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PRACTICAL GEOMETRY

PRACTICAL GEOMETRY. Introduction: Practical Geometry will enable the Student to identify and classify geometric figures and apply geometric properties to practical situations, special reasoning and visualization skill will be enhanced by using Geometric Models and a variety of technologies.

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PRACTICAL GEOMETRY

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  1. PRACTICAL GEOMETRY Introduction:Practical Geometry will enable the Student to identify and classify geometric figures and apply geometric properties to practical situations, special reasoning and visualization skill will be enhanced by using Geometric Models and a variety of technologies.

  2. The Bisectors of Angles of a given triangle. Given: ABC is a triangle Required : To draw the bisectors of angles A, B and C Construction: 1. Draw the ABC 2. With point B as a centre draw an arc of any radius, intersecting the sides BC and BA at point L and M ANGLES BISECTORS OF  ABC 3. Take point L as a centre and draw an arc of any radius. 4. Now take M as a centre and with the same radius draw the another arc, which cuts the previous arc at point P 5. Join point P to B and produce it. Thus BP is the bisector of <B 6. Similarly, repeat steps (2) to (5) to draw CQ and AR the bisectors of <C and <A respectively. Hence BP, CQ and AR are the required bisectors of angles of a  arc concurrent. Note: The angles bisectors intersect at a point I, which is called in centre and the bisectors of angles of a  are concurrent. • How to draw the right bisectors of sides of a triangle ABC. Given: A  ABC Required: To draw right bisectors of the sides AB, BC and CA

  3. CONSTRUCTION: 1. Draw the  ABC. 2. To draw right bisector of the side AB, with B as a centre and radius more than half of AB, draw arcs on either sides of AB. 3. Now with A as a centre and with the same radius, draw arcs on either sides of AB cutting previous arcs at P and Q. 4. Join P and Q Thus PQ is the right bisector of AB, 5. Similarly, repeat steps (2) to (4) and draw ST and LM, the right bisectors of BC and AC respectively. Hence PQ, ST and LM arc the required right bisector of the sides AB, BC and AC respectively of  ABC. Note: Right bisectors of ABC pass through O. So they arc concurrent. RIGHT BISECTORS OF  ABC

  4. COVERAGE OF THE LESSON ACCORDING TO ITS SYNOPSIS: 1. Construction of Circum Circle Construct the PQR in which mPQ = mQR = 8cm, m<Q =60°. Draw the “Circum Circle” of the  PQR. Circle: A set of points of a plane which are equidistant from a given (fixed) point is called a CIRCLE: The fixed point is called the “Centre” of the Circle. Circum Circle: A circle which passes through the three vertices of a triangle is called “Circum Circle” (or circumscribed circle) and the centre is called “Circum Centre” of the circle. Given: A  PQR, in which mPQ =mQR = 8cm and m<Q=60° Required: Draw the Circum circle of  PQR.

  5. CONSTRUCTION: • Draw a line segment QR = 8cm. • At point Q draw an <RQX = 60° • From QX, cut off PQ = 8 cm. • Join P to R. Thus  PQR is constructed. • Draw the right bisector LL’ of side QR. • Draw the right bisector MM’ of side PR. Let them intersect each other at point “O”. • With Centre “O” and radius equal to OR, draw a circle, which passes through the vertices of  PQR. Hence it is our required circum circle is constructed (constructed) CIRCUM CIRCLE OF  PQR

  6. 2. CONSTRUCTION OF ESCRIBED CIRCLE : 2. Construction of Escribed Circle Question: Construct  ABC in which mAB = 7.5cm mBC = 7.5cm, m<B=30° Draw the escribed circle opposite to Vertex B. Given: • ABC, in which mAB = mBC = 7.5cm, and m<B=30° Required: Draw the escribed circle opposite to Vertex B

  7. CONSTRUCTION: • Draw a line segment BC = 7.5cm. • At point B, draw an < CBX=30° • From BX, cut off AB = 7.5 cm. • Join A to C. Thus  ABC is constructed. • Produce (extend) BC beyond C to BY • Produce BA to BX. Making two exterior angles <ACY and <CAX. • Draw the bisector AA’ and CC’ of <CAX and <ACY respectively let them intersect each other at point “O”. • From point “O” draw OD perpendicular on BY. • With centre O and radius equal to OD draw a circle. Which touching BA, BY and AC. This is called the escribed circle opposite to vertex B ESCRIBED CIRCLE OPPOSITE TO VERTEX “B” OF  ABC ESCRIBED CIRCLE: A circle which touches one side of a triangle externally and the other two sides produced internally.

  8. CONSTRUCTION OF INCIRCLE Question: Construct the  ABC, in which m<B = 60°, mBC = 7.5cm, mAB = 7cm. Draw the in circle (inscribed circle) of ABC. Solution Given: A  ABC, in which m<B = 60°, mBC = 7.5cm and mAB = 7cm. Required: Draw the in circle of ABC. In-Circle: The circle which touches the three sides of a triangle is called in-circle or inscribed circle of the triangle and its centre is called “In-Centre”.

  9. CONSTRUCTION: • Draw a line segment BC = 7.5cm • At point “B” draw an <CBX = 60°. • From BX, cut of AB = 7cm. • Join A to C. Thus  ABC is constructed. • Draw the bisector BB’ of >B. • Draw the bisector CC’ of >C (et them intersect each other at point O. • From point “O” draw OD perpendicular on BC. • With Centre “O” and radius equal to OD draw a circle. The circle touches the three sides of ABC Hence it is our required in-circle is constructed.

  10. 4. CONSTRUCTION OF MEDIANS OF A TRIANGLE Question: Construct a  ABC, such that mAB = 6.8cm, mBC = 8cm and mAC = 10.0cm. Draw the medians of ABC. Median of a Triangle: Line segments joining, the vertices of a triangle to the mid points of the opposite sides are called medians of the triangle. Given: A triangle ABC, in which mAB = 6.8cm, mBC = 8cm and mAC = 10.0cm. Required: Draw the medians of ABC.

  11. CONSTRUCTION: • Draw a line segment BC = 8cm • From point B draw an arc of radius 6.8cm. • From point C draw an arc of radius 10cm which intersects the previous arc at point A. • Join A to B and A to C. Thus  ABC is constructed. • Draw the mid point D of Side BC • Draw the mid point E of Side AC • Draw the mid point F of Side AB • Join A to D, B to E and C to F. Thus AD, BE and CF are our required medians of ABC, which meet in a point O. It may be noted that medians of every triangle are concurrent (i.e. meet in one point) and their point of concurrency called “centraid”, divides each of them in the ratio of 2:1. By actual measurement it can be prove that mAO = mBO = mCO = 2 mOD mOE mOF 1

  12. MEDIANS OF  ABC

  13. CONSTRUCTION OFALTITUDES OF A TRIANGLE Question: Take any triangle ABC and draw its altitudes. Given: A  ABC Required: To draw altitudes of the  ABC Altitude: A perpendicular from the vertex of a triangle to the opposite side is called an altitude of the triangle.

  14. CONSTRUCTION: 1. Draw a  ABC 2. Take a point A as centre and draw an arc of suitable radius, which cuts BC at points D and E. 3. From D as centre, draw an arc of radius more than ½ mDE. 4. Again from point E draw another arc of same radius, cutting first arc at point F. 5. Join the points A and F, such that AF intersects BC at point P. Then AP is the altitude of the  ABC from the vertex A. 6. Similarly, repeat the steps (2) to (5) and draw BQ and CR, the altitudes of  ABC from the vertices B and C respectively. Hence AP, BQ and CR are the required altitudes of  ABC, which meet in a point O. The points P, Q and R are called the feet of these altitudes. Note: It may be remember that the altitudes of a triangle are always concurrent. The point of concurrency is called the “ORTHO CENTRE”

  15. ALTITUDES OF  ABC

  16. SUMMARY OF THE LESSON: • A circle which passes through the vertices of a triangle is called circum-circle. • A circle which touches the three sides of a triangle is called In-Circle. • The line segments joining the vertices of a triangle with the mid-points of the opposite sides are called “Medians” of the triangle. • The point of intersection of three medians is called “Centroid” and medians are concurrent. • The line segments which are perpendiculars from the vertices to the opposite sides of a triangle are called its “Altitudes”. • The point of intersection of three altitudes is called “Ortho Centre” and they are concurrent. • Construction of unique triangle is possible when the sum of measures of two sides is greater than the third side.

  17. ACTIVITIES (M.C.Qs) 1) The sum of measures of all the angles of a triangle is __________ degree. (a) 180 (b) 360 (c) 90 (d) 270 2) Two lines are said to be perpendicular on each other if they form a/an _____ angle. (a) acute (b) Obtuse (c) Reflex (d) Right 3) The right bisectors of a triangle are ________ (a) Concurrent (b) Not concurrent (c) Concurrent and passes through the same point (d) None 4) The bisectors of angles of a triangle are and the point of concurrency is called __________ (a) Incentre (b) Ortho Centre (c) Circum Centre (d) Centroid. 5) A circle which passes through the vertices of a triangle is called __________ . (a) In-circle (b) Circum-Cricle (c) Escribed Circle

  18. Page -2- 6) A circle which touches the three sides of a triangle is called __________ (a) Circum circle (b) Escribed circle (c) In-Circle 7) The point of intersection of three medians is called _____ . (a) Ortho Centre (b) Centroid (c) None (d) In-Centre 8) The point of intersection of three altitudes is called ________ (a) Centroid (b) Ortho Centre (c) In-Centre (d) Curcum Centre 9) The line segments joining the vertices of a triangle with the mid-points of the opposite sides are called __________ (a) Altitudes (b) Medians (c) Perpendiculars (d) None 10) The line segments which are perpendiculars from the vertices to the opposite sides of a triangle are called ________ (a) Medians (b) Altitudes (c) Bisectors (d) None

  19. MCQs Answer Key: 1- (a) 180 2. (d) Right 3. (c) Concurrent and passes through the same point. 4- (a) In-Centre 5. (b) Circum Circle 6. (c) In-Circle 7. (b) Centroid 8. (b) Ortho centre 9. (b) Medians 10. (b) Altitudes

  20. THANK YOU

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