2-9th Class Teaching Theorems1

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Published on May 31, 2013

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PowerPoint Presentation: S.JASMINE SUGIRTHA 1 Teaching Theorems IX Maths Teaching OF Theorems MATHS PowerPoint Presentation: Angles Theorem: I Vertically opposite angles are equal in measure. Theorem: II The measure of the three angles of a triangle sum to 180  Theorem: III An exterior angle of a triangle equals the sum of the two . interior opposite angles in measure. Theorem:IV If two sides of a triangle are equal in measure, then the . . . . angles opposite these sides are equal in measure. Theorem:V Opposite sides and opposite angles of a parallelogram are . . . respectively equal in measure. Theorem:VI A diagonal bisects the area of a parallelogram. Theorem: VII The measure of the angle at the centre of the circle is . . . twice the measure of the angle at the circumference, . standing on the same arc. Theorem:VIII A line through the centre of a circle perpendicular to a . . . . chord bisects the chord. S.JASMINE SUGIRTHA 2 MATHS PowerPoint Presentation: Angles An angle is formed when two lines meet. The size of the angle measures the amount of space between the lines. In the diagram the lines ba and bc are called the ‘arms’ of the angle, and the point ‘b’ at which they meet is called the ‘vertex’ of the angle. An angle is denoted by the symbol  .An angle can be named in one of the three ways: a c b . . Amount of space Angle S.JASMINE SUGIRTHA 3 MATHS PowerPoint Presentation: 1 . Three letters a b c . . Using three letters, with the centre at the vertex. The angle is now referred to as : abc or cba. S.JASMINE SUGIRTHA 4 MATHS PowerPoint Presentation: 2. A number c b . . 1 a Putting a number at the vertex of the angle. The angle is now referred to as 1. S.JASMINE SUGIRTHA 5 MATHS PowerPoint Presentation: 3. A capital letter b . . B a c Putting a capital letter at the vertex of the angle. The angle is now referred to as B. S.JASMINE SUGIRTHA 6 MATHS PowerPoint Presentation: Right angle A quarter of a revolution is called a right angle. Therefore a right angle is 90  . Straight angle A half a revolution or two right angles makes a straight angle. A straight angle is 180  . Measuring angles We use the symbol to denote a right angle. S.JASMINE SUGIRTHA 7 MATHS PowerPoint Presentation: Acute, Obtuse and reflex Angles Any angle that is less than 90  is called an acute angle . An angle that is greater than 90  but less than 180  is called an obtuse angle . An angle greater than 180  is called a reflex angle . S.JASMINE SUGIRTHA 8 MATHS PowerPoint Presentation: Angles on a straight line Angles on a straight line add up to 180  . A + B = 180 . Angles at a point Angles at a point add up to 360  . A+ B + C + D + E = 360  A B A B D E C S.JASMINE SUGIRTHA 9 MATHS PowerPoint Presentation: Pairs of lines: Consider the lines L and K : . p L K L intersects K at p written : L K = {p}  Intersecting S.JASMINE SUGIRTHA 10 MATHS PowerPoint Presentation: Parallel lines L K L is parallel to K Written: L  K Parallel lines never meet and are usually indicated by arrows. Parallel lines always remain the same distance apart. S.JASMINE SUGIRTHA 11 MATHS PowerPoint Presentation: Perpendicular L is perpendicular to K Written: L  K The symbol is placed where two lines meet to show that they are perpendicular L K S.JASMINE SUGIRTHA 12 MATHS PowerPoint Presentation: Parallel lines and Angles 1.Vertically opposite angles When two straight lines cross, four angles are formed. The two angles that are opposite each other are called vertically opposite angles . Thus a and b are vertically opposite angles. So also are the angles c and d. From the above diagram: A B C D A+ B = 180  …….. Straight angle B + C = 180  ……... Straight angle A + C = B + C ……… Now subtract c from both sides A = B S.JASMINE SUGIRTHA 13 MATHS PowerPoint Presentation: 2. Corresponding Angles The diagram below shows a line L and four other parallel lines intersecting it. The line L intersects each of these lines. L All the highlighted angles are in corresponding positions. These angles are known as corresponding angles . If you measure these angles you will find that they are all equal. S.JASMINE SUGIRTHA 14 MATHS PowerPoint Presentation: In the given diagram the line L intersects two parallel lines A and B. The highlighted angles are equal because they are corresponding angles. The angles marked with are also corresponding angles . A B L . . Remember: When a third line intersects two parallel lines the corresponding angles are equal. S.JASMINE SUGIRTHA 15 MATHS PowerPoint Presentation: 3. Alternate angles The diagram shows a line L intersecting two parallel lines A and B. The highlighted angles are between the parallel lines and on alternate sides of the line L. These shaded angles are called alternate angles and are equal in size. Remember the Z shape. A B L S.JASMINE SUGIRTHA 16 MATHS PowerPoint Presentation: Theorem: I Vertically opposite angles are equal in measure. Given: To prove : Construction: Proof: Straight angle Straight angle   1=  2 Label angle 3  1=  2 Intersecting lines L and K, with vertically opposite angles 1 and 2.  1+  3=180   2+  3=180  Q.E.D. L K 1 2   1+  3=  3+  2 .....Subtract  3 from both sides 3 S.JASMINE SUGIRTHA 17 MATHS PowerPoint Presentation: Theorem: II The measure of the three angles of a triangle sum to 180 . Given: To Prove:  1+  2+  3=180  Construction: Proof:  1=  4 and  2=  5 Alternate angles  1+  2+  3=  4+ 5+3 But  4+  5 +3=180 Straight angle   1+  2+  3=180  The triangle abc with 1,2 and 3. 4 5 a b c 1 2 3 Q.E.D. Draw a line through a, Parallel to bc. Label angles 4 and 5 . S.JASMINE SUGIRTHA 18 MATHS PowerPoint Presentation: Theorem: III An exterior angle of a triangle equals the sum of the two interior opposite angles in measure. Given: A triangle with interior opposite angles 1 and 2 and the exterior angle 3. To prove: 1+ 2= 3 Construction: Label angle 4 Proof: 1+ 2+ 4=180  3+ 4=180  Three angles in a triangle  1+ 2+ 4= 3+ 4 Straight angle  1+ 2= 3 a b c 3 1 2 4 Q.E.D. S.JASMINE SUGIRTHA 19 MATHS PowerPoint Presentation: a b c 1 2 Theorem:IV If to sides of a triangle are equal in measure, then the angles opposite these sides are equal in measure. Given: The triangle abc, with ab = ac and base angles 1 and 2. To prove:  1  =  2  Construction: Draw ad, the bisector of bac. Label angles 3 and 4. Proof:  ab  =  ac  given 3 = 4 construction  ad  =  ad  common SAS  1 = 2 Corresponding angles d 3 4 abd acd  Consider abd and acd: Q.E.D. S.JASMINE SUGIRTHA 20 MATHS PowerPoint Presentation: Theorem:V Opposite sides and opposite angles of a parallelgram are respectively equal in measure. Given: Parallelogram abcd a b c d To prove: Construction: Join a to c. Label angles 1,2,3 and 4. Proof: 1= 2 and 3= 4 Alternate angles  ac  =  ac  common ASA   ab  =  dc  and  ad  =  bc  Corresponding sides And abc = adc Corresponding angles Similarly, bad = bcd 1 2 3 4  ab  =  dc  ,  ad  =  bc  abc = adc, bad = bcd Consider abc and adc : abc adc  Q.E.D. S.JASMINE SUGIRTHA 21 MATHS PowerPoint Presentation: Theorem:VI A diagonal bisects the area of a parallelogram. a b c d Given: Parallelogram abcd with diagonal [ac]. To prove: Area of abc = area of adc. Proof: ab = dc Opposite sides ad = bc Opposite sides ac = ac Common SSS Consider abc and adc: abc adc  area abc = area adc  Q.E.D. S.JASMINE SUGIRTHA 22 MATHS PowerPoint Presentation: Theorem: VII The measure of the angle at the centre of the circle is twice the measure of the angle at the circumference, standing on the same arc. Given: Circle, centre o, containing points a, b and c. To prove: boc = 2 bac Construction: Join a to o and continue to d. Label angles 1,2,3,4 and 5. Proof: d a b c . o 1= 2 + 3 Exterior angle But 2 = 3  1 = 2 2 Similarly, 5 = 2 4  1+ 5 = 2 2 + 2 4  1 + 5 = 2(2 + 4) i.e. boc = 2 bac 1 2 3 4 5 Consider aob: Q.E.D. Base angles in an isosceles 23 MATHS PowerPoint Presentation: Deduction 1: All angles at the circumference on the same arc are equal in measure. To prove: bac = bdc Proof: 3 = 2 1 Angle at the centre is twice the angle on the circumference (both on the arc bc) 3 = 2 2 Angle at the centre is twice the angle on the circumference (both on arc bc)  2 1 = 2 2  1 = 2 i.e. bac = bdc Q.E.D. a b c d 3 1 2 . o S.JASMINE SUGIRTHA 24 MATHS PowerPoint Presentation: Deduction 2: An angle subtended by a diameter at the circumference is a right angle. To prove: bac = 90  Proof: 2 = 2 1 Angle at the centre is twice the angle on the circumference (both on the arc bc) straight line. But 2 = 180   2 1 = 180   1 = 90  i.e. bac = 90  Q.E.D. a b c o 2 1 . S.JASMINE SUGIRTHA 25 MATHS PowerPoint Presentation: Deduction 3: The sum of the opposite angles of a cyclic quadrilateral is 180  . To prove: bad + bcd = 180  3 = 2 1 Proof: Angle at the centre is twice the angle on the circumference. (both on minor arc bd) 4 = 2 2 Angle at the centre is twice the angle on the circumference. (Both on the major arc bd)  3 + 4 = 2 1 + 2 2 But 3 + 4 = 360  Angles at a point  2 1 + 2 2 = 360   1 + 2 = 180  i.e. bad + bcd = 180  Q.E.D. . o 4 a b c d 3 2 1 S.JASMINE SUGIRTHA 26 MATHS PowerPoint Presentation: a b c L d ∟ ∟ . Theorem:VIII A line through the centre of a circle perpendicular to a chord bisects the chord. Given: Circle, centre c, a line L containing c, chord [ab], such that L  ab and L  ab = d. To prove: ad = bd Construction: Label right angles 1 and 2. Proof: 1 = 2 = 90  Given ca = cb Both radii cd = cd common R H S Corresponding sides Consider cda and cdb: cda cdb  ad = bd  1 2 Q.E.D. S.JASMINE SUGIRTHA 27 MATHS PowerPoint Presentation: b a c d e f 2 1 2 3 1 3 Theorem: If two triangles are equiangular, the lengths of the corresponding sides are in proportion. Given : Two triangles with equal angles. To prove: |df| |ac| = |de| |ab| |ef| |bc| = Construction: On ab mark off ax equal in length to de. On ac mark off ay equal to df and label the angles 4 and 5. Proof: 1 = 4 [xy] is parallel to [bc] |ay| |ac| = |ax| |ab| As xy is parallel to bc. |df| |ac| = |de| |ab| Similarly |ef| |bc| = x y 4 5 Q.E.D. S.JASMINE SUGIRTHA 28 MATHS PowerPoint Presentation: Theorem: In a right-angled triangle, the square of the length of the side opposite to the right angle is equal to the sum of the squares of the other two sides. Q.E.D. c b a c b a b a c b a c 1 2 3 4 5 To prove that angle 1 is 90 º Proof: 3+ 4+ 5 = 180 º ……Angles in a triangle But 5 = 90 º => 3+ 4 = 90 º => 3+ 2 = 90 º ……Since 2 = 4 Now 1+ 2+ 3 = 180 º ……Straight line => 1 = 180 º - ( 3+ 2 ) => 1 = 180 º - ( 90 º ) ……Since 3+ 2 already proved to be 90º => 1 = 90 º S.JASMINE SUGIRTHA 29 MATHS THANK U: End MATHS S.JASMINE SUGIRTHA 30 THANK U THANK U

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