Theorems on Area. Introduction Axioms of Area. Congruence area axiom. Addition area axiom
|
|
- Alexia Hicks
- 6 years ago
- Views:
Transcription
1 3 Theorems on rea Introduction We know that Geometry originated from the need of measuring land or recasting/refixing its boundaries in the process of distribution of certain land or field among different people. You may recall that the part of the plane enclosed by a simple closed curve is called a planar region corresponding to that figure and the magnitude (or measure) of this plane region is called the area of that figure. This magnitude of measure is always expressed with the help of a number (in some unit) such as 5 cm, 3 m, 3 5 hectares etc. So, we can say that the area of a simple closed plane figure is a number (in some unit) associated with the part of the plane enclosed by the figure. In earlier classes, we have learnt some formulae for finding the areas of some different simple closed plane figures such as triangle, rectangle, square, parallelogram etc. In this chapter, we shall consolidate the knowledge about these formulae by studying some relationship between the areas of these geometric figures under the condition when they lie on the same base (or equal bases) and between the same parallel lines. 3. xioms of rea ongruence area axiom If two figures are congruent, then the areas enclosed by these figures are equal. Thus, if two figures and are congruent, then area enclosed by = area enclosed by. This is known as congruence area axiom. In the adjoining figure, Δ ΔR. So, area of Δ = area of ΔR. Note that by area of Δ we mean the area of the region enclosed by the triangle. ddition area axiom If a planar region R consists of two non-overlapping planar regions R and R, then area of region R = area of region R + area of region R. This is known as addition area axiom. In the adjoining figure, planar regions R and R are nonoverlapping. If R is the total region i.e. the region made up of regions R and R, then area of region R = area of region R + area of region R. R R R R
2 3. qual Figures Two figures are called equal if and only if they have equal area. as two congruent figures have equal area, therefore, two congruence figures are always equal figures. However, the converse may not be true i.e. two equal figures may not be congruent. For example, consider the two right triangles and R given below: R cm 3 cm 6 cm 4 cm rea of = 6 cm = 6 cm. rea of R = 4 3 cm = 6 cm. area of = area of R and R are equal figures. learly, these triangles are not congruent. 3.3 Theorems on rea Theorem 3. diagonal of a parallelogram divides it into two triangles of equal areas. Given. is a parallelogram and is its one diagonal. To prove. rea of = area of. roof. In and. =. pp. sides of gm.. =. pp. sides of gm. 3. = 3. ommon SSS rule of congruency. 5. rea of = area of ongruence area axiom. Theorem 3. arallelograms on the same base and between the same parallel lines are equal in area. Given. Two parallelograms and F on the same base and between the same parallel lines and. To prove. rea of gm = area of gm F. F Theorems on rea 55
3 roof. In ΔF and Δ. F =. orres. s, and is a transversal.. F =. orres. s, F and is a transversal. 3. = 3. pp. sides of gm. 4. ΔF Δ 4. S rule of congruency. 5. rea of ΔF = area of Δ 5. ongruent figures have equal area. 6. area of ΔF + area of quad. F = area of ΔF + area of quad. F 7. area of gm = area of gm F 6. dding same area on both sides. 7. ddition area axiom. orollary. parallelogram and a rectangle on the same base and between the same parallel lines are equal in area. roof. Let a parallelogram and a rectangle F be on the same base and between the same parallel lines and F (as shown in the adjoining figure). We want to prove that area of gm = area of rect. F. Since a rectangle is also a parallelogram, therefore, area of gm = area of rect. F (Theorem 3.). F Hence, a parallelogram and a rectangle on the same base and between the same parallel lines are equal in area. orollary. rea of a parallelogram = base height. roof. y corollary, we have area of parallelogram = area of rectangle F (i) (see figure of corollary ) lso area of rectangle F = (ii) From (i) and (ii), we get area of parallelogram = = base height. Hence, the area of a parallelogram = base height. orollary 3. arallelograms with equal bases and between the same parallels are equal in area. roof. Let and FGH be two parallelograms with equal bases i.e. = F and between the same parallel lines and H. From, draw M and from, draw N HG, then M = N ( MN is a rectangle (why?)), so opp. sides are equal i.e. M = N) H N G M y corollary, area of parallelogram = base height = M 56 Understanding IS mathematics Ix F = F N ( = F and M = N) = area of parallelogram FGH. Hence, parallelograms with equal bases and between the same parallel lines are equal in area.
4 onverse of theorem 3. The converse of the above theorem 3. is also true. In fact, we have: Theorem 3.3 arallelograms on the same base and having equal areas lie between same parallel lines. Given. Two parallelograms and F on the same base and area of gm = area of gm F. To prove. F. F onstruction. From, draw M and from, draw N. roof. M N. rea of gm = M. rea of a gm = base height.. rea of gm F = N. rea of a gm = base height. 3. M = N M = N 3. area of gm = area of gm F (given) 4. M N 4. dm and N are both perpendicular to the same line. 5. MN is a parallelogram. 5. Two sides M and N of quad. MN are equal and parallel. 6. MN i.e. F 6. y definition of a gm. orollary. arallelograms on equal bases and having equal areas have equal corresponding altitudes. (In the above proof of the theorem, we obtained M = N, so gm and gm F have equal corresponding altitudes.) orollary. arallelograms on equal bases and having equal areas lie between same parallel lines. Theorem 3.4 rea of a triangle is half that of a parallelogram on the same base and between the same parallel lines. Given. triangle and a parallelogram on the same base and between the same parallel lines and. F To prove. rea of Δ = area of gm. onstruction. Through, draw F to meet (produced) at F. Theorems on rea 57
5 roof.. F is a parallelogram. y construction.. area of Δ = area of gm F 3. area of gm = area of gm F 4. area of Δ = area of gm. be is a diagonal of gm F, and a diagonal divides it into two triangles of equal areas. 3. arallelograms on the same base and between the same parallels are equal in area. 4. From and 3. orollary. rea of a triangle = base height roof. Let be a triangle with base and N, then height of Δ = N. Through and draw lines parallel to and to meet at, then is a parallelogram. Thus, Δ and parallelogram are on the same base and between same parallel lines and. N rea of Δ = = = area of parallelogram (Theorem 3.4) N (area of gm = base height) base height. orollary. median of a triangle divides it into two triangles of equal areas. roof. Let be any triangle and be one of its medians (shown in the adjoining figure). We need to show that area of Δ = area of Δ. From, draw N. Now, area (Δ) = base corresponding height = N = N N = area of Δ. ( is mid-point of, so = ) Hence, a median of a triangle divides it into two triangles of equal area. orollary 3. rea of a trapezium = (sum of parallel sides) height. roof. Let be a trapezium in which. Join. From, draw N and from, draw M (produced). M Then N = M = height of trapezium = h (say). area of trapezium = area of Δ + area of Δ) N = N + 58 Understanding IS mathematics Ix M
6 = ( h + h) = ( + ) h = (sum of parallel sides) height. Theorem 3.5 Triangles on the same base (or equal bases) and between the same parallel lines are equal in area. Given. Two triangles and on the same and between the same parallel lines and l. To prove. rea of Δ = area of Δ. onstruction. Through, draw and F to meet line l at and F respectively. F l roof.. is a parallelogram.. (given), (onst.). F is a parallelogram. F (given), F (onst.) 3. area of gm = area of gm F. 3. arallelogram on same base and between same parallel lines. 4. area of Δ = area of gm 4. area of a Δ is half that of a gm on the same base and between same parallels. 5. area of Δ = area of gm F 5. area of a Δ is half that of a gm on the same base and between same parallels. 6. rea of Δ = area of Δ 6. From 3, 4 and 5. onverse of theorem 3.5 The converse of the above theorem 3.5 is also true. In fact, we have: Theorem 3.6 Triangles on the same base (or equal bases) and having equal areas lie between the same parallel lines. Given. Two triangles and on the same base, and area of Δ = area of Δ. To prove.. onstruction. From and, draw perpendiculars M and N on respectively. roof. M N. rea of Δ = M. rea of a triangle = base height.. rea of Δ = N. Same as above. 3. M = N 3. rea of Δ = area of Δ (given) 4. M = N 4. From 3, cancelling. Theorems on rea 59
7 5. M N 6. MN is a parallelogram. 5. M and N are both perpendiculars to the same line. 6. Two sides M and N of quad. MN are equal and parallel. 7. MN i.e. 7. y definition of a gm. orollary. Triangles on the same base (or equal bases) and having equal areas have equal corresponding altitudes. (In the above proof of the theorem, we obtained M = N. So, Δ and Δ have equal corresponding altitudes.) orollary. If two triangles lie between the same parallels (i.e. have equal altitudes), then the ratio of their areas equals the ratio of their bases. orollary 3. If two triangles have equal bases, then the ratio of their areas equals the ratio of their altitudes. Illustrative xamples xample. and are any two points lying on the sides and respectively of a parallelogram. Show that area of Δ = area of Δ. Solution. Given a parallelogram, and and are points lying on the sides and respectively as shown in the adjoining figure. s Δ and gm are on the same base and between the same parallels and, area of Δ = area of gm (i) lso, as Δ and gm are on the same and between the same parallels and, area of Δ = area of gm (ii) From (i) and (ii), we get area of Δ = area of Δ. xample. In the adjoining figure, is a rectangle with sides = 8 cm and = 5 cm. ompute (i) area of parallelogram F (ii) area of FG. Solution. (i) rea of gm F = area of rectangle = (8 5) cm = 40 cm. (ii) rea of FG = (on the same base and between the same parallels and ) area of gm F (on the same base F and between the same parallels F and G) F G = 40 cm = 0 cm. 60 Understanding IS mathematics Ix
8 xample 3. point is taken on the side of a parallelogram. and are produced to meet at F. rove that area of ΔF = area of quad. F. Solution. Join and F. s triangles F and F have same base F and are between the same parallels and F ( ), area of ΔF = area of ΔF (i) s diagonal divides gm into two triangles of equal area, area of Δ = area of Δ n adding (i) and (ii), we get (ii) area of ΔF + area of Δ = area of ΔF + area of Δ area of ΔF = area of quad. F. F xample 4. The diagonals of a parallelogram intersect at. straight line through meets at and the opposite side at. rove that area of quad. = area of gm. roof.. rea of = area of gm. diagonal divides a gm into two s of equal area. In and. =. lt. s. 3. = 3. Vert. opp. s. 4. = 4. iagonals bisect each other S rule of congruency. 6. rea of = area of 6. ongruence area axiom. 7. area of + area of quad. 7. dding same area on both sides. = area of + area of quad. 8. area of quad. = area of 8. ddition area axiom. 9. area of quad. 9. From 8 and. = area of gm... xample 5. is a trapezium with, and diagonals and meet at. rove that area of = area of. Theorems on rea 6
9 roof... Given.. rea of = area of. s on the same base and between the same parallels and are equal in area. 3. area of + area of 3. ddition area axiom. = area of + area of 4. rea of = area of subtracting same area from both sides. xample 6. The diagonals and of a quadrilateral intersect at. If =, prove that the triangles and are equal in area. roof.. is median of. = (given).. rea of = area of. median divides a into two s of equal area. 3. is median of 3. = (given) 4. rea of = area of 5. area of + area of = area of + area of 6. rea of = area of median divides a into two s of equal area. 5. dding and 4 6. ddition area axiom. xample 7. In quadrilateral, M is mid-point of the diagonal. rove that area of quad. M = area of quad. M. roof. 6 Understanding IS mathematics Ix. M is median of. M is mid-point of (given).. rea of M = area of M. median divides a into two s of equal area. 3. M is median of 3. M is mid-point of (given). 4. rea of M = area of M 4. median divides a triangle into two s of equal area. 5. area of M + area of M = area of 5. dding and 4. M + area of M 6. rea of quad. M = area of quad. M ddition area axiom. M
10 xample 8. In the adjoining figure, is mid-point of and is any point on side of Δ. If meets in, then prove that area of Δ = area of Δ. Solution. Join. Triangles and are on the same base and between same parallels and, area of Δ = area of Δ (i) s is mid-point of, so is a median of Δ. Since a median divides a triangle into two triangles of equal area, area of Δ = area of Δ area of Δ + area of Δ = area of Δ area of Δ + area of Δ = area of Δ (using (i)) area of Δ = area of Δ. xample 9. In the adjoining figure, is any pentagon. drawn parallel to meets produced at and drawn parallel to meets produced at. rove that area of = area of Δ. Solution. Δ and Δ are on the same base and between same parallels. area of Δ = area of Δ (i) Δ and Δ are on the same base and between same parallels, area of Δ = area of Δ (ii) lso, area of Δ = area of Δ (iii) n adding (i), (iii) and (ii), we get area of Δ + area of Δ + area of Δ = area of Δ + area of Δ + area of Δ area of = area of Δ. xample 0. The diagonals and of a quadrilateral intersect at in such a way that area of = area of. rove that is a trapezium. Solution. raw M and N. s M and N are both perpendiculars to, therefore, M N. Given area of = area of area of + area of M N = area of + area of (adding same area on both sides) area of = area of M = N dm = N. Thus M N and M = N, therefore, MN is a parallelogram d MN i.e.. Hence, is a trapezium. Theorems on rea 63
11 xample. rove that area of a rhombus = product of diagonals. Solution. Let be a rhombus, and let its diagonals intersect at. Since the diagonals of a rhombus cut at right angles, and. s area of a triangle = base height, area of = (i) and area of = (ii) n adding (i) and (ii), we get area of + area of = + area of rhombus = = = ( + ) product of diagonals. xample. is a trapezium with. line parallel to intersects at X and at Y. rove that: area of X = area of Y. Solution. Join X. s triangles X and X have same base X and are between the same parallels ( given, so, X ), area of X = area of X (i) s triangles Y and X have same base and are between the same parallels (XY given), area of Y = area of X (ii) From (i) and (ii), we get area of X = area of Y. xample 3. XY is a line parallel to side of a triangle. If and F meet XY at and F respectively, show that area of Δ = area of ΔF. Solution. s Δ and gm Y have the same base and are between the same parallels (given), area of Δ = area of gm Y (i) > >> > X >> Y s ΔF and gm XF have the same base F and are between the same parallels F (given), X Y F area of ΔF = area of gm XF (ii) ut gm Y and gm XF have the same base and are between the same parallels (XY given), area of gm Y = area of gm XF area of gm Y = area of gm XF area of Δ = area of ΔF (using (i) and (ii)) 64 Understanding IS mathematics Ix
12 xample 4. In the adjoining figure, the side of a parallelogram is produced to any point. line through and parallel to meets produced at and then parallelogram R is completed. Show that area of gm = area of gm R. Solution. Join and. s is a diagonal of gm, area of gm = area of Δ s is a diagonal of gm R, area of gm R = area of Δ Now, triangles and have the same base and are between the same parallels, area of Δ = area of Δ (i) (ii) area of Δ area of Δ = area of Δ area of Δ (subtracting same area from both sides) area of Δ = area of Δ area of Δ = area of Δ area of gm = area of gm R (using (i) and (ii)) xample 5. In the adjoining figure, RS and XYZ are two parallelograms of equal area. rove that SX is parallel to YR. Solution. Join XR, SY. Given area of gm SR = area of gm XYZ. Subtract area of gm SX from both sides. area of gm XR = area of gm SZY area of XR = area of SY (because diagonal divides a gm into two equal areas) dding area of YR to both sides, we get area of XYR = area of SYR. lso the s XYR and SYR have the same base YR, therefore, these lie between the same parallels sx is parallel to YR. xample 6. and F are mid-points of the sides and respectively of a triangle. If F and meet at, prove that area of = area of quad. F. Solution. Join F. s and F are mid-points of and respectively, F. rea of = area of F. (Triangles on the same base and between same parallels) area of area of = area of F area of area of = area of F (i) s F is mid-point of, area of F = area of F ( median divides a triangle into two triangles of equal area). area of F area of F = area of F area of [using (i)] area of = area of quad. F (from figure) Z S Theorems on rea R R X Y F R 65
13 xample 7. In the adjoining figure,, F and F are parallelograms. Show that area of Δ = area of ΔF. Solution. s is a parallelogram, Similarly, In Δ and ΔF, ad = = F and = F. (opp. sides of a gm) ad =, = F and = F Δ ΔF (by SSS rule of congruency) area of Δ = area of ΔF (congruent figures have equal areas) F xample 8. Triangles and are on the same base with, on opposite sides of. If area of = area of, prove that bisects. Solution. Let and intersect at. raw M and N. Given area of = area of M = N am = N. In M and N, M = N (vert. opp. s) M = N (each angle = 90 ) am = N 66 Understanding IS mathematics Ix (proved above) M N (S rule of congruency) a = (c.p.c.t.) Hence, bisects. xample 9. In the adjoining figure, is a parallelogram and is produced to a point such that =. If intersects at, show that area of Δ = area of Δ. Solution. Join. s triangles and have same base and are between the same parallels ( i.e. ), area of Δ = area of Δ (i) In quad., (, opp. sides of gm ) ad = (given) is a parallelogram, so its diagonals and bisect each other i.e. = and =. In Δ and Δ, = a = = Δ Δ (vert. opp. s) area of Δ = area of Δ (ii) From (i) and (ii), we get area of Δ = area of Δ. M N
14 xample 0. In the adjoining figure, is a parallelogram. is mid-point of and meets the diagonal at. If area of = 0 cm, calculate (i) : (ii) area of (iii) area of parallelogram. Solution. (i) Since is mid-point of, =. but = ( is a gm) = () In and, = (vert. opp. s) and = (alt. s) ~ b = = d (using ()) : = :. (ii) : = : : = : 3 () since the bases, of s, lie along the same line, and these triangles have equal heights, therefore, area of b = = 3 area of b (using ()) area of = 3 area of = (3 0) cm = 30 cm. (iii) rea of = area of ( median of a triangle divides it into two triangles of equal areas) = ( 30) cm = 60 cm. area of gm = area of. ( diagonal divides a gm into two triangles of equal areas) = ( 60) cm = 0 cm. xample. is a triangle whose area is 50 cm. and F are mid-points of the sides and respectively. rove that F is a trapezium. lso find its area. Solution. Since and F are mid-points of the sides and respectively, ef and F =. s F, F is a trapezium. From, draw M. Let M meet F at N. Since F, N = MN. ut MN = 90 ( M ) so N = 90 i.e. N F. lso, as is mid-point of and N M, N is mid-point of M. N M F Now, area of F = F N = ( M) = ( M) = (area of ) 4 4 = 4 (50 cm ) = 5 cm. rea of trapezium F = area of area of F = 50 cm 5 cm = 37 5 cm. Theorems on rea 67
15 xample. rove that the area of the quadrilateral formed by joining the mid-points of the adjacent sides of a quadrilateral is half the area of the given quadrilateral. Given. quadrilateral, and RS is the quadrilateral formed by joining mid-points of the sides R,, and respectively. To prove. rea of quad. RS = area of quad.. onstruction. Join and R. roof.. rea of R = area of. median divides a triangle into two triangles of equal area.. rea of SR = area of R. Same as in. 3. rea of SR = 4 area of 3. From and. 4. rea of = 4 area of 4. s in area of SR + area of 5. dding 3 and 4. = (area of + area of ) 4 6. area of SR + area of 6. ddition area axiom. = area of quad area of S + area of R 7. Same as in 6. = area of quad area of S + area of + area of R 8. dding 6 and 7. + area of SR = area of quad. S 9. area of S + area of + area of R + area of SR + area of quad. RS = area of quad. 0. area of quad. RS = area of quad ddition area axiom. 0. Subtracting 8 from 9. xercise 3. rove that the line segment joining the mid-points of a pair of opposite sides of a parallelogram divides it into two equal parallelograms.. rove that the diagonals of a parallelogram divide it into four triangles of equal area. 3. (a) In the figure () given below, is median of and is any point on. rove that (i) area of = area of 68 Understanding IS mathematics Ix (ii) area of = area of.
16 (b) In the figure () given below,. rove that (i) area of = area of (ii) area of = area of. () () Hint: (b) (i) rea of = area of, add area of to both sides. 4. (a) In the figure () given below, is a parallelogram and is any point in. rove that, area of + area of = area of. (b) In the figure () given below, is any point inside a parallelogram. rove that (i) area of + area of = area of gm. (ii) area of + area of = area of gm. () () Hint: (b) (i) Through, draw a straight line parallel to. 5. If, F, G and H are mid-points of the sides,, and respectively of a parallelogram, prove that area of quad. FGH = area of gm. Hint: Join HF. H = and F = H = F and H F, so FH is a gm. rea of FH = area of gm FH. 6. (a) In the figure () given below, is a parallelogram., are any two points on the sides and respectively. rove that area of = area of. (b) In the figure () given below, RS and RS are parallelograms and X is any point on the side R. Show that area of ΔXS = area of gm RS. X S R () () Theorems on rea 69
Prove that a + b = x + y. Join BD. In ABD, we have AOB = 180º AOB = 180º ( 1 + 2) AOB = 180º A
bhilasha lasses lass- IX ate: 03- -7 SLUTIN (hap 8,9,0) 50 ob no.-947967444. The sides and of a quadrilateral are produced as shown in fig. rove that a + b = x + y. Join. In, we have y a + + = 80º = 80º
More informationExercise. and 13x. We know that, sum of angles of a quadrilateral = x = 360 x = (Common in both triangles) and AC = BD
9 Exercise 9.1 Question 1. The angles of quadrilateral are in the ratio 3 : 5 : 9 : 13. Find all the angles of the quadrilateral. Solution Given, the ratio of the angles of quadrilateral are 3 : 5 : 9
More informationIntroduction Circle Some terms related with a circle
141 ircle Introduction In our day-to-day life, we come across many objects which are round in shape, such as dials of many clocks, wheels of a vehicle, bangles, key rings, coins of denomination ` 1, `
More informationTriangles. Exercise 4.1
4 Question. xercise 4. Fill in the blanks using the correct word given in brackets. (i) ll circles are....(congruent, similar) (ii) ll squares are....(similar, congruent) (iii) ll... triangles are similar.
More information1 st Preparatory. Part (1)
Part (1) (1) omplete: 1) The square is a rectangle in which. 2) in a parallelogram in which m ( ) = 60, then m ( ) =. 3) The sum of measures of the angles of the quadrilateral equals. 4) The ray drawn
More informationGEOMETRY. Similar Triangles
GOMTRY Similar Triangles SIMILR TRINGLS N THIR PROPRTIS efinition Two triangles are said to be similar if: (i) Their corresponding angles are equal, and (ii) Their corresponding sides are proportional.
More information2. A diagonal of a parallelogram divides it into two congruent triangles. 5. Diagonals of a rectangle bisect each other and are equal and vice-versa.
QURILTERLS 1. Sum of the angles of a quadrilateral is 360. 2. diagonal of a parallelogram divides it into two congruent triangles. 3. In a parallelogram, (i) opposite sides are equal (ii) opposite angles
More informationUnit 2 Review. Determine the scale factor of the dilation below.
Unit 2 Review 1. oes the graph below represent a dilation? Why or why not? y 10 9 8 7 (0, 7) 6 5 4 3 (0, 3.5) 2 1 (5, 7) (5, 3.5) -10-9 -8-7 -6-5 -4-3 -2-1 0 1 2 3 4 5 6 7 8 9 10-1 F -2 (5, 0) -3-4 -5-6
More informationChapter 19 Exercise 19.1
hapter 9 xercise 9... (i) n axiom is a statement that is accepted but cannot be proven, e.g. x + 0 = x. (ii) statement that can be proven logically: for example, ythagoras Theorem. (iii) The logical steps
More informationTheorem 1.2 (Converse of Pythagoras theorem). If the lengths of the sides of ABC satisfy a 2 + b 2 = c 2, then the triangle has a right angle at C.
hapter 1 Some asic Theorems 1.1 The ythagorean Theorem Theorem 1.1 (ythagoras). The lengths a b < c of the sides of a right triangle satisfy the relation a + b = c. roof. b a a 3 b b 4 b a b 4 1 a a 3
More informationCommon Core Readiness Assessment 4
ommon ore Readiness ssessment 4 1. Use the diagram and the information given to complete the missing element of the two-column proof. 2. Use the diagram and the information given to complete the missing
More informationMT - GEOMETRY - SEMI PRELIM - II : PAPER - 4
017 1100 MT.1. ttempt NY FIVE of the following : (i) In STR, line l side TR S SQ T = RQ x 4.5 = 1.3 3.9 x = MT - GEOMETRY - SEMI RELIM - II : ER - 4 Time : Hours Model nswer aper Max. Marks : 40 4.5 1.3
More informationName: Class: Date: 5. If the diagonals of a rhombus have lengths 6 and 8, then the perimeter of the rhombus is 28. a. True b.
Indicate whether the statement is true or false. 1. If the diagonals of a quadrilateral are perpendicular, the quadrilateral must be a square. 2. If M and N are midpoints of sides and of, then. 3. The
More informationChapter 1. Some Basic Theorems. 1.1 The Pythagorean Theorem
hapter 1 Some asic Theorems 1.1 The ythagorean Theorem Theorem 1.1 (ythagoras). The lengths a b < c of the sides of a right triangle satisfy the relation a 2 + b 2 = c 2. roof. b a a 3 2 b 2 b 4 b a b
More informationCircles-Tangent Properties
15 ircles-tangent roperties onstruction of tangent at a point on the circle. onstruction of tangents when the angle between radii is given. Tangents from an external point - construction and proof Touching
More information(1) then x y z 3xyz (1/2) (1/2) 8. Given, diameter of the pillar, d 50 cm m (1/2)
Sample Question Paper (Detailed Solutions) Mathematics lass th. p( x) x 6x. Let the angle x Then, supplement angle 80 x and complement angle ccording to the question, Supplement angle 0 x omplement angles
More information1 = 1, b d and c d. Chapter 7. Worked-Out Solutions Chapter 7 Maintaining Mathematical Proficiency (p. 357) Slope of line b:
hapter 7 aintaining athematical Proficienc (p. 357) 1. (7 x) = 16 (7 x) = 16 7 x = 7 = 7 x = 3 x 1 = 3 1 x = 3. 7(1 x) + = 19 = 7(1 x) = 1 7(1 x) 7 = 1 7 1 x = 3 1 = 1 x = x 1 = 1 x = 3. 3(x 5) + 8(x 5)
More informationCircles. Exercise 9.1
9 uestion. Exercise 9. How many tangents can a circle have? Solution For every point of a circle, we can draw a tangent. Therefore, infinite tangents can be drawn. uestion. Fill in the blanks. (i) tangent
More informationPythagoras Theorem and Its Applications
Lecture 10 Pythagoras Theorem and Its pplications Theorem I (Pythagoras Theorem) or a right-angled triangle with two legs a, b and hypotenuse c, the sum of squares of legs is equal to the square of its
More informationCLASS IX GEOMETRY MOCK TEST PAPER
Total time:3hrs darsha vidyalay hunashyal P. M.M=80 STION- 10 1=10 1) Name the point in a triangle that touches all sides of given triangle. Write its symbol of representation. 2) Where is thocenter of
More informationCOURSE STRUCTURE CLASS -IX
environment, observance of small family norms, removal of social barriers, elimination of gender biases; mathematical softwares. its beautiful structures and patterns, etc. COURSE STRUCTURE CLASS -IX Units
More informationMT EDUCARE LTD. MATHEMATICS SUBJECT : Q L M ICSE X. Geometry STEP UP ANSWERSHEET
IS X MT UR LT. SUJT : MTHMTIS Geometry ST U NSWRSHT 003 1. In QL and RM, LQ MR [Given] LQ RM [Given] QL ~ RM [y axiom of similarity] (i) Since, QL ~ RM QL M L RM QL RM L M (ii) In QL and RQ, we have Q
More informationDownloaded from
Triangles 1.In ABC right angled at C, AD is median. Then AB 2 = AC 2 - AD 2 AD 2 - AC 2 3AC 2-4AD 2 (D) 4AD 2-3AC 2 2.Which of the following statement is true? Any two right triangles are similar
More informationLos Angeles Unified School District Periodic Assessments. Geometry. Assessment 2 ASSESSMENT CODE LA08_G_T2_TST_31241
Los Angeles Unified School District Periodic Assessments Assessment 2 2008 2009 Los Angeles Unified School District Periodic Assessments LA08_G_T2_TST_31241 ASSESSMENT ODE 1100209 The test items contained
More informationEC and AB because AIA are congruent Substituting into the first equation above
4.1 Triangles Sum onjectures uxillary line: an extra line or segment that helps you with your proof. Page 202 Paragraph proof explaining why the Triangle Sum onjecture is true. onjecture: The sum of the
More informationB C. You try: What is the definition of an angle bisector?
US Geometry 1 What is the definition of a midpoint? The midpoint of a line segment is the point that divides the segment into two congruent segments. That is, M is the midpoint of if M is on and M M. 1
More informationUNIT 1 VECTORS INTRODUCTION 1.1 OBJECTIVES. Stucture
UNIT 1 VECTORS 1 Stucture 1.0 Introduction 1.1 Objectives 1.2 Vectors and Scalars 1.3 Components of a Vector 1.4 Section Formula 1.5 nswers to Check Your Progress 1.6 Summary 1.0 INTRODUCTION In this unit,
More informationMAHESH TUTORIALS. Time : 1 hr. 15 min. Q.1. Solve the following : 3
S.S.. MHESH TUTRILS Test - II atch : S Marks : 30 Date : GEMETRY hapter : 1,, 3 Time : 1 hr. 15 min..1. Solve the following : 3 The areas of two similar triangles are 18 cm and 3 cm respectively. What
More informationName: Date: Period: 1. In the diagram below,. [G.CO.6] 2. The diagram below shows a pair of congruent triangles, with and. [G.CO.
Name: Date: Period: Directions: Read each question carefully and choose the best answer for each question. You must show LL of your work to receive credit. 1. In the diagram below,. [G.CO.6] Which statement
More information9. Areas of Parallelograms and Triangles
9. Areas of Parallelograms and Triangles Q 1 State true or false : A diagonal of a parallelogram divides it into two parts of equal areas. Mark (1) Q 2 State true or false: Parallelograms on the same base
More informationA part of a line with two end points is called line segment and is denoted as AB
HTR 6 Lines and ngles Introduction In previous class we have studied that minimum two points are required to draw a line. line having one end point is called a ray. Now if two rays originate from a point,
More informationName: GEOMETRY: EXAM (A) A B C D E F G H D E. 1. How many non collinear points determine a plane?
GMTRY: XM () Name: 1. How many non collinear points determine a plane? ) none ) one ) two ) three 2. How many edges does a heagonal prism have? ) 6 ) 12 ) 18 ) 2. Name the intersection of planes Q and
More informationNCERT SOLUTIONS OF Mensuration Exercise 2
NCERT SOLUTIONS OF Mensuration Exercise 2 1 Question 1 The shape of the top surface of a table is a trapezium. Find its area if its parallel sides are 1 m and 1.2 m and perpendicular distance between them
More information8. Quadrilaterals. If AC = 21 cm, BC = 29 cm and AB = 30 cm, find the perimeter of the quadrilateral ARPQ.
8. Quadrilaterals Q 1 Name a quadrilateral whose each pair of opposite sides is equal. Mark (1) Q 2 What is the sum of two consecutive angles in a parallelogram? Mark (1) Q 3 The angles of quadrilateral
More informationCBSE Class IX Syllabus. Mathematics Class 9 Syllabus
Mathematics Class 9 Syllabus Course Structure First Term Units Unit Marks I Number System 17 II Algebra 25 III Geometry 37 IV Co-ordinate Geometry 6 V Mensuration 5 Total 90 Second Term Units Unit Marks
More informationMathematics Class 9 Syllabus. Course Structure. I Number System 17 II Algebra 25 III Geometry 37 IV Co-ordinate Geometry 6 V Mensuration 5 Total 90
Mathematics Class 9 Syllabus Course Structure First Term Units Unit Marks I Number System 17 II Algebra 25 III Geometry 37 IV Co-ordinate Geometry 6 V Mensuration 5 Total 90 Second Term Units Unit Marks
More information2013 ACTM Regional Geometry Exam
2013 TM Regional Geometry Exam In each of the following choose the EST answer and record your choice on the answer sheet provided. To insure correct scoring, be sure to make all erasures completely. The
More informationProperties of Quadrilaterals
Properties of Quadrilaterals 1 Proving Properties of Parallelograms Given: ABC is a parallelogram Prove: AB C and A BC A B C Statements Reasons 1. 1. Given 2. AB C and A BC 2. 3. AB CB and CB AB 3. 4.
More information3 = 1, a c, a d, b c, and b d.
hapter 7 Maintaining Mathematical Proficienc (p. 357) 1. (7 x) = 16 (7 x) = 16 7 x = 7 7 x = 3 x 1 = 3 1 x = 3. 7(1 x) + = 19 7(1 x) = 1 7(1 x) = 1 7 7 1 x = 3 1 1 x = x 1 = 1 x = 3. 3(x 5) + 8(x 5) =
More informationMATH 243 Winter 2008 Geometry II: Transformation Geometry Solutions to Problem Set 1 Completion Date: Monday January 21, 2008
MTH 4 Winter 008 Geometry II: Transformation Geometry Solutions to Problem Set 1 ompletion Date: Monday January 1, 008 Department of Mathematical Statistical Sciences University of lberta Question 1. Let
More informationClass IX Chapter 8 Quadrilaterals Maths
Class IX Chapter 8 Quadrilaterals Maths Exercise 8.1 Question 1: The angles of quadrilateral are in the ratio 3: 5: 9: 13. Find all the angles of the quadrilateral. Answer: Let the common ratio between
More informationClass IX Chapter 8 Quadrilaterals Maths
1 Class IX Chapter 8 Quadrilaterals Maths Exercise 8.1 Question 1: The angles of quadrilateral are in the ratio 3: 5: 9: 13. Find all the angles of the quadrilateral. Let the common ratio between the angles
More informationAREAS OF PARALLELOGRAMS AND TRIANGLES
AREAS OF PARALLELOGRAMS AND TRIANGLES Main Concepts and Results: The area of a closed plane figure is the measure of the region inside the figure: Fig.1 The shaded parts (Fig.1) represent the regions whose
More informationAlg. Exercise (1) Department : Math Form : 1 st prep. Sheet. [1] Complete : 1) Rational number is 2) The set of integer is.. 3) If. is rational if x.
airo Governorate Nozha irectorate of Education Nozha Language Schools Ismailia Road epartment : Math Form : 1 st prep. Sheet [1] omplete : lg. Exercise (1) 1) Rational number is ) The set of integer is..
More informationTRIANGLES CHAPTER 7. (A) Main Concepts and Results. (B) Multiple Choice Questions
CHAPTER 7 TRIANGLES (A) Main Concepts and Results Triangles and their parts, Congruence of triangles, Congruence and correspondence of vertices, Criteria for Congruence of triangles: (i) SAS (ii) ASA (iii)
More informationVIII - Geometric Vectors
MTHEMTIS 0-NY-05 Vectors and Matrices Martin Huard Fall 07 VIII - Geometric Vectors. Find all ectors in the following parallelepiped that are equialent to the gien ectors. E F H G a) b) c) d) E e) f) F
More informationChapter (Circle) * Circle - circle is locus of such points which are at equidistant from a fixed point in
Chapter - 10 (Circle) Key Concept * Circle - circle is locus of such points which are at equidistant from a fixed point in a plane. * Concentric circle - Circle having same centre called concentric circle.
More informationName Geometry Common Core Regents Review Packet - 3. Topic 1 : Equation of a circle
Name Geometry Common Core Regents Review Packet - 3 Topic 1 : Equation of a circle Equation with center (0,0) and radius r Equation with center (h,k) and radius r ( ) ( ) 1. The endpoints of a diameter
More informationBRILLIANT PUBLIC SCHOOL, SITAMARHI (Affiliated up to +2 level to C.B.S.E., New Delhi) Affiliation No
RILLINT PULI SHOOL, SITMRHI (ffiliated up to + level to..s.e., New elhi) ffiliation No. - 049 SE oard Level IX S..- II Maths hapterwise Printable Worksheets with Solution Session : 04-5 Office: Rajopatti,
More informationAdding auxiliary lines: Solutions
dding auxiliary lines: Solutions http://topdrawer.aamt.edu.au/geometric-reasoning/misunderstandings/evealingthe-invisible/dding-auxilliary-lines 1. im: To prove b = a + c onstruction: roduce E to meet
More informationCOURSE STRUCTURE CLASS IX Maths
COURSE STRUCTURE CLASS IX Maths Units Unit Name Marks I NUMBER SYSTEMS 08 II ALGEBRA 17 III COORDINATE GEOMETRY 04 IV GEOMETRY 28 V MENSURATION 13 VI STATISTICS & PROBABILITY 10 Total 80 UNIT I: NUMBER
More information= Find the value of n.
nswers: (0- HKM Heat Events) reated by: Mr. Francis Hung Last updated: pril 0 09 099 00 - Individual 9 0 0900 - Group 0 0 9 0 0 Individual Events I How many pairs of distinct integers between and 0 inclusively
More informationWhich statement is true about parallelogram FGHJ and parallelogram F ''G''H''J ''?
Unit 2 Review 1. Parallelogram FGHJ was translated 3 units down to form parallelogram F 'G'H'J '. Parallelogram F 'G'H'J ' was then rotated 90 counterclockwise about point G' to obtain parallelogram F
More informationMT - MATHEMATICS (71) GEOMETRY - PRELIM II - PAPER - 6 (E)
04 00 Seat No. MT - MTHEMTIS (7) GEOMETRY - PRELIM II - (E) Time : Hours (Pages 3) Max. Marks : 40 Note : ll questions are compulsory. Use of calculator is not allowed. Q.. Solve NY FIVE of the following
More informationPRACTICE QUESTIONS CLASS IX: CHAPTER 4 LINEAR EQUATION IN TWO VARIABLES
PRACTICE QUESTIONS CLASS IX: CHAPTER 4 LINEAR EQUATION IN TWO VARIABLES 1. Find the value of k, if x =, y = 1 is a solution of the equation x + 3y = k.. Find the points where the graph of the equation
More informationTOPIC-1 Rational Numbers
TOPI- Rational Numbers Unit -I : Number System hapter - : Real Numbers Rational Number : number r is called a rational number, if it can be written in the form p/q, where p and q are integers and q 0,
More informationEXERCISE Two angles of a quadrilateral are 70 and 130 and the other two angles are equal. Find the measure of these two angles.
CHPTER 3 UNIT-5 EXERCISE 3.5.2 1. Two angles of a quadrilateral are 70 and 130 and the other two angles are equal. Find the measure of these two angles. ns: Let and be x, C=70 and D = 130 + + C + D = 360
More informationAreas of Polygons and Circles
Chapter 8 Areas of Polygons and Circles Copyright Cengage Learning. All rights reserved. 8.1 Area and Initial Postulates Copyright Cengage Learning. All rights reserved. Area and Initial Postulates Because
More informationNozha Directorate of Education Form : 2 nd Prep. Nozha Language Schools Ismailia Road Branch
Cairo Governorate Department : Maths Nozha Directorate of Education Form : 2 nd Prep. Nozha Language Schools Sheet Ismailia Road Branch Sheet ( 1) 1-Complete 1. in the parallelogram, each two opposite
More informationREVISED vide circular No.63 on
Circular no. 63 COURSE STRUCTURE (FIRST TERM) CLASS -IX First Term Marks: 90 REVISED vide circular No.63 on 22.09.2015 UNIT I: NUMBER SYSTEMS 1. REAL NUMBERS (18 Periods) 1. Review of representation of
More informationMathematics. Sample Question Paper. Class 9th. (Detailed Solutions) 2. From the figure, ADB ACB We have [( 16) ] [( 2 ) ] 3.
6 Sample Question Paper (etailed Solutions) Mathematics lass 9th. Given equation is ( k ) ( k ) y 0. t and y, ( k ) ( k ) 0 k 6k 9 0 4k 8 0 4k 8 k. From the figure, 40 [ angles in the same segment are
More informationAREAS OF PARALLELOGRAMS AND TRIANGLES
15 MATHEMATICS AREAS OF PARALLELOGRAMS AND TRIANGLES CHAPTER 9 9.1 Introduction In Chapter 5, you have seen that the study of Geometry, originated with the measurement of earth (lands) in the process of
More informationTERMWISE SYLLABUS SESSION CLASS-IX SUBJECT : MATHEMATICS. Course Structure. Schedule for Periodic Assessments and CASExam. of Session
TERMWISE SYLLABUS SESSION-2018-19 CLASS-IX SUBJECT : MATHEMATICS Course Structure Units Unit Name Marks I NUMBER SYSTEMS 08 II ALGEBRA 17 III COORDINATE GEOMETRY 04 IV GEOMETRY 28 V MENSURATION 13 VI STATISTICS
More informationPart (1) Second : Trigonometry. Tan
Part (1) Second : Trigonometry (1) Complete the following table : The angle Ratio 42 12 \ Sin 0.3214 Cas 0.5321 Tan 2.0625 (2) Complete the following : 1) 46 36 \ 24 \\ =. In degrees. 2) 44.125 = in degrees,
More informationGeometry Problem Solving Drill 13: Parallelograms
Geometry Problem Solving Drill 13: Parallelograms Question No. 1 of 10 Question 1. Mr. Smith s garden has 4 equal sides. It has 2 pairs of parallel sides. There are no right angles. Choose the most precise
More informationProofs. by Bill Hanlon
Proofs by Bill Hanlon Future Reference To prove congruence, it is important that you remember not only your congruence theorems, but know your parallel line theorems, and theorems concerning triangles.
More informationC.B.S.E Class X
SOLVE PPER with SE Marking Scheme..S.E. 08 lass X elhi & Outside elhi Set Mathematics Time : Hours Ma. Marks : 80 General Instructions : (i) ll questions in both the sections are compulsory. (ii) This
More informationObjectives To find the measure of an inscribed angle To find the measure of an angle formed by a tangent and a chord
1-3 Inscribed ngles ommon ore State Standards G-.. Identify and describe relationships among inscribed angles, radii, and chords. lso G-..3, G-..4 M 1, M 3, M 4, M 6 bjectives To find the measure of an
More informationDefinitions. (V.1). A magnitude is a part of a magnitude, the less of the greater, when it measures
hapter 8 Euclid s Elements ooks V 8.1 V.1-3 efinitions. (V.1). magnitude is a part of a magnitude, the less of the greater, when it measures the greater. (V.2). The greater is a multiple of the less when
More informationSOLUTIONS SECTION A [1] = 27(27 15)(27 25)(27 14) = 27(12)(2)(13) = cm. = s(s a)(s b)(s c)
1. (A) 1 1 1 11 1 + 6 6 5 30 5 5 5 5 6 = 6 6 SOLUTIONS SECTION A. (B) Let the angles be x and 3x respectively x+3x = 180 o (sum of angles on same side of transversal is 180 o ) x=36 0 So, larger angle=3x
More informationTRIPURA BOARD OF SECONDARY EDUCATION. SYLLABUS (effective from 2016) SUBJECT : MATHEMATICS (Class IX)
TRIPURA BOARD OF SECONDARY EDUCATION SYLLABUS (effective from 2016) SUBJECT : MATHEMATICS (Class IX) Total Page- 10 MATHEMATICS COURSE STRUCTURE Class IX HALF YEARLY One Paper Time: 3 Hours Marks: 90 Unit
More informationMathematics. Sample Question Paper. Class 10th. (Detailed Solutions) Sample Question Paper 13
Sample uestion aper Sample uestion aper (etailed Solutions) Mathematics lass 0th 5. We have, k and 5 k as three consecutive terms of an. 8 Their common difference will be same. 5 i.e. k k k k 5 k 8 k 8
More informationXIV GEOMETRICAL OLYMPIAD IN HONOUR OF I.F.SHARYGIN The correspondence round. Solutions
XIV GEOMETRIL OLYMPI IN HONOUR OF I.F.SHRYGIN The correspondence round. Solutions 1. (L.Shteingarts, grade 8) Three circles lie inside a square. Each of them touches externally two remaining circles. lso
More informationCalculators are NOT permitted.
THE 0-0 KEESW STTE UIVERSITY HIGH SHOOL THETIS OETITIO RT II In addition to scoring student responses based on whether a solution is correct and complete, consideration will be given to elegance, simplicity,
More informationEXTRA HOTS SUMS CHAPTER : 1 - SIMILARITY. = (5 marks) Proof : In ABQ, A ray BP bisects ABQ [Given] AP PQ = AB. By property of an angle...
MT EDURE LTD. EXTR HOTS SUMS HTER : 1 - SIMILRITY GEOMETRY 1. isectors of and in meet each other at. Line cuts the side at Q. Then prove that : + Q roof : In Q, ray bisects Q [Given] Q y property of an
More informationIncoming Magnet Precalculus / Functions Summer Review Assignment
Incoming Magnet recalculus / Functions Summer Review ssignment Students, This assignment should serve as a review of the lgebra and Geometry skills necessary for success in recalculus. These skills were
More informationYear 9 Term 3 Homework
Yimin Math Centre Year 9 Term 3 Homework Student Name: Grade: Date: Score: Table of contents 5 Year 9 Term 3 Week 5 Homework 1 5.1 Geometry (Review)................................... 1 5.1.1 Angle sum
More informationEuclidian Geometry Grade 10 to 12 (CAPS)
Euclidian Geometry Grade 10 to 12 (CAPS) Compiled by Marlene Malan marlene.mcubed@gmail.com Prepared by Marlene Malan CAPS DOCUMENT (Paper 2) Grade 10 Grade 11 Grade 12 (a) Revise basic results established
More informationQ.2 A, B and C are points in the xy plane such that A(1, 2) ; B (5, 6) and AC = 3BC. Then. (C) 1 1 or
STRAIGHT LINE [STRAIGHT OBJECTIVE TYPE] Q. A variable rectangle PQRS has its sides parallel to fied directions. Q and S lie respectivel on the lines = a, = a and P lies on the ais. Then the locus of R
More informationSolutions to Exercises in Chapter 1
Solutions to Exercises in hapter 1 1.6.1 heck that the formula 1 a c b d works for rectangles but not for 4 parallelograms. b a c a d d b c FIGURE S1.1: Exercise 1.6.1. rectangle and a parallelogram For
More informationTOPICS IN GEOMETRY: THE GEOMETRY OF THE EUCLIDEAN PLANE. 1. Introduction
TOPIS IN GEOMETRY: THE GEOMETRY OF THE EULIEN PLNE TUGHT Y PIOTR PRZYTYKI. NOTES Y YLN NT. Note. These course notes are based on a course taught by Piotr Przytycki at McGill University in the fall of 2016.
More information1983 FG8.1, 1991 HG9, 1996 HG9
nswers: (1- HKMO Heat Events) reated by: Mr. Francis Hung Last updated: 6 February 017 - Individual 1 11 70 6 1160 7 11 8 80 1 10 1 km 6 11-1 Group 6 7 7 6 8 70 10 Individual Events I1 X is a point on
More informationHonors Geometry Mid-Term Exam Review
Class: Date: Honors Geometry Mid-Term Exam Review Multiple Choice Identify the letter of the choice that best completes the statement or answers the question. 1. Classify the triangle by its sides. The
More informationGeometry. Midterm Review
Geometry Midterm Review Class: Date: Geometry Midterm Review Multiple Choice Identify the choice that best completes the statement or answers the question. 1 A plumber knows that if you shut off the water
More informationGeometric Structures
Geometric Structures 3 Hour Exams & Final Fall 2003 Exam I Page 1 Exam II Page 6 Exam III Page 10 Final Exam Page 14 Oklahoma State University Terry Williams, Instructor Geometric Structures Exam I Fall
More informationUnit 1: Introduction to Proof
Unit 1: Introduction to Proof Prove geometric theorems both formally and informally using a variety of methods. G.CO.9 Prove and apply theorems about lines and angles. Theorems include but are not restricted
More informationCBSE OSWAAL BOOKS LEARNING MADE SIMPLE. Published by : 1/11, Sahitya Kunj, M.G. Road, Agra , UP (India) Ph.: ,
OSWAAL BOOKS LEARNING MADE SIMPLE CBSE SOLVED PAPER 2018 MATHEMATICS CLASS 9 Published by : OSWAAL BOOKS 1/11, Sahitya Kunj, M.G. Road, Agra - 282002, UP (India) Ph.: 0562 2857671, 2527781 email: contact@oswaalbooks.com
More informationQUESTION BANK ON STRAIGHT LINE AND CIRCLE
QUESTION BANK ON STRAIGHT LINE AND CIRCLE Select the correct alternative : (Only one is correct) Q. If the lines x + y + = 0 ; 4x + y + 4 = 0 and x + αy + β = 0, where α + β =, are concurrent then α =,
More informationQuestions. Exercise (1)
Questions Exercise (1) (1) hoose the correct answer: 1) The acute angle supplements. angle. a) acute b) obtuse c) right d) reflex 2) The right angle complements angle whose measure is. a) 0 b) 45 c) 90
More information3. MATHEMATICS (CODE NO. 041) The Syllabus in the subject of Mathematics has undergone changes from time to time in accordance with growth of the subject and emerging needs of the society. The present
More informationKENDRIYA VIDYALAYA SANGATHAN, HYDERABAD REGION
KENDRIYA VIDYALAYA SANGATHAN, HYDERABAD REGION SAMPLE PAPER 02 F PERIODIC TEST III EXAM (2017-18) SUBJECT: MATHEMATICS(041) BLUE PRINT : CLASS IX Unit Chapter VSA (1 mark) SA I (2 marks) SA II (3 marks)
More informationCLASS IX : CHAPTER - 1 NUMBER SYSTEM
MCQ WORKSHEET-I CLASS IX : CHAPTER - 1 NUMBER SYSTEM 1. Rational number 3 40 is equal to: (a) 0.75 (b) 0.1 (c) 0.01 (d) 0.075. A rational number between 3 and 4 is: (a) 3 (b) 4 3 (c) 7 (d) 7 4 3. A rational
More information6 CHAPTER. Triangles. A plane figure bounded by three line segments is called a triangle.
6 CHAPTER We are Starting from a Point but want to Make it a Circle of Infinite Radius A plane figure bounded by three line segments is called a triangle We denote a triangle by the symbol In fig ABC has
More information0809ge. Geometry Regents Exam Based on the diagram below, which statement is true?
0809ge 1 Based on the diagram below, which statement is true? 3 In the diagram of ABC below, AB AC. The measure of B is 40. 1) a b ) a c 3) b c 4) d e What is the measure of A? 1) 40 ) 50 3) 70 4) 100
More informationSimilarity of Triangle
Similarity of Triangle 95 17 Similarity of Triangle 17.1 INTRODUCTION Looking around you will see many objects which are of the same shape but of same or different sizes. For examples, leaves of a tree
More informationGeometry Honors Review for Midterm Exam
Geometry Honors Review for Midterm Exam Format of Midterm Exam: Scantron Sheet: Always/Sometimes/Never and Multiple Choice 40 Questions @ 1 point each = 40 pts. Free Response: Show all work and write answers
More informationMA 460 Supplement: Analytic geometry
M 460 Supplement: nalytic geometry Donu rapura In the 1600 s Descartes introduced cartesian coordinates which changed the way we now do geometry. This also paved for subsequent developments such as calculus.
More information12.1 Triangle Proportionality Theorem
Name lass Date 12.1 Triangle Proportionality Theorem ssential Question: When a line parallel to one side of a triangle intersects the other two sides, how does it divide those sides? Resource Locker xplore
More informationUNIT 1: SIMILARITY, CONGRUENCE, AND PROOFS. 1) Figure A'B'C'D'F' is a dilation of figure ABCDF by a scale factor of 1. 2 centered at ( 4, 1).
EOCT Practice Items 1) Figure A'B'C'D'F' is a dilation of figure ABCDF by a scale factor of 1. 2 centered at ( 4, 1). The dilation is Which statement is true? A. B. C. D. AB B' C' A' B' BC AB BC A' B'
More informationPage 1
nswers: (008-09 HKMO Heat Events) reated by: Mr. Francis Hung Last updated: 8 ugust 08 08-09 Individual 00 980 007 008 0 7 9 8 7 9 0 00 (= 8.) Spare 0 9 7 Spare 08-09 Group 8 7 8 9 0 0 (=.) Individual
More information