Geometry
Category : 9th Class
Geometry
In this chapter, we will learn about introduction to euclid’s geometry, lines and angles, triangles, quadrilaterals, areas of parallelograms and triangles and circles.
Axioms
Axioms or postulates are the assumptions which are obvious universal truths. They are not proved.
Theorems
Theorems are statements which are proved using definitions, axioms, previously proved statements and deductive reasoning.
Euclid’s Axioms
Euclid’s Postulates
Point
A point is a fine dot. For example, P is a point as shown in the figure.
\[\bullet P\]
Line Segment
A line segment is a straight path between two given points. For example in the shown figure PQ is a straight path between the pints P and Q and so is called a line segment \[\overline{PQ}\]
A line segment has a definite length.
Ray
A ray is a line segment extending indefinitely in one direction. A ray has no definite length. For example, in the shown figure \[\overrightarrow{PQ}\] is representing a ray having one and point P.
Line
A line is a ne is obtained on extending a line segment indefinitely in both the directions. In the shown figure, 0\[\overleftrightarrow{PQ}\] is represented as a line. A line has no end points so a line has no definite lengths.
Angle
An angle is generated when two rays originated from the same end point. In the shown figure POQ is the angle formed by two rays \[\]and \[\]. Here O is called the vertex of the \[\angle \]POQ and PO and OQ are called the arms of the angle POQ.
Note:
(i) If a ray stands on a line, then so formed adjacent angles are supplementary and its converse.
(ii) The vertically opposite angles formed by two intersecting lines are equal.
Parallel Lines and a Transversal
If a transversal intersects two parallel lines, then
(i) Each pair of alternate interior angles is equal and conversely.
(ii) Each pair of corresponding angles is equal and conversely.
(iii) Each pair of interior angles on the same side of the transversal is supplementary and conversely.
Note: Lines parallel to the same line are parallel to each other.
Triangle
A closed figure formed by three line segments is called a triangle. A triangle has three sides, three angles and three vertices. Two figures of same shape and same size are called congruent figures. Two circles of the same radii are congruent. Two triangles ABC and PQR are congruent \[(ie.\,\,\Delta \,\,ABC\cong \Delta \,\,PQR)\]under the correspondence \[A\leftrightarrow P,\,\,B\leftrightarrow Q\,\,and\,\,C\leftrightarrow R\]
Note:
(i) Sum of all the three interior angles of a triangle is\[180{}^\circ \].
(ii) The exterior angle of a triangle is equal to the sum of the corresponding two interior opposite angles.
Congruent Figures
Two geometrical figures are said to be congruent if they have same shape and size. e.g. two angles are said to be congruent if they have same measures similarly two line segments are said to be congruent if they have same lengths.
Congruency of Triangles
Two triangles ABC and DEF are said to be congruent if and only if, AB = DE, BC = EF, CA =FD, \[\angle \]A = \[\angle \]D, \[\angle \]B = \[\angle \]E and \[\angle \]C = \[\angle \]F.
Criteria for Congruency of triangles
There are generally four criteria for the congruency of triangles which are given below.
S-S-S Criteria
Two triangles are said to be congruent if the three sides of one triangle are equal to the corresponding three sides of the other.
S-A-S Criteria
Two triangles are said to be congruent if the two sides and included angle of one triangle is equal to the other.
A-S-A Criteria
If the two angles and the side included by the angles are equal to the corresponding two angles and included side of the other triangle, then the two triangles are congruent.
R-H-S Criteria
This criteria is for a right-angled triangle. If one side and hypotenuse of a right-angled triangle is equal to the corresponding side and hypotenuse of other right-angled triangle then two right angled triangles are said to be congruent.
Some Important Results
Similarity of Triangles
When two geometrical shapes resembles same but need not to be equal in size are called similar figures. Let us observe the following examples:
(i) Any two line segments are always similar.
(ii) Two circles of different radius are always similar.
(iii) For rectilinear figures if all the corresponding angles of a polygon are equal and the ratio of their corresponding sides are also equal, then they are said to be similar.
The criteria of similarity of two triangles are
(i) A-A-A criterion (ii) S-S-S criterion
(iii) S-A-S criterion
The ratio of the areas of two similar triangles is equal to the ratio of the,
Quadrilaterals
A Plane figure bounded by four line segments is called a quadrilateral.
Properties of a Quadrilateral
(i) Opposite sides are equal.
Or (ii) Opposite angles are equal.
Or (iii) Diagonals bisect each other.
Or (iv) A pair of opposite sides is equal and parallel.
Results on Areas of Parallelograms and Triangles
Circle
Circle is the locus of a point which moves in a plane in such a way that its distance from a fixed point is always constant. We know that the fixed point is called centre and the fixed distance is called its radius. Also,
\[D\text{ }=\text{ }2r\],\[C=2\pi r\], where D is diameter, C is circumference of circle and r is radius.
Terms Related to a Circle
Following are some terms which are very useful in solving the problems related to circles.
Secant
When a line intersects a circle at two distinct points, it is called a secant of the circle. In the following figure line m is a secant of the circle.
Tangent
A line which touches the circle at exactly one point is called a tangent to the circle. In the following figure PQR is a tangent of the circle.
Concentric Circles
Circles are said to be concentric if and only if they have the same centre and different radii.
Concurrent Arc
A continuous piece of circumference of a circle is called an arc and two arcs are said to be concurrent if they subtend equal angles at the centre.
Here, \[Arc\,\,\overset\frown{RNS}\equiv Arc\overset\frown{\,\,PMQ}\]
Properties of Circles
Note:
(i) The sum of either pair of opposite angles of a cyclic quadrilateral is\[180{}^\circ \].
(ii) If the sum of a pair of opposite angles of a quadrilateral is 180°, the quadrilateral is cyclic.
(iii) Radius of incircle and circumcircle of an equilateral triangle of side a \[a=\frac{a}{2\sqrt{3}}\] and \[\frac{a}{\sqrt{3}}\] respectively.
Example:
In the figure given below the height of two poles are 10 metres and 15 metres. If the poles are 30 metres apart then the height of point of intersection of the lines joining the top of each pole from opposite foot of the other pole is:
(a) 18m (b) 8m
(c) 6m (d) 9 m
(e) None of these
Answer (c)
In the given figure, \[\Delta CAB\sim \Delta CEF\Rightarrow \frac{EF}{10}\]\[=\frac{FC}{30}\Rightarrow FC=3EF\]
Similarly in \[\Delta ACD\sim \Delta AFE,\] so we have \[\frac{EF}{15}=\frac{AF}{30}\Rightarrow AF=2EF\]
From above equations, we get \[5EF=30\Rightarrow EF=6m\]
Example:
In an equilateral triangle show that the centroid and circumcenter coincide.
Solution:
Given: An equilateral triangle PQR in which M, L and N are the mid points of sides QR, RP and PQ respectively.
To prove: The centroid and circumcentre are coincident in \[\Delta PQR\]
Construction: Draw medians PM, QL and RN
Proof: Let G be the centroid of \[\Delta PQR\] is the point of intersection of medians QL, RN and MP.
In \[\Delta QNR\] and \[\Delta QLR\]we have,
\[\angle Q=\angle R=60{}^\circ \] and QN=RL and QR=RQ
\[\Rightarrow \] (By SAS)
\[\Rightarrow \]RN=QL (CPCT) … (i)
Similarly, in \[\Delta RPN\cong \Delta PRM\]
\[\Rightarrow \]RN = PM (CPCT) ... (ii)
From (i) and (ii), we get
RN = PM = QL
\[\Rightarrow \]\[\frac{2}{3}\] RN \[\frac{2}{3}\] PM = \[\frac{2}{3}\] QL
\[\Rightarrow \] GR= GP = QG \[\Rightarrow \] G is equidistant from the vertices.
\[\Rightarrow \] G is circumcentre of\[\Delta PQR\].
Hence the centroid and circumcentre of \[\Delta PQR\] are coincident.
Example:
In the figure given below.
\[\frac{AP}{PC}=\frac{3}{4}\] and \[\frac{BR}{RP}=\frac{3}{2}\] and BQ = 15 cm. Find AQ
Solution:
Given: \[\frac{AP}{PC}=\frac{3}{4}\], \[\frac{BR}{RP}=\frac{3}{2}\] and BQ = 15cm
Draw PS || CQ which meets AB at S.
Applying basic proportionality theorem, we get
\[\frac{BQ}{QS}=\frac{BR}{RP}\Rightarrow \frac{15}{QS}=\frac{3}{2}\Rightarrow \]QS=10 cm.
Similarly in\[\Delta AQC\], we get
\[\frac{AS}{SQ}=\frac{BR}{PC}\Rightarrow\frac{AS}{10}=\frac{3}{2}\Rightarrow AS=\frac{30}{4}\]= 7.5 cm.
\[\therefore \] AQ = AS + QS = 10 + 7.5 = 17.5 cm.
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