Theorems on Area. Introduction Axioms of Area. Congruence area axiom. Addition area axiom

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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

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

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

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