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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
Advanced-Geometry 2 of 41 ©2014, www.learningforknowledge.com/glg
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
Advanced-Geometry 3 of 41 ©2014, www.learningforknowledge.com/glg
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.
Advanced-Geometry 4 of 41 ©2014, www.learningforknowledge.com/glg
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
Advanced-Geometry 5 of 41 ©2014, www.learningforknowledge.com/glg
∴ ∠ 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
Advanced-Geometry 6 of 41 ©2014, www.learningforknowledge.com/glg
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
Advanced-Geometry 7 of 41 ©2014, www.learningforknowledge.com/glg
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
Advanced-Geometry 8 of 41 ©2014, www.learningforknowledge.com/glg
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
Advanced-Geometry 9 of 41 ©2014, www.learningforknowledge.com/glg
10. In the figure below write the value of xº in terms of aº and bº.
Ans: xº = ����(���)
�
aº
xº
bº
xº
Advanced-Geometry 10 of 41 ©2014, www.learningforknowledge.com/glg
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
Advanced-Geometry 11 of 41 ©2014, www.learningforknowledge.com/glg
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
Advanced-Geometry 12 of 41 ©2014, www.learningforknowledge.com/glg
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
Advanced-Geometry 13 of 41 ©2014, www.learningforknowledge.com/glg
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°
Advanced-Geometry 14 of 41 ©2014, www.learningforknowledge.com/glg
∠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
Advanced-Geometry 15 of 41 ©2014, www.learningforknowledge.com/glg
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
Advanced-Geometry 16 of 41 ©2014, www.learningforknowledge.com/glg
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
Advanced-Geometry 17 of 41 ©2014, www.learningforknowledge.com/glg
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
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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
Advanced-Geometry 19 of 41 ©2014, www.learningforknowledge.com/glg
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
Advanced-Geometry 20 of 41 ©2014, www.learningforknowledge.com/glg
∠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.
Advanced-Geometry 21 of 41 ©2014, www.learningforknowledge.com/glg
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
Advanced-Geometry 22 of 41 ©2014, www.learningforknowledge.com/glg
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
Advanced-Geometry 23 of 41 ©2014, www.learningforknowledge.com/glg
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
Advanced-Geometry 24 of 41 ©2014, www.learningforknowledge.com/glg
∴ ∠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
Advanced-Geometry 25 of 41 ©2014, www.learningforknowledge.com/glg
∴ ∠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
Advanced-Geometry 26 of 41 ©2014, www.learningforknowledge.com/glg
∴ 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°
Advanced-Geometry 27 of 41 ©2014, www.learningforknowledge.com/glg
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°
Advanced-Geometry 28 of 41 ©2014, www.learningforknowledge.com/glg
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
Advanced-Geometry 29 of 41 ©2014, www.learningforknowledge.com/glg
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
Advanced-Geometry 30 of 41 ©2014, www.learningforknowledge.com/glg
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
Advanced-Geometry 31 of 41 ©2014, www.learningforknowledge.com/glg
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
Advanced-Geometry 32 of 41 ©2014, www.learningforknowledge.com/glg
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
Advanced-Geometry 33 of 41 ©2014, www.learningforknowledge.com/glg
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
Advanced-Geometry 34 of 41 ©2014, www.learningforknowledge.com/glg
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
Advanced-Geometry 35 of 41 ©2014, www.learningforknowledge.com/glg
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
Advanced-Geometry 36 of 41 ©2014, www.learningforknowledge.com/glg
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
Advanced-Geometry 37 of 41 ©2014, www.learningforknowledge.com/glg
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
Advanced-Geometry 38 of 41 ©2014, www.learningforknowledge.com/glg
∴ 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.
Advanced-Geometry 39 of 41 ©2014, www.learningforknowledge.com/glg
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
Advanced-Geometry 40 of 41 ©2014, www.learningforknowledge.com/glg
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
Advanced-Geometry 41 of 41 ©2014, www.learningforknowledge.com/glg
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