ℎ�)

Or, A = �

�× 6 × (3 + 2)

∴ A = 15

Answer: 15cm2

Example 2:

Figure below shows a pair of parallel lines PQ and RS. Vertices A and D of triangles ABC

and DEF are on line PQ. Base sides BC and EF are on line RS. Given BC=3.7cms and

EF=4.3cms and the distance between PQ and RS is h=4cms, find the areas of ∆ ABC and ∆

DEF.

Working:

Base of ∆ABC b1 = 3.7 cms

Area of ∆ABC = �

�M�ℎ

= �

�3.7 × 4 = 7.4

Base of ∆DEF b2 = 4.3 cms

Area of ∆ABC = �

�M�ℎ =

�

�4.3 × 4 = 8.6

Answer: 7.4 cm2, 8.6 cm2

x x x

x x x x x

x x x x

C

3.7cms 4.3cms

4cm

s

Q P

R S

A

B

D

E F


Example 3: Find the area of the following figure.

Working: Area of the figure:

= 8 × (19 + 19 + 12) + 30 × 12

= 400 + 360

= 760

Answer: 760 sq. mm.

Example 4: l1 and l2 are the lengths of a pair of parallel sides of a trapezium. They are

separated by a distance h. Show that the area of the trapezium is equal to �

�(R� + R�)ℎ .

Working: Consider a trapezium ABCD, where AB∥CD

as shown below. AE and BF are perpendicular to CD.

Consider triangles ADE and BCF.

Area of ∆ ADE = �

�(BS)ℎ

Area of ∆ BCF = �

�(FT)ℎ

Area of rectangle ABFE = l1 ⨉ h

∴ Area of trapezium ABCD = Area of (∆ADE + ∆BCF + Rectangle ABFE)

= �

�(BS)ℎ +

�

�(FT)ℎ + R�ℎ

= �

�ℎ(BS + FT) + R�ℎ

But DE + CF = R� − R�

∴ Area of trapezium ABCD = �

�ℎ(R� − R�) + R�ℎ

= �

�R�ℎ −

�

�R�ℎ + R�ℎ

= �

�(R� + R�)ℎ

Answer: Area of trapezium = �

�(R� + R�)ℎ

8mm

19mm 19mm

30mm

12mm

h h

A B

C D E F

l1

l2


Example 5: Figure drawn below shows a ∆ABC. Line DE is parallel to BC. AB measures

6cms and AD measures 2cms. Find the ratio of the areas of trapezium DECB and ∆ADE.

Working: ∆ADE is similar to ∆ABC (three corresponding

angles are equal)

∴ Area of ∆��

∆�:V=

��J

�:J

= �J

�J=

W

;

Or ∆��

∆�:V− 1 =

W

;− 1

∆��∆�:V

∆�:V=

W�;

;=

�

;

But Area of (∆ABC - ∆ADE) is equal to Area of trapezium DECB.

∴ �XY�Z[\X�]Y^_`a:V��

�XY�Z[∆�:V=

�

;

Answer: Ratio of areas is equal to 5:4

EXERCISES

1. Find the Area of a Rhombus of side length 6.3cms and height 4.1cms.

Ans: 25.83 cm2.

2. A kite has an area of 35cm2. Its longest diagonal measures 7cms. Find the length of its

shortest diagonal.

Ans: 5 cms.

3. ∆PQR is a right-angle triangle. ST is parallel to QR. ∠PQR=90°, PS=5cms, PQ=7cms

and QR=4cms. Find the areas of ∆PQR and ∆PST.

Ans: 14 cm2, 7.15 cm2.

4. ABCDEF is a regular hexagon of side length 3.5cms. Diagonal AE measures

6.06cms. Find the area of the hexagon and length of FC.

Ans: 31.81 cm2 , 7 cms.

A

B C

D E

A B

C

D E

F


5. PQRS is a square. Show that the area of the square is equal to �

�G� , where d is the

length of the diagonals PR or QS. Given PR and QS intersect at T.

Hint: Find area of ∆PTQ.

6. Find the area of the following figure.

Ans: 2880 mm2

7. Figure below shows a rectangle ABCD and parallelogram EBFD. Find the areas of

ABCD, EBFD, ∆BFC and ∆AED.

Ans: 2592 mm2, 1944 mm2,

∆BFC ≡ ∆AED = 324 mm2.

8. Find the area of the following figure.

Ans: 1162.5 mm2.

9. Area of a rectangle is 800mm2. Length of the rectangle is twice that of its width. Find

the dimensions of the rectangle.

Ans: 40mm ⨉ 20mm..

18mm 18mm

60mm

70mm

36mm

18mm A B

C D

E

F

36mm

72mm

15mm

15mm

30mm

60mm

40mm 50mm


10. An area is fenced by two parallel sides, each 180m long and separated by a distance

of 120m. Two semi-circular fences enclose the remaining sides. Find the fenced area.

Ans: 32,904 sq.m.

Solids

Solids occupy 3 dimensional space. They have an outer surface and volume. Solids have been

named after their shape.

For example, a Cube is a solid with 6 surfaces which are in the shape of

squares and are of the same size.

Surface Area

Surface area is the amount of two dimensional space taken up by the surfaces of a solid.

For example, consider a cardboard box. Surface area of the box is the area of cardboard

required to make it. This box can be made from a single sheet of cardboard cut in the profile

of rectangles having a common edge. A diagram showing a flat piece of cardboard that can

be folded to the shape of a box is called a Net.

Sum of the areas of rectangles shown in the figure is

the total surface area of the solid.

Volume

Volume is the amount of space occupied by a solid or liquid in a container. It is measured in

cubic units of length. A practical unit for volume is cm3.

A cube having all its sides equal to 1cm has a volume of 1cm3. Volume of

any solid is measured in multiples or submultiples of cm3.

Net Box

1cm

1cm 1cm


Mensuration Table

This table lists the formulae for connecting Volume (V) and Surface Area (SA) to the

dimensions of solids of regular shapes.

1. Cube

Edge length = l

V = R × R × R = R�

SA = 6 × R� = R�

2. Cuboid

SA = 2(RM + Mℎ + Rℎ)

V = l b h

l = length, b = breadth, h = height

3. Triangular Prism

SA = b h + 3 l b

V = CSA ⨉ l (cross-sectional area ⨉ length)

Height of ∆ = h

∴ V = �

�MℎR

Note: V = CSA ⨉ l is true for any prism having a regular polygon

as its cross-section CS.

4. Cylinder

r = Radius of cylinder

h = Height of cylinder

SA = 4 r h + 2 4 r2

V = 4 r2 h

l

l

b

h

l

b

h

h

r


5. Cone

r = Radius of base

h = Vertical height

l = Slant Height

SA = 4 r l + 4 r2

V = �

�78�ℎ

6. Pyramid

h = Height of pyramid

a = Area of base

SA = Sum of the areas of surfaces

V = �

�bℎ

7. Sphere

r = Radius of sphere

SA = 4 4 r2

V = ;

�78�

Example 1: Dimensions of a cuboid are as follows.

l = 8 cms

b = 5 cms

h = 3 cms

Find its volume and surface area.

Working: Volume of cuboid V = l ⨉ b ⨉ h

= 8 ⨉ 5 ⨉ 3

l

r

h

h

r


∴ V = 120 cm3

Surface Area = ( l b + b h + h l ) ⨉ 2

= 2 (8 ⨉ 5 + 5 ⨉ 3 + 3 ⨉ 8 )

= 2 ⨉ 79

∴ SA = 158 cm2

Answer: V = 120 cm3, SA = 158 cm2

Example 2: A hexagonal prism has a side of length 20cms and base area of 8cm2 and a

volume of 84cm3. Find its height.

Working: Volume of prism V = CSA ⨉ h

CSA = Base Area = 8

∴ V = 8 ⨉ h

Or h = c

�=

�;

�

∴ h = 10.5 cms

Answer: Height = 10.5 cms.

Example 3: A cube has a volume of 27cm3. Find its surface area.

Working: d = R�

27 = R�

∴ R = √27e = 3

Surface Area f6 = 6R�

= 6 × 3� = 6 × 9

∴ SA = 54

Answer: SA = 54 cm2

Example 4: Draw a Net diagram for a cylinder of 5cms radius and 9cms height. Find its

volume, curved surface area and total surface area.


Working:

Curved surface area = 24r h

= 2 4⨉5⨉9 = 904

Area of two circles = 78� + 78� = 278�

= 275� = 507

Total surface area = 278ℎ + 278�

= 907 + 507 = 1407

Volume d = 78�ℎ = 7 × 5� × 9 = 7 × 25 × 9 = 2257

Answer: V = 225 , Curved SA = 904, Total SA = 1404 .

Example 5: An ice cream cone has a diameter of 6cms and depth of 5cms. Ice cream in the

cone is topped by a hemispherical scoop. Find the weight of the ice cream if 1cc of it weighs

1.05gms.

Working: Volume of cone = �

�78�ℎ

= �

�× 3.14 × 3� × 5

= 3 × 15.70

= 47.10 cm3

Volume of hemisphere = ;

�

hXe

�

= ;

�×

�.�;×�e

�

= 2 × 3.14 × 9 = 6.28 × 9 = 56.52 cm3

Total Volume = Volume of cone + Volume of hemisphere

= 47.10 + 56.52 = 103.62 cm3

Hence of weight of ice cream = 1.05 ⨉ 103.62

= 108.80 gms.

Answer: 108.80 gms.

r

24 r h

6cms

5cms


EXECISES

AdvGeoSAVol

1. A cone has the following dimensions.

Base radius = 3cms, Height = 4cms, Slant Height = 5cms.

Find its surface area and volume.

Ans: 75.36 cm2, 37.68 cm3

2. Small cubes of volume 0.729cm3 are packed inside a box of size 9cm x 18cm x 4.5cm.

How many of them together will exactly fit inside the box?

Ans: 1000

3. Volume of a cuboid is 385mm3. Surface area of its largest side is 77mm2. Find the

height of the cuboid and its surface area.

Ans: 5 mm, 254 mm2

4. Find the volume of a sphere whose surface area is 314cm2.

Ans: 523.33 cm3

5. A hollow cylinder has an outside diameter of 20cms, inside diameter of 14 cms and

length of 10 cms. What is its volume? Find the surface area of the cylinder.

Ans: 1601.40 cm3, 527.52 cm2

6. A cone 8cms high and 6cms diameter is cut into two parts as shown below.

Find the volumes of the both the parts.

Ans: 11.92 cm3, 214.16 cm3

3cms

Part 1

Part 2


7. A spherical ball of 1331mm3 volume is melted and recast into the shape of a cube

without wasting any material. What is the surface area of the cube?

Ans: 726 mm2

8. A spherical ball of 113cm3 volume is to be gold-plated at a cost of $14 per cm2 of

surface area. What is the cost of plating?

Ans: $3,400/-

9. Water is stored in a cuboid trough of base area 240cm2. 15% of the volume of water

had evaporated in a day. Amount of water evaporated was 360cc. What was the depth

of water before and after evaporation?

Ans: 10 cms, 8,5 cms

10. Distance from Earth’s equator to North pole is 10,000km. Find the surface area and

volume of Earth.

Ans: 5 ⨉108 km2, 103 km3