CHAPTER – 4

GEOMETRY
In CAT and other similar MBA Entrance Exams, the Two lines which make an angle of 90° with each othe r are
problems relating to Geometry cover mostly triangles, said to be PERPENDICULAR to each other.
quadrilaterals and circles. Even though polygons with
more than four sides are also covered, the emphasis on If a line l passes through the mid-point of a line segment
such polygons is not as much as it is on triangles and AB, then the line l is said to be the BISECTOR of segment
circles. In this chapter, we will look at some properties as AB.
well as theorems and riders on parallel lines, angles,
triangles (including congruence and similarity of If a line l is drawn passing through the vertex of an angle
triangles), quadrilaterals, circles and polygons. dividing the angle into two equal parts, then the line l is said
to be the ANGLE BISECTOR of the angle. Any point on the
ANGLES AND LINES angle bisector of an angle is EQUIDISTANT from the two
arms of the angle.
An angle of 90° is a right angle; an angle less tha n 90° is
acute angle; an angle between 90° and 180° is an ob tuse If a line l is perpendicular to a line segment and passes
angle; and angle between 180° and 360°is a reflex angle. through the mid-point of AB, then the line l is said to be
the PERPENDICULAR BISECTOR of the segment AB.
The sum of all angles made on one side of a straight line
AB at a point O by any number of lines joining the line AB Any point on the perpendicular bisector of a line segment
at O is 180°. In Fig. 4.01 below, the sum of the an gles u, is EQUIDISTANT from the two end points of the segment.
v, x, y and z is equal to 180°.
Fig. 4.04
When any number of straight lines join at a point, the sum
of all the angles around that point is 360°. In P
Fig. 4.02 below, the sum of the angles u, v, w, x, y and z
is equal to 360°.
R
Fig.4.01

A B

z
y u Q
x v
In Fig. 4.04, line PQ is the perpendicular bisector of the
line segment AB. Any point R on the perpendicular
Fig. 4.02 bisector of AB will be equidistant from A and B, i.e.,
RA = RB. In particular, PA = PB, QA = QB.

PARALLEL LINES
u
z v When a straight line cuts two or more lines at distinct
points, then the cutting line is called the TRANSVERSAL.
y w When a straight line XY cuts two parallel lines PQ and RS
x [as shown in Fig. 4.05], the following relations hold
between the various angles that are formed. [M and N are
Two angles whose sum is 90° are said to be the points of intersection of XY with PQ and RS
complementary angles and two angles whose sum is 180° respectively].
are said to be supplementary angles.
Fig. 4.05
When two straight lines intersect, vertically opposite angles X
are equal. In Fig. 4.03 given below, ∠AOB and ∠COD are
vertically opposite angles and ∠BOC and ∠AOD are P Q
vertically opposite angles. So, ∠AOB = ∠COD and ∠BOC M
= ∠AOD. N
Fig. 4.03 R S

A B
Y
(a) Alternate angles are equal, i.e.
∠PMN = ∠MNS and ∠QMN = ∠MNR
o
(b) Corresponding angles are equal, i.e.
∠XMQ = ∠MNS; ∠QMN = ∠SNY;
D C
∠XMP = ∠MNR; ∠PMN = ∠RNY

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(c) Sum of the interior angles on the same side of the scalene triangle is one in which no two sides are equal,
transversal is equal to 180°, i.e. and hence no two angles are equal.
∠QMN + ∠MNS = 180°and ∠PMN + ∠MNR = 180°
In an isosceles triangle, the unequal side is called the
(d) Sum of the exterior angles on the same side of the BASE. The angle where the two equal sides meet is called
transversal is equal to 180°, i.e. the VERTICAL ANGLE. In an isosceles triangle, the
∠XMQ + ∠SNY = 180°; and ∠XMP + ∠RNY = 180° perpendicular drawn to the base from the vertex opposite
If three or more parallel lines make intercepts on a the base (i.e., the altitude drawn to the base) bisects the
transversal in a certain proportion, then they make base as well as the vertical angle. That is, the altitude
intercepts in the same proportion on any other transversal drawn to the base will also be the perpendicular bisector
as well. In Fig. 4.06, the lines AB, CD and EF are parallel of the base as well as the angle bisector of the vertical
and the transversal XY cuts them at the points P, Q and angle. It will also be the median drawn to the base.
R. If we now take a second transversal, UV, cutting the
three parallel lines at the points J, K and L, then we have In an equilateral triangle, the perpendicular bisector, the
PQ/QR = JK/KL. median and the altitude drawn to a particular side coincide
and that will also be the angle bisector of the opposite
Fig. 4.06 angle. If a is the side of an equilateral triangle, then its
altitude is equal to 3 a / 2
X U
Sum of any two sides of a triangle is greater than the third
A P J B side. Therefore, the difference of any two sides of a
Q K triangle is less than the third side.
C D
R L If the sides are arranged in the ascending order of their
E F length, then the angles opposite the sides (in the same
order) will also be in ascending order (i.e., greater side
Y V has the greater angle opposite to it). Therefore, if the sides
are arranged in descending order of their length, the
If three or more parallel lines make equal intercepts on angles opposite the sides (in the same order) will also be
one transversal, they make equal intercepts on any other in descending order (i.e., smaller angle has smaller side
transversal as well. opposite to it).
TRIANGLES There can be only one right angle or only one obtuse
angle in any triangle. There can also not be one right
Fig. 4.07
angle and an obtuse angle both present in the same
triangle.
X
The hypotenuse is the side opposite the right angle in a right-
angled triangle. In a right-angled triangle, the hypotenuse is
Y Z the longest side. In an obtuse angled triangle, the side
opposite the obtuse angle is the longest side.
Fig. 4.08
The sum of the three angles of a triangle is 180°
(Right-angled triangle)
The exterior angle of a triangle at each vertex is equal to A
the sum of the interior angles at the other two vertices.
(Exterior angle is the angle formed at any vertex, by one
side and the extended portion of the second side at that
vertex).

A line perpendicular to a side and passing through the

midpoint of the side is said to be the perpendicular bisector B C
of the side. It is not necessary that the perpendicular
bisector of a side should pass through the opposite vertex In a right-angled triangle, the square of the hypotenuse
in a triangle in general. (the side opposite the right angle) is equal to the sum of the
squares of the other two sides. In Fig. 4.08, AC2 = AB2 +
The perpendicular drawn to a side from the opposite BC2. This result is known as Pythagoras theorem.
vertex is called the altitude to that side.
Fig. 4.09
The line joining the midpoint of a side with the opposite (Acute-angled triangle)
vertex is called the median drawn to that side. A median A
divides the triangle into two equal halves as far as the area
is concerned.
An equilateral triangle is one in which all the sides are
equal (and hence, all angles are equal, i.e., each of the
angles is equal to 60°). A triangle in which two si des are
equal is called an isosceles triangle. In an isosceles C
B D
triangle the angles opposite the equal sides are equal. A

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In an acute angled triangle, the square of the side GEOMETRIC CENTRES OF A TRIANGLE
opposite the acute angle is less than the sum of the
CIRCUMCENTRE
squares of the other two sides by a quantity equal to twice
the product of one of these two sides and the projection Fig 4.13
of the second side on the first side.
In Fig. 4.09, AC2 = AB2 + BC2 − 2 BC.BD A

Fig. 4.10
R
(Obtuse angled triangle)

A S R
R

B C

The three perpendicular bisectors of a triangle meet at a

C B D point called Circumcentre of the triangle. This is normally
denotes as S. The circumcentre of a triangle is equidistant
In an obtuse angled triangle, the square of the side from its vertices and the distance of circumcentre from
opposite the obtuse angle is greater than the sum of the each of the three vertices is called circumradius
squares of the other two sides by a quantity equal to twice (represented by R) of the triangle. The circle drawn with
the product of one of the sides containing the obtuse angle the circumcentre as centre and circumradius as radius is
and the projection of the second side on the first side. called the cirumcircle of the triangle. It passes through all
three vertices of the triangle. (refer to Fig. 4.13)
In Fig. 4.10, AC2 = AB2 + BC2 + 2BC.BD
INCENTRE AND EXCENTRES
Fig. 4.11
Fig. 4.14
A
A

x° x°
x x
E
r
F r
B D C
Ι
In any triangle, the internal bisector of an angle bisects the z r y
z y
opposite side in the ratio of the other two sides.
B D C
In triangle ABC, if AD is the angle bisector of angle A, then
BD/DC = AB/AC. This is called the Angle Bisector
Theorem (refer to Fig. 4.11). The bisectors of the three angles of a triangle meet at a
point called incentre of the triangle. This is normally
Fig. 4.12 denoted as I. The incentre is equidistant from the three
sides of the triangle i.e., the perpendiculars drawn from
A
the incentre to the three sides are equal in length and this
length is called the inradius (represented by r) of the
triangle. The circle drawn with the incentre as centre and
the inradius as radius is called the incircle of the triangle.
It touches all three sides of the triangle.

1
In Fig. 4.14, ∠BIC = 90° + ∠A where I is the incentre.
2
B D C
∠CIA = 90° + 1 ∠B ; and ∠AIB = 90° + 1 ∠C .
2 2
In ∆ABC, if AD is the median from A to side BC
(meeting BC at its mid point D), then 2(AD2 + BD2) = AB2 If the bisector of one angle and the bisectors of the
+ AC2. This is called the Apollonius Theorem. This is external angles at the other two vertices are drawn, they
useful in calculating the lengths of the three medians meet at a point called Excentre. There are three excentres
given the lengths of the three sides of the triangle for any triangle - one corresponding to the bisector of each
(refer to Fig. 4.12). angle.

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ORTHOCENTRE Fig. 4.17
A
Fig 4.15
A
• G
E
F
B D C
O
Hence, in equilateral triangle ABC shown in Fig. 4.17, AD
is the median, altitude, angle bisector and perpendicular
B D C bisector. G is the centroid which divides the median in the
ratio of 2 : 1. Hence, AG = 2/3 AD and GD = 1/3 AD.
The three altitudes of a triangle meet at a point called
Orthocentre. This is normally denoted as O (refer to Fig. But since AD is also the perpendicular bisector and angle
4.15). bisector and since G is the circumcentre as well as the
∠BOC = 180°– A; ∠COA = 180°– B; incentre, AG is the circumradius and GD is the inradius of
∠AOB = 180°– C. the equilateral triangle ABC. Since AD is also the altitude,
its length is equal to 3 a / 2 where a is the side of the
CENTROID equilateral triangle. Hence, the circumradius of the
Fig. 4.16 equilateral triangle,
A  3  . a =
R= 2 a/ 3
3  2 

F E  
G and the inradius (r) = 1  3 .a 
3  2 

= a/2 3
B D C
Since the radii of the circumcircle and the incircle of an
The three medians of a triangle meet at a point called the equilateral triangle are in the ratio 2 : 1, the areas of the
Centroid. This is normally denoted by G (refer to circumcircle and the incircle of an equilateral triangle will
Fig. 4.16). be in the ratio of 4 : 1.

Important points about geometric centres of a When the three medians of a triangle (i.e., the medians to
triangle the three sides of a triangle from the corresponding
opposite vertices) are drawn, the resulting six triangles
Please note the following important points pertaining to are equal in area and hence the area of each of these
the geometric centres of a triangle ABC. In an acute triangles is equal to one-sixth of the area of the original
angled triangle, the circumcentre lies inside the triangle. triangle.
In a right-angled triangle, the circumcentre lies on the Fig. 4.18
hypotenuse of the triangle (it is the midpoint of the A
hypotenuse). In an obtuse angled triangle, the
circumcentre lies outside the triangle.
In an acute angled triangle, the orthocentre lies inside the
F E
triangle. In a right-angled triangle, the vertex where the
right angle is formed (i.e., the vertex opposite the G
hypotenuse) is the orthocentre. In an obtuse angled
triangle, the orthocentre lies outside the triangle.
B D C
In a right-angled triangle the length of the median drawn
to the hypotenuse is equal to half the hypotenuse. This
median is also the circumradius of the right-angled In Fig 4.18, AD, BE and CF are the medians drawn to the
triangle. three sides. The three medians meet at the centroid G.
The six resulting triangles AGF, BGF, BGD, CGD, CGE
The centroid divides each of the medians in the ratio and AGE are equal in area and each of them is equal to
2 : 1, the part of the median towards the vertex being twice 1/6th of the area of triangle ABC.
in length to the part towards the side.
The inradius is less than half the smallest altitude of the SIMILARITY OF TRIANGLES
triangle.
Two triangles are said to be similar if the three angles of
In an isosceles triangle, the centroid, the orthocentre, the one triangle are equal to the three angles of the
circumcentre and the incentre, all lie on the median to the second triangle. Similar triangles are alike in shape.
base. The corresponding angles of two similar triangles are equal
but the corresponding sides are only proportional and not
In an equilateral triangle, the centroid, the orthocentre, necessarily equal.
the circumcentre and the incentre, all coincide.
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 Fig. 4.19 3. AC2 = CB·CD.

A D CONGRUENCE OF TRIANGLES
Two triangles will be congruent if at least one of the
following conditions is satisfied:
– Three sides of one triangle are respectively equal to
B C E F the three sides of the second triangle (normally
referred to as the S-S-S rule, i.e., the side-side-side
congruence).
For example, in Fig 4.19, ∠A = ∠D, ∠B = ∠E and
∠C = ∠F. Therefore ∆ABC is similar to ∆DEF. We write – Two sides and the included angle of one triangle are
∆ABC ∼ ∆DEF. (Note: We should not write any of the respectively equal to two sides and the included
following ∆ABC ∼ DFE, ∆ABC ∼ ∆ EFD. ∆ABC ∼ ∆EDF, angle of the second triangle (normally referred to as
∆ABC ∼ ∆FDE or ∆ABC ∼ ∆FED. The relation should the S-A-S rule, i.e., side-angle-side congruence).
indicate which pair of angles are equal.) The ratios of the – Two angles and the included side of a triangle are
corresponding sides are equal. respectively equal to two angles and the included side
i.e. AB = BC = CA of the second triangle (normally referred to as the A-
DE EF FD S-A rule, i.e., angle-side-angle congruence). (Note:
Hence AAS is also a sufficient condition for
By "corresponding sides", we mean that if we take a side congruence.)
opposite to a particular angle in one triangle, we should
consider the side opposite to the equal angle in the – Two right-angled triangles are congruent if the
second triangle. In this case, since AB is the side opposite hypotenuse and one side of one triangle are
to ∠C in ∠ABC, and since ∠C = ∠F, we have taken DE respectively equal to hypotenuse and one side of the
second triangle.
which is the side opposite to ∠F in ∆DEF.
In two congruent triangles,
Two triangles are similar if,
- the three angles of one triangle are respectively equal - the corresponding sides (i.e., sides opposite to
to the three angles of the second triangle, or corresponding angles) are equal.
- two sides of one triangle are proportional to two sides - the corresponding angles (angles opposite to
of the other and the included angles are equal, or corresponding sides) are equal.
- if the three sides of one triangle are proportional to - the areas of the two triangles will be equal.
the three sides of another triangle.
Some more useful points about triangles
In two similar triangles, Fig. 4.21
(a) Ratio of corresponding sides = Ratio of heights
A
(altitudes) = Ratio of the lengths of the medians
= Ratio of the lengths of the angle bisectors = Ratio
of inradii = Ratio of circumradii = Ratio of perimeters.
(b) Ratio of areas = Ratio of squares of corresponding P Q
sides
Fig. 4.20
A B C
A line drawn parallel to one side of a triangle divides the
other two sides in the same proportion. For example, in
Fig. 4.21, PQ is drawn parallel to BC in ∆ABC. This will
divide the other two sides AB and AC in the same ratio,
i.e., AP/PB = AQ/QC.
B D C
Fig. 4.22
In a right-angled triangle, the altitude drawn to the A
hypotenuse divides the given triangle into two similar
triangles, each of which is in turn similar to the original
m m
triangle. In triangle ABC in Fig. 4.20, ABC is a right-angled
triangle where ∠A is a right angle. AD is the perpendicular
drawn to the hypotenuse BC. The triangles ABD, CAD and P Q
CBA are similar because of the equal angles given below.
n n
In triangle ABC, ∠A = 90°. If ∠B = θ, then ∠C = 90° – θ.
In triangle ABD, ∠ADB = 90°. We already know that B C
∠B = θ; hence ∠BAD = 90° – θ. Conversely, a line joining two points lying on different
In triangle ADC, ∠ADC = 90°. We already know that sides of a triangle, dividing the respective sides in the
∠C = 90° – θ; hence ∠CAD = θ. same ratio, is parallel to the third side. In Fig. 4.22, P
We can write down the relationship between the sides in divides AB in the ratio m : n and Q divides AC in the ratio
these three triangles. The important relationships that m : n. Now, the line joining P and Q is parallel to the third
emerge out of this exercise are : m
side BC. Also, the length of PQ will be equal to
1. AD2 = BD·DC; m+n
2. AB2 = BC·BD; times the length of BC.

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We can say that a line drawn through a point on a side of TRAPEZIUM
the triangle parallel to a second side will cut the third side
Fig. 4.26
in the same ratio in which the first side is divided.
The line joining the midpoints of two sides of a triangle is D A
parallel to the third side and it is half the third side.
Fig. 4.23
A D m
P Q

R ↑ S
T n
↓
C E F B
If one side of a quadrilateral is parallel to the opposite
B C side, then it is called a trapezium. The two sides other
than the parallel sides in a trapezium are called the
Two triangles having the same base and lying between oblique sides.
the same pair of parallel lines have their areas equal
In Fig. 4.26, ABCD is a trapezium, in which AD is parallel
(Fig. 4.23).
to BC.
AD is parallel to BC. Hence, area of ∆ABC = area of ∆DBC
If the midpoints of the two oblique sides are joined, it is equal
QUADRILATERALS in length to the average of the two parallel sides, i.e., in Fig.
4.26, PQ = 1/2 [AD + BC]
Any four-sided closed figure is called a quadrilateral. By
imposing certain conditions on the sides and/or angles of In general, if a line is drawn in between the two parallel
a quadrilateral, we can get the figures trapezium, sides of the trapezium such that it is parallel to the parallel
parallelogram, rhombus, rectangle, square. sides and also divides the distance between the two
parallel sides in the ratio m : n (where the portion closer
Fig. 4.24 to the shorter of the two parallel sides is m), the length of
A D the line is given by :
E m n
(Longer side) + (Shorter side), where shorter
m+n m+n
side and longer side refer to the shorter and longer of the
F two parallel sides of the trapezium.
In Fig. 4.26, RS is the line parallel to AD and BC and the
ratio of the distances DT and TE is m : n.
B C mBC nAD
The length of RS is given by +
The sum of the four angles of a quadrilateral is equal to m+n m+n
360°. The perpendiculars drawn to a diagonal (in a
quadrilateral) from the opposite vertices are called
PARALLELOGRAM
"offsets". In Fig. 4.24, BE and DF are the offsets drawn to Fig. 4.27
the diagonal AC.
A B
If the four vertices of a quadrilateral lie on the circumference
of a circle (i.e., if the quadrilateral can be inscribed in a
circle) it is called a cyclic quadrilateral (refer to Fig. 4.25).
In a cyclic quadrilateral, the sum of opposite angles = 180°
i.e., in Fig. 4.25, A + C = 180° and B + D = 180°.
D E C
Fig. 4.25
A A quadrilateral in which both pairs of opposite sides are
parallel is called a parallelogram.
In a parallelogram
- both pairs of opposite sides are equal
- both pairs of opposite angles are equal
- Sum of any two adjacent angles is 180°.
B D - Each diagonal divides the parallelogram into two
congruent triangles.
- The diagonals bisect each other.
C
E Conversely, if in a quadrilateral, if any one of the
Also, in a cyclic quadrilateral, any exterior angle is equal following conditions holds, then it is a parallelogram.
to the interior angle at the opposite vertex i.e., in Fig. 4.25, - both pairs of opposite sides are equal
∠DCE is equal to ∠BAD. - both pairs of the opposite angles are equal
- the diagonals bisect each other
Now, we will look at different quadrilaterals and their - a pair of opposite sides are parallel and equal
properties.
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If two adjacent angles of a parallelogram are equal, then RECTANGLE
all four angles will be equal and each in turn will be equal
to 90°. Then the figure will be a rectangle. A rectangle is also a special type of parallelogram and
hence all properties of a parallelogram apply to rectangles
If any two adjacent sides of a parallelogram are equal, also. A rectangle is a parallelogram in which two adjacent
then all four sides are equal to each other and the figure angles are equal. Therefore, each of the angles is equal
is a rhombus. to 90°.
The diagonals of a rectangle are equal (and, of course,
Fig. 4.28 bisect each other).
A B
When a rectangle is inscribed in a circle, the diagonals
become diameters of the circle.
If a and b are the two adjacent sides of a rectangle, then
P
the diagonal is given by a2 +b2 .
D C If a rectangle and a triangle are on the same base and
between the same parallels, then the area of the triangle
If any point inside a parallelogram is taken and is joined
will be equal to half the area of the rectangle.
to all the four vertices the four resulting triangles will be
such that the sum of the areas of opposite triangles is SQUARE
equal. In Fig. 4.28, P is a point inside the parallelogram
ABCD and it is joined to the four vertices of the A square is a rectangle in which all four sides are equal
parallelogram by the lines PA, PB, PC and PD (or a rhombus in which all four angles are equal, i.e., all
respectively. Then Area of triangle PAB + Area of triangle are right angles) Hence, the diagonals are equal and they
PCD = Area of triangle PBC + Area of triangle PAD = Half bisect at right angles. So, all the properties of a rectangle
the area of parallelogram ABCD. as well as those of a rhombus hold good for a square.
Diagonal = 2 × Side
If there is a parallelogram and a triangle with the same
When a square is inscribed in a circle, the diagonals
base and between the same parallel lines, then the area
become the diameters of the circle.
of the triangle will be half that of the parallelogram.
When a circle is inscribed in a square, the side of the
If there is a parallelogram and a rectangle with the same square is equal to the diameter of the circle.
base and between the same parallel lines, then the areas
The largest rectangle that can be inscribed in a given
of the parallelogram and the rectangle will be the same.
circle will be a square.
The figure formed by joining the midpoints of the sides of
any quadrilateral taken in order, is a parallelogram.
POLYGON
RHOMBUS Any closed figure with three or more sides is called a
polygon.
Fig. 4.29
A polygon, which is such that for the line say ℓ containing
B
any one side, all the other sides are on the same side of ℓ
is called a convex polygon. It there is at least one side for
which the other sides lie on both sides of ℓ, the polygon is
A C said to be a concave polygon. A convex polygon is one in
P which each of the interior angles is less than 180°. It can
be noticed that any straight line drawn cutting a convex
polygon passes through only two sides of the polygon, as
shown in the figure below.
D Fig. 4.30 (a)
A rhombus is a parallelogram in which all sides are equal.
(Note: If in a parallelogram, one pair of adjacent sides are
equal, it is a rhombus).
Since a rhombus is a parallelogram, all the properties of a
parallelogram apply to a rhombus. Further, in a rhombus,
the diagonals bisect each other perpendicularly.
Convex Polygon
Conversely, any quadrilateral where the two diagonals In a concave polygon, it is possible to draw lines passing
bisect each other at right angles is a rhombus. through more than two sides, as shown in the figure below.
The four triangles that are formed by the two bisecting Fig. 4.30 (b)
diagonals with the four sides of the rhombus are all
congruent. In Fig. 4.29, the four triangles PAB, PBC, PCD
and PAD are congruent.
Side of a rhombus
= 1 4 ( Sum of squares of the diagonals)
Concave Polygon
nd
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A regular polygon is a convex polygon in which all sides A diameter of a circle is a straight line passing through the
are equal and all angles are equal. A regular polygon can centre of the circle and joining two points on the circle.
be inscribed in a circle. The centre of the circumscribing A circle is symmetric about any diameter.
circle (the circle in which the polygon is inscribed) of a
regular polygon is called the centre of the polygon. A chord (e.g. AB in Fig. 4.32) is a point joining any two
The names of polygons with three, four, five, six, seven, points on the circumference of a circle. A diameter is the
eight, nine and ten sides are respectively triangle, quad- longest chord in a circle.
rilateral, pentagon, hexagon, heptagon, octagon,
A secant is a line intersecting a circle in two distinct points
nonagon and decagon.
and extending outside the circle also.
The sum of interior angles of a convex polygon is equal to
(2n − 4) right angles where n is the number of the sides of A line that touches the circle at only one point is a tangent
the polygon. to the circle (R1R2 is a tangent touching the circle at the
If each of the sides of a convex polygon is extended, the point R in Fig. 4.32).
sum of the external angles thus formed is equal to 4 right
Angles (i.e., 360°). If PAB and PCD are two secants (in Fig. 4.31), then
PA.PB = PC.PD
In a regular polygon of n sides, if each of the interior
2n − 4 If PAB and PCD are secants and PT is a tangent to the
angles is d°, then d = ( 90 ° ) and each exterior
n circle at T (in Fig. 4.31), then PA.PB = PC.PD = PT².
360 °
angle = . Two tangents can be drawn to the circle from any point
n
outside the circle and these two tangents are equal in
It will be helpful to remember the interior and exterior length. In Fig. 4.32, P is the external point and the two
angles of the following regular polygons: tangents PX and PY are equal.
Interior Exterior
A tangent is perpendicular to the radius drawn at the
Regular pentagon 108° 72°
Regular hexagon 120° 60° point of tangency (In Fig. 4.32, R1R2 ⊥ OR). Conversely,
if a perpendicular is drawn to the tangent at the point of
Regular octagon 135° 45°
tangency, it passes through the centre of the circle.
If the centre of a regular polygon (with n sides) is joined
with each of the vertices, we get n identical triangles A perpendicular drawn from the centre of the circle to
inside the polygon. In general, all these triangles are a chord bisects the chord (In Fig. 4.32, OC, the
isosceles triangles. Only in case of a regular hexagon, perpendicular from O to the chord AB bisects AB) and
all these triangles are equilateral triangles, i.e., in a regular conversely, the perpendicular bisector of a chord passes
hexagon, the radius of the circumscribing circle is equal to through the centre of the circle.
the side of the hexagon.
Two chords that are equal in length will be equidistant
A line joining any two non-adjacent vertices of a polygon from the centre, and conversely two chords which are
is called a diagonal. A polygon with n sides will have equidistant from the centre of the circle will be of the same
n(n - 3) length.
diagonals.
2
One and only one circle passes through any three given
CIRCLES non-collinear points.
Fig. 4.31
Fig. 4.33
B P
A
X Y
C
P D
T Q
A circle is a closed curve drawn such that any point on the When there are two intersecting circles, the line joining the
curve is equidistant from a fixed point. The fixed point is centres of the two circles will perpendicularly bisect the
called the centre of the circle and the distance from the line segment joining the points of intersection. In Fig. 4.33,
centre to any point on the circle is called the radius of the the two circles with centres X and Y respectively intersect
circle. at the two points P and Q. The line XY (the line joining the
Fig. 4.32 centres) bisects PQ (the line segment joining the two
R1 points of intersection).
X
Two circles are said to touch each other if a common
tangent can be drawn touching both the circles at the
90° O same point. This is called the point of contact of the two
R B circles. The two circles may touch each other internally
P
(as in Fig. 4.34) or externally (As in Fig. 4.35). When two
C circles touch each other, then the point of contact and
A the centres of the two circles are collinear, i.e., the point
Y of contact lies on the line joining the centres of the
R2
two circles.
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 The transverse common tangent is at the point of contact
Fig. 4.34
(Refer to Fig. 4.35).
Fig. 4.38

If two circles touch internally, the distance between the Two circles which are non-intersecting and non-enclosing
centres of the two circles is equal to the difference in the (i.e. one does not lie inside the other) have four common
radii of the two circles. When two circles touch each other tangents - two direct and two transverse common tangents
externally, then the distance between the centres of the (Refer to Fig. 4.38).
two circles is equal to the sum of the radii of the two
circles. If r1 and r2 are the radii of the two non-intersecting non-
Fig. 4.35 enclosing circles,

Length of the direct common tangent

= (Distance between centre)2 - (r1 - r2 )2

Length of transverse common tangent

= (Distance between centre)2 - (r1 + r2 )2

A line which is tangent to two circles is called a common Two circles are said to be concentric if they have the same
tangent. In general, for two circles, there can be anywhere centre. For two concentric circles, the circle with the
from zero to four common tangents drawn depending on smaller radius lies completely within the circle with the
the position of one circle in relation to the other. bigger radius.

A common tangent which cuts the line joining the centres ARCS AND SECTORS
of the two circles, not between the centres but rather
beyond either centre, or if the centres of the two circles lie Fig. 4.39
on the same side of the common tangent, the tangent is
called a direct common tangent. Instead, if they lie on X D Y
opposite sides, it is called a transverse common tangent.
Fig. 4.36
O

A • B
C
An arc is a connected part of a circle. In Fig. 4.39, ACB is
If two circles are such that one lies completely inside the called minor arc and ADB is called major arc. In general,
other (without touching it), then there will not be any if we talk of an arc AB, we refer to the minor arc. AOB is
common tangent to these circles (refer to Fig. 4.36). called the angle formed by the arc AB (at the centre of the
circle).
Two circles touch each other internally (i.e., one circle lies
inside the other), then there is only one common tangent The angle subtended by an arc at the centre is double the
possible and it is drawn at the point of contact of the two angle subtended by the arc at any point X (or Y) on the
circles (refer to Fig. 4.34). remaining part of the circle.
Fig. 4.37
In Fig. 4.39, ∠AOB = 2∠AXB = 2∠AYB
Angles in the same segment are equal. In Fig. 4.39,
∠AXB = ∠AYB.
Fig. 4.40

Two intersecting circles have two common tangents. Both R S

these are direct common tangents and the two
intersecting circles do not have a transverse common
U
tangent (refer to Fig. 4.37).

Two circles touching each other externally have three

common tangents. Out of these, two are direct common
Q T P
tangents and one is a transverse common tangent.
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The angle between a tangent and a chord through the AREAS OF PLANE FIGURES
point of contact of the tangent is equal to the angle made
by the chord in the alternate segment (i.e., segment of the Mensuration is the branch of geometry that deals with the
circle on the side other than the side of location of the measurement of length, area and volume. We have
angle between the tangent and the chord). This is looked at properties of plane figures till now. Here, in
normally referred to as the "alternate segment theorem." addition to areas of plane figures, we will also look at
In Fig. 4.40, PQ is a tangent to the circle at the point T and surface areas and volumes of "solids." Solids are objects,
TS is a chord drawn at the point of contact. Considering which have three dimensions (plane figures have only two
∠PTS which is the angle between the tangent and the dimensions).
chord, the angle TRS is the angle in the "alternate
segment". So, ∠PTS = ∠TRS. Let us briefly look at the formulae for areas of various
Similarly, ∠QTS = ∠TUS. plane figures and surface areas and volumes of various
solids.
Fig. 4.41
TRIANGLES
The area of a triangle is represented by the symbol ∆. For
C B any triangle, the three sides are represented by a, b and
90°
c and the angles opposite these sides represented by A,
B and C respectively.
(i) For any triangle in general,
A
(a) When the lengths of three sides a, b, c are given,
We have already seen in quadrilaterals, that the opposite Area = s(s - a)(s - b)(s - c) , where
angles of a cyclic quadrilateral are supplementary and a+b+c
that the external angle of a cyclic quadrilateral is equal to s=
2
the interior opposite angle. This is called Hero's formula.
The angle in a semicircle (or the angle the diameter (b) When base (b) and altitude (height) to that base
subtends in a semicircle) is a right angle. The converse of are given,
the above is also true and is very useful in a number of Area = ½ ( base)(altitude) = ½ bh
cases - in a right angled triangle, a semi-circle with the (c) Area = ½ absinC = ½ bcsinA = ½ casinB
hypotenuse as the diameter can be drawn passing
through the third vertex (Refer to Fig. 4.41). abc
(d) Area = , where R is the circumradius of the
4R
Fig. 4.42 triangle.
B (e) Area = rs, where r is the inradius of the triangle
and s, the semi-perimeter.
A Out of these five formulae, the first and the second
θ are the most commonly used and are also more
important from the examination point of view.
o
(ii) For a right angled triangle,
Area = (½) Product of the sides containing the right
angle
The region bounded by an arc and the two radii at the two
end points of the arc is called sector. (iii) For an equilateral triangle
3 a2
In Fig. 4.42, the shaded figure AOB is called the minor Area = , where a is the side of the triangle
4
sector. The rest of the circle is the major sector.
3a
Fig. 4.43 The height of an equilatera l triangle = .
2

(iv) For an isosceles triangle

b 4

A B
4a

b
2

Area = − , where a is length of either of the

⋅
O two equal sides and b is the third side.

QUADRILATERALS
(i) For any quadrilateral
The region bounded by an arc and the line segment
Area of the quadrilateral = ½ (One diagonal) (Sum of
joining the endpoints of the arc is called a segment of a
the offsets drawn to that diagonal)
circle (or circular region). In fig 4.43, the shaded region is
a minor segment. The rest of the circular region is a major For example, for the quadrilateral ABCD shown in Fig.
segment. 4.24, area of quadrilateral ABCD = ½ x AC x (BE + DF)

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(ii) For a cyclic quadrilateral where the lengths of the four Examples:
sides are a, b, c and d,
Area = (s - a) (s - b) (s - c) (s - d) , where s is the semi- 4.01. Calculate the value of x in the figure.
perimeter, i.e., s = (a + b + c + d)/2
(iii) For a trapezium
Area = ½(Sum of parallel sides) Distance between 40°
170° 20°
them
= (½) (AD + BC) (AF) (refer to Fig. 4.26)
(iv) For a parallelogram x
(a) Area = Base (Height)
(b) Area = Product of two sides (Sine of included 25°
angle)
(v) For a rhombus
Area = ½ (Product of the diagonals) Sol: The angles at a point add up to 360°.
Perimeter = 4 x Side of the rhombus Hence (20° + 40° + 25° + 170°) + x = 360°
∴ x = 360° − 255° = 105°
(vi) For a rectangle
Area = Length (Breadth) 4.02. If l, m and n are lines parallel to each other,
Perimeter = 2(ℓ + b), where ℓ and b are the length and calculate the measures of the angles α and β in
the given figure. Given that angle ABC is 100°
the breadth of the rectangle respectively
and angle DBC is 75°. Also A, E; D, B; and C, F
(vii) For a square are points on l, m and n respectively.
(a) Area = Side2 E
(b) Area = (½)(Diagonal 2) A l
•
α
[diagonal = 2 (side), β
Perimeter = 4(Side)]
• m
(viii) For a polygon D B
(a) Area of a regular polygon = ½ (Perimeter)
(Perpendicular distance from the centre of the • n
C F
polygon to any side)
(The centre of a regular polygon is equidistant Sol: As ∠DBC = 75°
from all its sides) ∠DBA = 100° −∠DBC = 100° − 75° = 25°
(b) For a polygon which is not regular, the area has α = ∠ DBA = 25°
to be found out by dividing the polygon into (As α and DBA are alternate angles, they are equal)
suitable number of quadrilaterals and triangles As interior angles on the same side of the
and adding up the areas of all such figures transversal are supplementary,
present in the polygon. α + β = 180° ⇒ ∴β = 180° − 25 = 155°

CIRCLE 4.03. How many degrees are there in an angle which

is equal to one-fourth of its supplement?
(i) Area of the circle = π r2, where r is the radius of the
circle Sol: Let the angle be x. Then its supplement is 180°− x.
Circumference = 2π r 1
Given, x = (180° − x) ⇒ 5x = 180° ⇒ x = 36°.
4
(ii) Sector of a circle
θ
Length of arc = ( 2 π r) 4.04. How many degrees are there in an angle which
360 °
θ

is equal to one-fifth of its complement?

π
r
2

( ) , where θ is the angle of the

360

Area =
° Sol: Let the angle be x. Then its complement is 90° − x.
sector in degrees and r is the radius of the circle. Given that, x = 1 (90° − x)
Area = (1/2)lr; l is length of arc and r is radius. 5
⇒ 6x = 90° ⇒ x = 15°
(iii) Ring : Ring is the space enclosed by two concentric
circles. 4.05. In a right-angled triangle ABC, find the length of
the altitude BD drawn from B to the hypotenuse
Area = πR2 – πr2 = π(R + r) (R – r), where R is the radius AC, given AB = 9 cm and BC = 12 cm.
of the outer circle and r is the radius of the inner circle.
Sol: Since the sides containing the right angle are
9 cm and 12 cm, the area of the triangle
ELLIPSE
= 1 ( 9 )(12 ) = 54 cm2.
2
Area = π ab where "a" is semi-major axis and "b" is semi-
As AB = 9 cm, BC = 12 cm, AC will be 15 cm.
minor axis.
Considering AC as the base and BD as the
Perimeter = π (a + b)
altitude, the area of ∆ABC
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 = 1 ( AC )( BD ) = 1 (15 )( BD ) cm = 54 cm 2 4.09. A and B leave a point at the same time.
2 2 A travels North at 18 km/hr and B travels West
at 24 km/hr. Find the distance between A and B
54(2) cm2 108 after two hours.
BD = = cm = 7.2 cm.
15 cm 15
Sol: Let the starting point be O. After 2 hours, let A be
at X, which will be 2 × 18 = 36 km from O and B
4.06. In the figure given below, ABC is a triangle in
be at Y, which will be 2 × 24 = 48 km from O in
which D is the midpoint of the side BC and E is
the direction perpendicular to OX. Using
the midpoint of AD. What is the ratio of the areas
Pythagoras theorem we can find the third side
of ∆BED and ∆ABC?
XY. XY = 36 2 + 48 2 = 60 km . Hence A and B
A
will be 60 km away from each other.

4.10. In the figure given below, if the length of major arc

AB = 154 cm, find the length of the radius OA and
E find the area of the minor sector AOB.
Sol:

B D C
O
Sol: Since AE = ED, BE is a median on AD and the 150°
median divides ∆ABD into halves, such that area A B
of ∆BED = Area of ∆AEB = 1 (Area of ∆ABD).
2 Angle of the major arc = 360° − 150° = 210°
Since AD is the median, it divides ∆ABC into Length of the major arc =
Angle of the arc
( 2 πr )
halves such that area of ∆ABD = area of ∆ADC = 360 °
1 210o  22 
(Area of ∆ABC).
2 ∴ (2) (OA) = 154 cm ⇒ OA = 42 cm.
o
Hence the area of ∆BED: Area of ∆ABC
360  7
Area of the minor sector
1 
( )
= 1  Area of ∆ABC  : Area of ∆ABC =
Angle of the sec tor
× πr 2
2 2  360 °
=
1
: 1 = 1 : 4. 150 22 2
4 =  (42 ) cm2 = 2310 cm2.
360 7 
4.07. Triangles ABC and DEF are similar. If∠A = ∠D, 4.11. In a circle with centre O, PT and PS are tangents
∠C = ∠F, BC = 3.4 cm, DE = 4.8 cm and AB = drawn to it from point P. If PT = 24 cm and OT =
1.6 cm, then find the length of EF. 10 cm, then find the length of PO.
Sol: Given that triangles ABC and DEF are similar. Sol: T
Hence the corresponding sides are proportional.
AB BC
∴We have, =
DE EF P
O
(BC) (DE) (3.4)(4.8) cm 2
Hence EF =
(AB) = (1.6) cm
= 10.2 cm.

S
4.08. In the figure given below, AD is perpendicular to
From the above figure
BC in triangle ABC. Find x and y.
∠OTP = 90°, (Radius makes an angle of 90° with
A the tangent at the point of tangency).
In triangle OTP,
PO2 = PT2 + OT2 = 242 + 102 cm2
10 cm 17 cm PO = 24 2 + 10 2 cm = 26 cm
x cm
4.12. In parallelogram ABCD, DP and CP are the angle
bisectors of ∠CDA and ∠DCB respectively.
B 6 cm D y cm C Find ∠CPD.
Sol: D C
Sol: From ∆ABD, x2 = AB2 − BD2
(By Pythagoras theorem)
•
x2 = 102 − 62 = 64 ⇒ x = 8. P
From ∆ADC y2 = AC2 − x2
(By Phythagoras theorem) A B
y = 17 2 − 8 2 = 225 = 15 Given DP is the angle bisector of ∠CDA
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 4.14. For a polygon of 12 sides, find the sum of the
⇒ ∠CDP = 1 ∠CDA ----- (1)
2 interior angles of the polygon.
Given CP is the angle bisector of ∠DCB
Sol: Sum of the interior angles of a polygon.
⇒ ∠DCP = 1 ∠DCB ---- (2). = (2n − 4) × 90° where n is the number of sides.
2
∴Sum of the interior angles of this polygon
In a parallelogram the sum of adjacent angles
= 180°. = [2(12) − 4] 90º = 1800º.
Therefore, ∠CDA + ∠DCB = 180°.
4.15. If one angle of a pentagon is 40º and all the other
From (1) and (2),
angles are equal, find the measure of each of the
2 ∠CDP + 2 ∠DCP = 180°
other angles.
⇒ ∠CDP + ∠DCP = 90°
In triangle ∠PDC, Sol: Sum of the interior angles of a pentagon.
∠CPD = 180° − (∠PDC + ∠PCD) = [(2 (5) − 4)] 90° = 540º
= 180° − 90° = 90° One angle is 40°, the sum of the remaining four
equal angles
4.13. The number of sides of a regular polygon is 18. = 540° − 40° = 500°
Find the interior angle of the polygon. 500 °
Each of the other four angles = = 125 °.
Sol: Exterior angle of a regular polygon 4
= 360°/n = 360°/18 = 20°.
The sum of the interior angle and the exterior
angle is 180°.
Interior angle of this polygon = 180° − 20°
= 160°.

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 Concept Review Questions
Directions for questions 1 to 35: For the Multiple Choice Questions, select the correct alternative from the given choices.
For the Non-Multiple Choice Questions, write your answer in the box provided.

1. Find the circumradius of a triangle whose sides are 9. In triangle PQR, if ∠P = 50°, find ∠QIR, where I is the
6 cm, 8 cm and 10 cm. (in cm) incentre of the triangle. (in degrees)

2. The sides of a triangle are 12 cm, 16 cm and 20 cm. 10. In ∆ABC, AB2 + AC2 < BC2 and ∠BAC= x° , then
Find the sum of the distances from the orthocentre to which of the following is true?
the vertices of the triangle. (in cm) (A) x < 90° (B) x = 90°
(C) x > 90° (D) None of these

11. The sides of a triangle are 4 cm, 6 cm and 8 cm. Its

3. In a triangle, the smallest altitude is 18 cm. circumcentre lies _____.
The inradius is always (A) inside the triangle
(A) > 9 cm (B) 9 cm (B) on the triangle
(C) < 9 cm (D) Either (A) or (B) (C) outside the triangle
(D) either inside or on the triangle
4. In an equilateral triangle, the orthocentre divides each
median into 2 parts. The larger part and the smaller 12. The point of concurrence of angle bisectors of a
part are in the ratio of triangle is called the ______ of the triangle.
(A) 3 : 2 (A) incentre (B) circumcentre
(B) 3 : 1 (C) centroid (D) orthocentre
(C) 2 : 1 13. In triangle PQR, S and T are points on PQ and PR
(D) None of these respectively, such that ST is parallel to QR. If PS = 8 cm,
SQ = 4 cm and TR = 6 cm, find PT. (in cm)
5. An equilateral triangle has a side of 18 cm. Find the
inradius of the triangle (in cm).
(A) 3 3 (B) 4 3 (C) 6 3 (D) 8 3
14. PQRS is a cyclic quadrilateral ∠P = 50° and ∠Q = 70°.
Find the respective measures of ∠R and ∠S.
6. In the above question, find the cirumradius of the
(A) 50°, 70° (B) 70°, 50°
triangle (in cm).
(C) 110°, 130° (D) 130°, 110°
(A) 3 3 (B) 4 3 (C) 6 3 (D) 8 3
15. Triangle PQR is right angled at Q. QS is an altitude to
7. A PR. Find QS if PS = 32 cm and SR = 18 cm. (in cm)

F E 16. ABCD is a trapezium. AB is parallel to CD.

G
AB = 12 cm and CD = 24 cm. Find the length of the
line EF which is parallel to AB and CD and divides the
B distance between them in the ratio 2 : 3. (in cm)
C
D
In the figure, AD, BE and CF are the medians and G
is the centroid. If the area of triangle ABC is
17. In triangle PQR, S and T are points on PQ and PR
24 cm2, find the area of triangle GFB (in cm2).
respectively such that ST is parallel to QR. PS = 4 cm,
SQ = 12 cm and ST = 3 cm. Find QR (in cm).
(A) 9.6 (B) 8.4 (C) 10.8 (D) 12
8. A 18. The number of diagonals of a regular decagon is

19. The radius of a circle is 15 cm. Find the length of a

chord which is 9 cm from the centre of the circle.
(in cm)
B D C

In the figure, AB = 10 cm, AC = 12 cm and DC = 8 cm. If

AD bisects ∠BAC, find BD (in cm). 20. P and Q are points on the circumference of a circle
2 1 2 with centre O. R is a point on the major arc PQ.
(A) 5 (B) 5 (C) 6 (D) 6 ∠POQ = 100°. Find ∠PRQ.
3 3 3
(A) 40° (B) 45° (C) 50° (D) 55°
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21. XY is the diameter of a circle. Z is a point on the 28.
circumference of the circle. XY = 26 cm and A
XZ = 24 cm. Find YZ (in cm).

O
50° C
22. Among a parallelogram, a rhombus and an isosceles
trapezium, how many are necessarily cyclic
quadrilaterals? B
In the figure, O is the centre of the circle. AC and BC
are tangents to the circles. If ∠ACB = 50°, find ∠AOB.
(A) 130° (B) 80° (C) 100° (D) 90°
23.
A
D 29. V

P Q
O X
R S
B M N
C

In the figure, ∠BCA = 45° and ∠BDC = 50°. Find ∠ABC T U

(in degrees). Z

W Y
In the figure, ∠XZT = 130°. PQ, RS and TU are
24. A regular hexagon has a side of 4 cm. Find its area parallel. VW and XY are parallel. Find ∠VOP.
(in cm2). (A) 125° (B) 130° (C) 120° (D) 135°
(A) 24 3 (B) 18 3
30. The midpoints of the sides of a quadrilateral of area
(C) 36 3 (D) None of these 200 sq.cm are joined. What type of quadrilateral is
formed and what is its area?
25. A right circular cone is cut parallel to its base at (A) Parallelogram, 100 sq.cm.
one-third of its height from the top. Find the ratio (B) Rectangle, 100 sq.cm.
of the volume of the smaller piece to that of the (C) Rectangle, 50 sq.cm.
original cone. (D) Parallelogram, 50 sq.cm.
(A) 1 : 26 (B) 1 : 27 (C) 2 : 25 (D) 1 : 8
31. Find the maximum number of common tangents
26. P that can be drawn to two non-intersecting and
Q non-enclosing circles.
R (A) 1 (B) 2 (C) 3 (D) 4
60°
O 32. Find the maximum number of common tangents that
can be drawn to two circles which touch each other
externally.

In the figure, O is the centre of the circle and

∠POR = 60°. Find ∠PQR (in degrees). 33. Find the maximum number of common tangents that
can be drawn to two circles which intersect each other.

27. X
34. A triangle has its circumcentre on one of its sides. It
can be _____.
(A) acute–angled
Y (B) obtuse–angled
(C) right–angled
(D) acute, obtuse or right-angled
Z 35. An equilateral triangle has a side of 12 cm. Find the
In the figure, XZ is the diameter of the circle. area (in cm2) of the triangle formed by the incentre,
∠XZY = 35°. Find ∠YXZ (in degrees). centroid and the circumcentre.

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 Exercise – 4(a)
Directions for questions 1 to 35: For the Multiple Choice Questions, select the correct alternative from the given choices.
For the Non-Multiple Choice Questions, write your answer in the box provided.

1. (i) In the given figure, lines m and n are parallel and (A) 10 3 cm
t is the transversal.
(B) 15 3 cm
t
1 (C) 20 3 cm
2
m (D) 24 3 cm
4 3

4. In the given figure, QS : SR = PV : VS = 4 : 3,

PT = 8 cm. Find PR (in cm).
P
5 6
n
8
7
T
V

Q R
∠8 is less than ∠3 by 90°. Consider the following S
statements:
Ι. ∠1 = ∠3 = ∠5 = ∠7 = 135°
ΙΙ. ∠2 = ∠4 = ∠6 = ∠8 = 45°
5. The total number of diagonals in a regular polygon is
20. What is the interior angle of the polygon?
Which of the following statements is/are true?
(A) 90° (B) 115° (C) 120° (D) 135°
(A) Only Ι (B) Only ΙΙ
(C) Both Ι and ΙΙ (D) Neither Ι nor ΙΙ 6. If each angle of a regular polygon of n sides is 144°,
find the value of n.
(ii) In the given figure, ℓ, m and n are parallel lines.
PQ is a transversal intersecting the three lines.
The perpendicular distances between (ℓ and m)
and (m and n) are in the ratio 3 : 4. If AC = 12 cm, 7. An equilateral triangle has a side of 36 cm. Find the
what is the length of AB (in cm)? circumradius of the triangle (in cm).
P (A) 6 3 (B) 8 3 (C) 12 3 (D) 16 3
A ℓ
8. How many scalene triangles with integral sides
m (in cm) have a perimeter of 20 cm?
C (A) 3 (B) 4 (C) 5 (D) 6
n
B 9. In ∆ABC, ∠ABC = 80°, ∠ACB = 40°, AP is the
Q
bisector of ∠BAC and AQ ⊥ BC. Find ∠PAQ.
(A) 10° (B) 15° (C) 20° (D) 35°

10. In triangle PQR, PA, QB and RC are the medians.

The centroid is at point G. What is the ratio of the area
2. In the given figure, PR || TS and PU || RS. Find ∠TPU of quadrilateral CQAG and the area of triangle PQR?
(in degrees). (A) 1 : 4 (B) 1 : 3 (C) 1 : 2 (D) 1 : 6
T
P U 11. In the given figure, QS = QT and RU = RT. ∠QPR =
36°. Find ∠STU.
60°
P
50°
Q
R S

S
U
3. ABC is a right-angled triangle, right-angled at B.
An equilateral triangle ABD is constructed on side AB. Q R
T
A line is drawn from D parallel to BC, meeting AB at
the point E. Find the length of DE, if AC = 41 cm and BC
= 9 cm. (A) 96° (B) 84° (C) 72° (D) 60°

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12. (i) A R
Q

P
S T

(A) 16 15 cm2, 9 15 cm2

(B) 8 15 cm2, 4.5 15 cm2
(C) 24 15 cm2, 13.5 15 cm2
(D) None of these
16. In the given figure (not to scale), PQRS is a rectangle
inscribed in a circle with centre O. PQ > PS. Areas of
the rectangle PQRS and the circle are in the ratio
2 5 : 3π and ∠OQP = ∠TQR. Find QR : RT.
B D C
In the given triangle ABC, AD is the bisector of
P Q
∠A. If AB = 12 cm, AC = 16 cm and BD = 3 cm,
what is the length of BC? .O
(A) 8 cm (B) 15 cm (C) 6 cm (D) 7 cm
S T R
(ii) ABC is a right-angled triangle, right angled at B.
The bisector of the external angle at A meets CB
produced at D. If AC = 5 cm, BC = 4 cm, find the
length of CD. (A) 1: 5 (B) 2 : 5 (C) 5 :1 (D) 5 :2
(A) 5 cm (B) 7 cm (C) 8 cm (D) 6 cm
17. In trapezium ABCD, AB and CD are the parallel sides.
The diagonals AC and BD intersect at X. The ratio of
13. A the area of ∆ AXB to the area of ∆ CXD is
25 : 49. Find the ratio of the area of ∆ AXD and the
P area of ABCD.
F E 35 5 1 1
(A) (B) (C) (D)
144 7 4 5
Q R
18. ABC is a triangle. D and E are points on AB and AC
B D C respectively. F is a point on BC. DF is parallel to AC
and EF is parallel to AB. The areas of the triangles
In the given figure, if the area of triangle ABC is EFC and ABC are 64 and 400 respectively. Find the
64 cm2, then what is the area of the triangle PQR, area of the quadrilateral ADFE.
where D, E and F are the midpoints of the sides of
∆ABC and P, Q and R are the midpoints of the sides
of ∆DEF (in cm2)?
19. In the given figure, AB is parallel to QR and the ratio
of the areas of triangles PAB and PQR is 1 : 4. If the
perimeter of PQR is 24 cm, what is the perimeter of
14. In the given figure, PS = 32 cm, PQ = 18 cm, PAB (in cm)?
P
SR = 25 cm and ∠QRP = ∠QSR.
S

Q A B

Q R
P

R
Find the ratio of the perimeters of the triangle PQR
20. PQR is a triangle in which PQ = PR = 26 cm. S is a
and PRS.
point on segment QR such that SR = 3 cm and
(A) 3 : 2 (B) 3 : 1 (C) 2 : 1 (D) 3 : 4
PS = 25 cm. Find the length of QS.
(A) 16 cm (B) 17 cm (C) 18 cm (D) 20 cm
15. In the given figure radii of two circles with centres
S and T respectively are in the ratio 5 : 3 and 21. In the right-angled triangle ABC, the right angle is at
PT = 12 cm. If R and Q are points of contacts of the B and BD is perpendicular to AC. The lengths of AD
tangent drawn from P to the circles with centres S and and CD are 5 and 6 respectively. Find AB2.
T respectively, then find the areas of the quadrilateral (A) 144 (B) 121
RQTS and the triangle PQT. (C) 132 (D) None of these
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22. In triangle ABC, AD and BE are medians and G is the 27. In the given figure, PT and PR are two tangents to the
centroid. ∠AGE = 30°, AD = 12, BE = 18. Find the circle at T and R respectively. If O is the centre of the
area of the triangle. circle and ∠TPR = 60°, find ∠TSR.
T

23. ABCD is a trapezium, in which AD and BC are

parallel. If the four sides AB, BC, CD and DA are 8, O S
P
10, 12 and 16 respectively, then find the magnitude
of the sum of the squares of the two diagonals.

R
24. In the given figure given, PQRS is a parallelogram. (A) 40° (B) 60°
Find y (in degrees). (C) 120° (D) None of these
P Q
28. Three circles with centres X, Y and Z with radius
x + 20° 30 cm each intersect each other as shown. AB = 12 cm,
CD = 6 cm and EF = 18 cm. Find the perimeter of
triangle XYZ (in cm).
y + 10°
x – 40°
S R
X
C A

25. D B
A D

Y F E Z

B C
In the given figure, ABCD is a parallelogram and the
lengths of AC and BD are 10 units and 12 units
respectively. The sum of the squares of AB and BC is 29. Two parallel chords of a circle of radius 25 cm have
_____. lengths of 30 cm and 48 cm. Find the distance
(A) 120 (B) 140 (C) 122 (D) 128 between the two chords (in cm).
26. (i) In the given figure, ABCD is a cyclic quadrilateral. (A) 6.5 or 13.5 (B) 13 or 27
If ∠CDX = 130°, then what is ∠AEC (in degrees)? (C) 19.5 or 40.5 (D) None of these

X 30. A, B, C and D are points on the circumference of a

E circle. ABD is equilateral and AC is the diameter of
D the circle. Find the ratio of the perimeter of ABCD and
the length of AC.
A 3 1
(A) 1+ (B) + 3
2 2
(C) 1+ 3 (D) 2 + 3
B
31.
P S R
C
Y•
X•
(ii) P
120
Q 160

T O Q
R
In the given figure, triangle PQR is right-angled at Q.
S QS is the altitude to side PR. Circles with centres X
and Y are inscribed in ∆PQS and ∆RQS respectively.
In the given figure, O is the centre of the right circle. Find XY2.
∠OSR = 20°. Find ∠PTS (in degrees).

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32. Triangle PQR has a perimeter of 24 cm. X is a point adjacent points by straight lines. Find the perimeter of
inside PQR which is equidistant from each vertex of the polygon so formed.
PQR. ∠QPX = 35° and ∠PRX = 25° (A) 30( 6 − 2 ) (B) 60( 6 − 2 )
Consider the following statements
Ι. PQ is less than 8 cm. (C) 60( 6 + 2 ) (D) 30( 6 + 2 )
ΙΙ. PR is more than 8 cm.
35. A
Which statement(s) is/are true?
(A) Only Ι (B) Only ΙΙ
(C) Both Ι and ΙΙ (D) Neither Ι nor ΙΙ E
F
33. The sides of a cyclic quadrilateral are 10 cm, 14 cm,
12 cm and 15 cm. If one of its diagonals is 15 cm,
then find the other diagonal (in cm). B D C
In the given figure, AB = 5 cm, BC = 6 cm, CA = 7 cm.
Find DC (in cm).
34. There are 12 uniformly spaced points on a circle with
radius 10. Each of these points is joined with the two

Exercise – 4(b)
Directions for questions 1 to 55: For the Multiple Choice diagonals. The perimeter of any of these triangles is
Questions, select the correct alternative from the given 60 cm. Find the length of the shorter diagonal (in cm).
choices. For the Non-Multiple Choice Questions, write
your answer in the box provided.

Very Easy / Easy… … ……… … 5. AB is a chord of the circle with centre O. M is the
midpoint of AB and P is a point on BA produced. If PB
1. ℓ1 is 8 cm and PA is 3 cm, then what is the difference of
x° PM2 and AM2 (in cm2)?
(A) 20 (B) 40 (C) 24 (D) 26
75°
6. In the given figure,
E

2x° C
ℓ2 P Q

In the given figure, ℓ1 || ℓ 2. What is the value of x? R S

T B
A
2. In the following figure, find ∠BZX (in degrees).
F
A PQ || AB, AB || RS and RC || EF. If ∠CRS = 50°, find
∠STA (in degrees).

P
120°
X 7. In the following figure, ∠RQZ = 2∠QRS and PQ || ST.
Find ∠RQZ.
Q
25° P
80°
40°
Y Z
Q B S
Z T
M
120°
Moderate… … … …… ..

3. In triangle ABC, ∠B – ∠A = ∠C – ∠B and ∠A : ∠C = R

3 : 5. Find the difference of the largest and the (A) 20° (B) 30° (C) 40° (D) 60°
smallest angles.
(A) 20° (B) 25° (C) 30° (D) 35° 8. If X is any point inside the triangle ABC and P is the
perimeter of triangle ABC, then AX + BX + CX is
4. A rhombus is divided into two triangles by its longer greater than _______.
diagonal. The perimeter of either of these triangles is 3P P
(A) 3P (B) 2P (C) (D)
100 cm. It is divided into four triangles by its 2 2
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9. In triangle ABC, AB = 12 cm and BC = 21 cm. What is 14. ABCD is a trapezium in which AB is parallel to DC.
the range of the perimeter (p) of ABC? The diagonals AC and BD intersect at the point E and
(A) 3 cm < p < 21 cm (B) 9 cm < p < 33 cm AB : CD = 2 : 3. If the area of triangle AEB = 64 sq.cm,
(C) 33 cm < p < 66 cm (D) 42 cm < p < 66 cm find the area of triangle CED (in cm2).

10. In the figure below, PQRS is a rectangle. QR = 2PQ

= 2QT and ∠RTQ = 90°. Find ∠QTP.
15. In the given figure ABCD is a rectangle. Find the
S R length of the line joining the midpoints of BF and EF.
A 24 B

T F
4
D 16 E 8 C
P Q
(A) 2 13
(A) 15° (B) 20° (C) 25° (D) 30°
(B) 3 3
11. In the figure AC = CD. Find ∠BCT (in degrees). (C) 13
A (D) None of the previous choices

16. In the given figure, AB = 30 cm, CD = 45 cm and BC

25°

= 15 cm. Find EF (in cm).

D

A
E
C
40° 35°

B 20° B F C
T
D
17. The sides of triangle ABC are a cm, b cm, and c cm.
The sides of triangle DEF are d cm, e cm and f cm.
12. In the figure, ∠EBC = 25°, ∠BAC = 35° and ∠AED a2 + b2 + c2 = 50
= 80°. Find ( ∠ABC + ∠EAD + ∠ADE) (in degrees). d2 + e2 + f 2 = 50
ad + be + cf = 50
A
What can be said about the two triangles?
35° (A) They have the same perimeter.
(B) They have the same area.
(C) Both (A) and (B)
80° D (D) Neither (A) nor (B)
B 25° E
18. The perimeters of similar triangles ABC and PQR are
C 32 cm and 40 cm respectively. If the area of triangle
PQR is 100 cm2, find the area of triangle ABC (in cm2).

13. A 19. In the given figure, PS is the median drawn to the side
QR of triangle PQR. If PQ = 11 cm, PR = 13 cm and
30° PS = 8 cm, what is the length of QS?
P
60° R
B D C
In the given figure, the lengths of BD and DC are in S
the ratio of 2 : 3 and BC = 5 cm. Find AC. Q
(A) 17 cm (B) 23 cm (A) 4 14 cm (B) 9 cm

(C) 29 cm (D) 21 cm (C) 3 15 cm (D) 10 cm

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20. In triangle ABC, AD ⊥ BC. D divides BC in the ratio 1 26. A
: 3 internally. If AB = 9 cm and AC = 21 cm, find BC.
(A) 12 5 cm (B) 15 5 cm
(C) 16 5 cm (D) 18 5 cm
45° D
21. In triangle ABC, AB = AC and D is a point on BC. If AB O
= 65 cm, AD = 63 cm and CD = 8 cm, find BD (in cm). B

22. C
A
In the parallelogram given, AB = 8 cm and BO = 6 cm
Find BC.
(A) 12 cm (B) 2 14 cm
(C) 2 7 cm (D) None of these
D C
B
27. The diagonal of a square ABCD is 22 2 cm. An
In triangle ABC, the lengths of AB, BC and AC are
equilateral triangle BCP is drawn on side BC. There
5 cm, 3 cm and 7cm respectively. D is a point on CB
is a point M on BC such that PM is parallel to AB. Find
extended such that AD is perpendicular to BC. What
the ratio of PM and AC.
is the length of AD?
(A) 3 :4 2 (B) 3 :2 2
(A) 6.25 cm (B) 5 3 cm
(C) 2 3 : 3 2 (D) None of these
(C) 4 3 cm (D) None of these
28. In a quadrilateral ABCD, AB is parallel to CD.
23. D and E are two points on the sides AB and AC of AB = 8 cm, BC = 10 cm, CD = 16 cm and AD = 10 cm.
triangle ABC respectively such that triangle ADE is Find the sum of the lengths of the diagonals.
similar to triangle ABC. DE divides triangle ABC into
(A) 4 57 cm (B) 5 53 cm
two parts of equal area. If AB = 12 cm, find BD.
(A) (6 − 2 ) cm (C) 6 47 cm (D) 7 41 cm

(B) (12 − 6 2 ) cm Directions for questions 29 and 30: These questions

are based on the information given below.
(C) (12 − 3 2 ) cm
(D) None of these A

Directions for questions 24 and 25: These questions

are based on the information given below. P
•O
A C E
R
D C
Q
B

B K D
In the given figure, AB and DC are equal chords of the
circle with centre O and OP = 3 cm, AP = 8 cm, OR = 5 cm.

J I H G F 29. What is the length of CR (in cm)?

In the figure above, AJ = 2IB

= 3JI = CH = 3IH 30. Find the length of QC (in cm).
= 3HG = EF = 3GF.

24. Find ∠CDK.

(A) 60° P
(B) 45° 31. In the given figure, PB,
(C) 30° PC and QR are tangents
(D) None of these to the circle at B, C and A
A respectively. If PB = 8 Q R
cm, find the perimeter
25. Find the ratio of the areas of quadrilaterals ABIJ and (in cm) of the triangle B C
EFHC. PQR.
(A) 5 : 12 (B) 7 : 16
(C) 3 : 8 (D) 11 : 24

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32. In the figure below, ∠QPT = 140º, PQ is tangent to 37. Find the ratio of the area of the square MKRT and the
the circle at Q and PR = QR. O is the centre of the area of the circle.
circle. Find ∠ROS (in degrees). (A) 1 : 4 (B) 7 : 22 (C) 7 : 44 (D) 14 : 33

Q 38. Two parallel chords PQ and RS are drawn on

opposite sides of the centre of a circle, of radius
17 cm. If RS is at a distance of 22 cm from PQ and
O PQ is 16 cm long, find RS.
140° (A) 4 15 cm (B) 10 10 cm

S (C) 8 15 cm (D) 9 10 cm
T P R

39. In the following figure, C is the point of contact of the

two circles and A, B are the points of contact of the
common tangent PQ with the two circles. Find ∠ACB.
33. In an isosceles triangle ABC, AC = AB, BD is the foot A B
of the perpendicular from B to AC and the lengths of P Q
BD and CD are 5 cm and 4 cm respectively. Find the
length of AD (in cm). C

(A) 60° (B) 90°

34. In the figure below, RS is a diameter, which when (C) 120° (D) Cannot be determined
extended meets the tangent drawn at P at the point
40. In a polygon, the sum of all its interior angles equals
Q. If ∠PQR = 30°, then ______.
the sum of all its exterior angles. Each interior angle
P is x°or its supplement where x ≠ 90. How many of the
angles must be x°?

R S Q
41. D
(A) PQ < RQ (B) PQ = RQ
(C) PQ > RQ (D) PQ ≥ RQ C A
E
35. A
O

D B B

C In the above figure, O is the center of the circle, OD

In the given figure, ABCD is a quadrilateral. If is perpendicular to AC. AB = 25 cm, CE = 12 cm. Find
AB = 10 cm, AD = 3 cm and CD = 4 cm, then what is the length of DE (in cm).
the length of BC?
(A) 11 cm (B) 12 cm
(C) 10 cm (D) Cannot be determined
42. ABCD is a cyclic quadrilateral in which AC is a
Directions for questions 36 and 37: These questions diameter of the circumcircle of the quadrilateral and
are based on the information given below. 3
AB= AD = (AC). Find ∠BCD.
In the figure, K is the centre of the circle, T is a point on
2
the circumference of the circle and MKRT is a square. (A) 60° (B) 120°
(C) 90° (D) 150°
U
M T
43. What is the sum of the squares of the sides of triangle
ABC (in cm2), if the sum of the squares of its medians
is 36 cm2?
K S
R

44. In triangle PQR, ∠R is an obtuse angle. The

36. Find the ratio of the perimeter of the square MKRT perpendicular drawn from P meets QR produced at S.
and the circumference of the circle. If QR = RS, QR = 8 cm and PQ = 20 cm, then find PR.

(A) 7 2 : 11 (B) 7 : 11 2 (A) 3 13 cm (B) 4 13 cm

(C) 7 : 2 (D) 7 : 22 (C) 6 13 cm (D) 7 13 cm

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45. In triangle ABC, what is the distance between the 51.
midpoint of BC and the foot of the perpendicular from C
A to BC, if the lengths BC, CA and AB are 5 cm, 7 cm
and 6 cm respectively? (in cm)

46. In triangle ABC, S is the point equidistant from A, B

and C. ∠BSC = 120°, and ∠BSA = 140°. Find ∠ABC
(in degrees). A B
In the figure, ABC is an equilateral triangle of side a
and each circle is of radius r. Find r in terms of a.
Directions for questions 47 and 48: These questions a(2 − 3 ) a(2 − 3 )
are based on the information given below. (A) (B)
4 2
ABCD is a rectangle divided into two rectangles AEFD a( 2 + 3 ) a(2 + 3 )
(C) (D)
and EBCF by drawing a line segment EF. AE = 28 cm, EB 16 32
= 12 cm, and BC = 10 cm. G, H and I lie within AEFD. J,
K and L lie within EBCF. The distances from each point in Difficult / Very Difficult … …… . .
one rectangle to each of the three points in other were
measured. The shortest of these distances was found to 52. S
be 41 cm.
R
47. Which of the following is always true?
(A) Distance between G and H is less than that
between J and I. Q
(B) (I, J) can represent the closest pair of points
among the 6 given points. P
(C) The closest pair of points among the 6 given T U V
points must be a pair of points in EBCF.
(D) None of the above In the figure above, PQ = QU = US = SV = VR = RT
= TP. Find ∠P.
48. Which of the following can be the measure of GI?
(A) 0.5 cm (B) 1.5 cm (C) 2.5 cm (D) 3.5 cm 2°
(A) 30° (B) 25
7
49. The centres of three circles which touch externally
form a triangle of sides 15 cm, 16 cm and 17 cm. Find 5°
the radius of the largest circle (in cm). (C) 25 (D) Cannot be determined
7

53. A D
50. In the figure below, AB and CD are two diameters of
O
the circle with centre J perpendicular to each other.
Chords EF and GH are perpendicular to AB and CD 120°
respectively at I and L respectively. Chords EF and B C
GH intersect at K. AI : IB = DL : LC = 1 : 3 and AB = 8
cm. Find GK (in cm). In the given parallelogram, AO is 5 units, BO is 7 units
C and angle BOC = 120°. The area of quadrilateral
E ABCD is _____.
(A) 35 sq.units (B) 70 sq.units
(C) 35 3 sq.units (D) 70 3 sq.units.

54. If ABCD is a cyclic quadrilateral and AD = 3 units, CD

A B = 4 units, BC = 5 units and AB = 6 units, then the
I J
product of AC and BD is _____.
G (A) 29 sq.units (B) 35 sq.units
K H
L (C) 39 sq.units (D) 41 sq.units

55. A, B, C and D are points on a circle, in that order such

F
( )
D that ∠CAB = 20° The angle bisectors of ∠CBD and
(A) 2 3 − 1 ∠BCA intersect at E and F is a point on BC such that
(B) 2(2 3 −1 ) EF is perpendicular to EC. If ∠BEF = ∠ABD, find
∠ACD (in degrees).
(C) 3 −1
(
(D) 3 3 − 1 )
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 Data Sufficiency … … …… .. 60. In the figure below, if BE = 10 cm,
and ∠A= 45°, then what is the length of AB?
Directions for questions 56 to 65: Each question is
followed by two statements, Ι and ΙΙ. Indicate your B
responses based on the following directives:

Mark (A) if the question can be answered using

one of the statements alone, but cannot be
10
answered using the other statement alone.
Mark (B) if the question can be answered using
either statement alone. 45o
Mark (C) if the question can be answered using A D E C
Ι and ΙΙ together but not using Ι or ΙΙ alone
Mark (D) if the question cannot be answered even Ι. ∠BCA = 45°.
using Ι and ΙΙ together. ΙΙ. ∠BEA = 60°.

56. In the figure below, ABC is a triangle, AD and BE are 61. PQRS is a rectangle and points A, B, C lie on PQ,
two of the medians. What is the area of the
QR, RS, respectively. What is the ratio of the area of
quadrilateral GDCE?
the triangle ABC to the area of rectangle PQRS?
A Ι. Point A divides PQ in the ratio 1 : 2.
ΙΙ. AC is parallel to PS

E 62. In the figure below two tangents from P meet the

G circle at A and B. What is the length of PO?
A
B C
D
Ι. AG : GE = 2 : 1
ΙΙ. AB = 10 cm and BC = 20 cm O P

57. What is the length of the median AM of the triangle

ABC, if ∠BAC = 90°? B
A
Ι. O is the centre of the circle whose radius is
10 cm.
ΙΙ. The circumradius of the triangle AOP is
14 cm.

B D M C 63. In the figure below, A and B are the points of contact of

the direct common tangent from P to the circles with
Ι. BC = 21 cm. centres O1 and O2 respectively. If r1,r2 are the radii of
ΙΙ. AD = 14 cm. the circles, find the length of AB.

58. In the figure below, ∠AED = 33°. If D, E and F are the A

mid-points of the sides ∆ABC, what is the measure of B
r1
∠EFB?
A r2
O1 O2 P

D 33° E

Ι. r1 = 15 cm, r2 = 9 cm
ΙΙ. ∠BPO = 30°
B C
F
64. The points O1 and O2 are the centres of two circles
Ι. ∠FEC = 78°
having radii 10 cm and 20 cm respectively. What is
ΙΙ. ∠BAC = 78°
the distance between the points O1 and O2?
Ι. The two circles have only one common tangent
59. Find the area of the triangle ABC.
ΙΙ. The two circles have three common tangents.
Ι. ∠ABC = 60°
ΙΙ. AB = 10 cm and AC = 10 cm.

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65. F

E A
D

C B

What is the measure of ∠BEA in the figure above if ∠ABD = ∠BCF = 90°?
Ι. ∠BDA = 30° ΙΙ. ∠CFB = 30°

Key

Concept Review Questions

1. 5 6. C 11. C 16. 16.8 21. 10 26. 150 31. D
2. 28 7. 4 12. A 17. D 22. 1 27. 55 32. 3
3. C 8. D 13. 12 18. 35 23. 85 28. A 33. 2
4. C 9. 115 14. D 19. 24 24. A 29. B 34. C
5. A 10. C 15. 24 20. C 25. B 30. A 35. 0

Exercise – 4(a)

1. (i) C 8. B 15. B 23. 528 30. C

(ii) 28 9. C 16. C 24. 110 31. 3200
2. 70 10. B 17. A 25. C 32. C
3. C 11. C 18. 192 26. (i) 50 33. 22
4. 18.5 12. (i) D 19. 12 (ii) 70 34. B
5. D (ii) D 20. B 27. B 35. 4
6. 10 13. 4 21. D 28. 144
7. C 14. D 22. 72 29. B

Exercise – 4(b)
1. 25 14. 144 27. B 40. 2 53. C
2. 125 15. A 28. A 41. 9 54. C
3. C 16. 18 29. 12 42. B 55. 20
4. 20 17. C 30. 8 43. 48 56. C
5. C 18. 64 31. 16 44. B 57. A
6. 130 19. B 32. 120 45. 1.3 58. B
7. C 20. A 33. 1.125 46. 50 59. C
8. D 21. 32 34. A 47. A 60. A
9. D 22. D 35. A 48. A 61. C
10. A 23. B 36. B 49. 9 62. A
11. 70 24. D 37. C 50. A 63. A
12. 170 25. C 38. C 51. B 64. B
13. D 26. B 39. B 52. C 65. A

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Tel : 040–40088400 Fax : 040–27847334 email : info@time4education.com website : www.time4education.com SM1002108/67