Areas of Parallelograms and Triangles Exercises

Lifeskills' Complete NCERT Solutions Class-IX Mathematics 103 Areas of Parallelograms and Triangles 9 Chapter In the Chapter O Area of a figure is a number (in some unit) associated with the part of the plane enclosed by that figure. Two congruent figures have equal areas but the converse need not be true. If a planar region formed by a figure T is made up of two non-overlapping planar regions formed by figures P and Q, then ar (T) = ar (P) + ar (Q), where ar (X) denotes the area of figure X. Two figures are said to be on the same base and between the same parallels, if they have a common base (side) and the vertices, (or the vertex) opposite to the common base of each figure lie on a line parallel to the base. Parallelograms on the same base (or equal bases) and between the same parallels are equal in area. Area of a parallelogram is the product of its base and the corresponding altitude. Parallelograms on the same base (or equal bases) and having equal areas lie between the same parallels. If a parallelogram and a triangle are on the same base and between the same parallels, then area of the triangle is half the area of the parallelogram. Triangles on the same base (or equal bases) and between the same parallels are equal in area. Area of a triangle is half the product of its base and the corresponding altitude. Triangles on the same base (or equal bases) and having equal areas lie between the same parallels. A median of a triangle divides it into two triangles of equal areas. O Area of a triangle = O Area of a Trapezium = O Area of a rhombus = O O O O O O O O O O O O O O 1 × Base × Height 2 1 × (Sum of the parallel sides) × (Distance between them) 2 1 × Product of the diagonals 2 The line segment joining the mid-points of two sides of a triangle is parallel to the third side. [Mid-point theorem] If a planar region formed by a figure T is made up of two non-overlapping planar regions formed by figures P and Q, then ar (T) = ar (P) + ar (Q). Figures on the same base and between the same parallels : Two figures are said to be on the same base and between the same parallels, if they have a common base (side) and the vertices (or the vertex) opposite to the common base of each figure lie on a line parallel to the base. TEXT BOOK QUESTION (SOLVED) EXERCISE 9.1 Q.1. Which of the following figures lie on the same base and between the same parallels. In such a case, write the common base and the two parallels. Ans. (i) ∆PCD and trapezium ABCD are on the same base DC and between the same parallels DC and AB. (ii) The two figures are not between the same parallels. (iii) ΔQRT and quad. RQTS are on the same base RQ and between the same parallels QR and PS. (iv) ΔPQR and quad. ABCD are not on the same base. (v) Quad. ABCD and ADQP are on the same base AD and between the same parallels AD and BQ. (vi) The figures are not on the same base. EXERCISE 9.2 Q.1. In Fig., ABCD is a parallelogram, AE ⊥ DC and CF ⊥ AD. If AB = 16 cm, AE = 8 cm and CF = 10 cm, find AD. 10. (AD) = 128 128 = 12.8 cm. 10 Q.2. If E,F,G and H are respectively the midpoints of the sides of a parallelogram ABCD, show AD = 1 ar (ABCD). 2 Ans. Given :A parallelogram ABCD in which E,F, G and H are mid-points of sides AB, BC, CD and DA respectively. that ar (EFGH) = Ans. Base DC = 16 cm as opposite AB = 16 cm, Height AE = 8 cm. ∴ Area of || gm ABCD = DC × AE = 16 × 8 = 128 cm2 ...(i) Again taking AD as the base and height = CF = 10 cm. Area of ||gm ABCD = AD × CF = AD × 10 = 10 AD cm2 ...(ii) Clearly from (i) and (ii) 1 ar (ABCD) 2 Proof : ∆EFG and ||gm BCGF are on the same base GE and between the same parallels GE and BC. To prove : ar (EFGH) = ∴ ar (FGH) = 1 ar (FBCH) 2 ...(i) and also ar (EFH) = 1 ar (AFHD) 2 ... (ii) Adding (i) and (ii), we get area (FGH) + ar (EFH) = 1 1 ar (FBCH) + ar (AFHD) 2 2 1 ar (ABCD) 2 Q.3. P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD. Show that : ar (APB) = ar (BQC). Ans. A parallelogram ABCD in which two points P and Q are lying on DC and Ad respectively. Also ∆BCQ and ΔABP are formed. To prove : ar (APB) = ar (BQC). Proof : ∆APB and parallelogram ABCD are on the same base AB and between the same parallels AB and DC. Hence, ar (EFGH) = ∴ ar (APB) = 1 ar (ABCD) ...(i) 2 1 ar (ABCD) ...(ii) 2 From (i) and (ii), we get Hence, ar (APB) = ar (BCQ) Proved. Q.4. In Fig., P is a point in the interior of a parallelogram ABCD. Show that Similarly, ar (BCQ) = 1 ar (ABCD) 2 (ii) ar (APD) + ar (PBC) = ar (APB) + ar (PCD) [Hint : Through P, draw a line parallel to AB.] (i) ar (APB) + ar (PCD) = Ans. Given : A parallelogram ABCD. P is a point inside it. To Prove : 1 ar (ABCD) 2 (ii) ar (APD) + ar (PBC) = ar (APB) + ar (PCD) Construction : Draw EF through P parallel to AB, and GH through P parallel to AD. Proof : In parallelogram EPGA, AP is a diagonal. ∴ area of ∆APG = area of ∆APF ...(i) In parallelogram BGPE, PB is a diagonal, ∴ area of ∆BPG = area of ∆EPB ...(ii) In parallelogram DHPF, DP is a diagonal, ∴ area of ∆DPH = area of ∆DPF ...(iii) In parallelogram HCEP, CP is a diagonal, ∴ area of ∆CPH = area of ∆CPE ...(iv) Adding (i), (ii), (iii) and (iv) area of ∆APG + area of ∆BPG + are of ∆DPH + area of ∆CPH = area of ∆APF + area of ∆EPB + area of ∆DPF + area of ∆CPE ⇒ [area of ∆APG + area of ∆BPG] + [area of ∆DPH + area of ∆CPH] = [area of ∆APB + area of ∆BPG] + [area of ∆EPB + area of ∆CPE] ⇒ area of ∆APB + area of ∆CPD = area of ∆APD + area of ∆BPC ...(v) But area of parallelogram ABCD = area of ∆APB + area of ∆CPD + area of ∆APD + area of ∆BPC ...(vi) From (v) and (vi), we get area of ∆APB + area of ∆PCD (i) ar (APB) + ar (PCD) = = 1 area of ABCD 2 or, ar (APB) + ar (PCD) = 1 ar (ABCD) 2 Proved. Also, from (v), ⇒ ar (APD) + ar (PBC) = ar (APB) + ar (CPD) Proved. Q.5. In Fig., PQRS and ABRS are parallelograms and X is any point on side BR. Show that (i) ar (PQRS) = ar (ABRS) (ii) ar (AX S) = 1 ar (PQRS) 2 Q.6. A farmer was having a field in the form of a parallelogram PQRS. She took any point A on RS and joined it to points P and Q. In how many parts the field is divided? What are the shapes of these parts? The farmer wants to sow wheat and pulses in equal portions of the field separately. How should she do it? Ans. Given : A parallelogram PQRS. A is any point on SR. A joined to P and Q. The field is divided into three parts (i) ∆ASP, (ii) ∆APQ and (iii) ∆AQR. All the parts are triangles. Ans. Given : Two parallelograms PQRS and ABRS and X is any point on side BR. To prove : (i) ar (PQRS) = ar (ABRS) 1 ar (PQRS) 2 Proof : (i) Parallelograms PQRS and ABRS are on the same base SR and between the same parallels SR and PB. So, ar (PQRS) = ar (ABRS) ...(i) (ii) ∆AXS are ||gm ABRS are on the same base AS and between the same parallels AS and RB. (ii) ar (∆AXS) = 1 ar (ABRS) 2 From (i) and (ii), we get So, ar (∆AXS) = ar (∆AXS) = 1 ar (PQRS. 2 ...(ii) ∆APQ and ||gm PQRS are on the same base PQ and between the same parallels, so ar (APQ) = 1 ar (ABC) 2 1 ar (PQRS) ...(ii) 2 From (i) and (ii), we get ar (APQ) = ar (ASP) + ar (AQR). Thus she should sow wheat in ∆APQ and pulses in other two ∆, or vice-versa. Proved. ∴ ar (ASP) + ar (AQR) EXERCISE 9.3 Q.1. In Fig., E is any point on median AD of a ∆ABC. Show that ar (ABE) = ar (ACE). Again, in ∆EBC, ED is the median, therefore, Area of ∆EBD = area of ∆ECD ...(ii) (Median divides the triangle into two equal parts) Subtracting (ii) from (i), we have area of ∆ABD – area of ∆EBD = area of ∆ACD – area of ∆ECD. ⇒ area of ∆ABE = area of ∆ACE ⇒ ar (ABE) = ar (ACE) Proved. Q.2. In a triangle ABC, E is the mid-point of Ans. Given : A triangle ABC, whose one median is AD. E is a point on AD. To prove : ar (ABE) = ar (ACE) Proof : Area of ∆ABD = Area of ∆ACD ...(i) (Median divides the triangle into two equal parts) 1 ar(ABC). 4 Ans. Given : In ∆ABC, E is the mid-point of median AD. median AD. Show that ar (BED) = To prove : ar (BED) = 1 ar (ABC) 4 Proof : In ∆ABC, AD is the median. We know that a median divides a triangle into two triangles of equal area. ∴ ar (ABD) = ar (∆COD) = ar (∆AOD) Proved. Q.4. In Fig., ABC and ABD are two triangles on the same base AB. If line-segment CD is bisected by AB at O, show that ar(ABC) = ar (ABD). = ar (ADC) 1 ar (ABC) ...(i) 2 Again in a similar manner, in ∆ABD, EB is the median. Thus ar (ABD) = 1 area (ABD) 2 From (i) and (ii), we get ∴ Area (BED = ar (BED) = ...(ii) 1 1 × ar (ABC) 2 2 1 ar (ABC). Proved. 4 Q.3. Show that the diagonals of a parallelogram divide it into four triangles of equal area. Hence, ar (BED) = Ans. Given : ∆ABC and ∆ABD are on the same base AB. The line CD is bisected by AB at O. i.e., OC = OD. To prove : ar (ABC) = ar (ABD) Proof : In ∆ACD, OA is the median ∴ ar (OAC) = ar (OAD) ...(i) Similarly, in ∆BCD, OB is the median ∴ ar (OBC) = ar (OBD) ...(ii) From (i) and (ii), we get ar (OAC) + ar (OBC) = ar (OAD) + ar (OBD) Hence, ar (ABC) = ar (ABD). Proved. Q.5. D, E and F are respectively the mid-points of the sides BC, CA and AB of a ∆ABC. Show that (i) BDEF is a parallelogram. (ii) ar (DEF) = Ans. Given : A ||gm ABCD in which diagonals interset each other at O. To Prove : ar (∆OAB) = ar (∆BOC) = ar (∆COD) = ar (∆AOD) Proof : We know that the diagonals of a parallelogram bisect each. ∴ In ∆ABC, OB is the median and we know that A's median divided a triangle of two triangles of equal area. ∴ ar (∆AOB) = ar (∆BOC) ...(i) Similarly, ar (∆BOC) = ar (∆COD) ...(ii) Also ar ∆COD) = ar (∆AOD) ...(iii) From (i), (ii) and (iii), we get ar (∆AOB) = ar (∆BOC) 1 ar (ABC) 4 1 ar (ABC) 2 Ans. Given : A ∆ABC in which D, E and F are respectively the mid-points of sides BC, CA and AB. To prove : (i) BDEF is a parallelogram (iii) ar (BDEF) = (ii) ar (DEF) = 1 ar (ABC) 4 (iii) ar (BDEF) = 1 ar (ABC) 2 Proof : (i) In ∆ABC, F is the mid-point of AB and E is the mid-point of AC. ∴ FE || BC or BD (mid-point theorem) ...(i) Again in DABC, E is the mid-point of AC and D is the mid-point of BC, ∴ DE || BE (Mid-point theorem) ...(ii) From (i) and (ii), we get BDEF is a parallelogram. (ii) In ||gm BDEF, ∆BDF ≅ ∆DEF [A diagonal of a ||gm divides it into two congruent triangles] Also ∆AFE ≅ ∆DEC ≅ ∆DEF ≅ ∆BDF ∴ ar (AFE) = ar (DEC) = ar(DEF) = ar (BDF)...(iii) perpendiculars to A. Proof : (i) In ∆DOL and ∆BMO ∠DLO = ∠BMO (each = 90°) OB = OD (given) ∠DOL = ∠BOM (Ver. opposite angles) ∴ ∆DOL = ∆BMO (AAS rule) Thus DL ≅ MB (CPCT) ...(i) Now ar (DOC) = ar (AOB) = 1 × CA × DL 2 ...(ii) 1 × CA × BM 2 ...(iii) 1 ar (BDEF). Proved 2 Q.6. In Fig., diagonals AC and BD of quadrilateral ABCD intersect at O such that OB = OD. If AB = CD, then show that: (i) ar (DOC) = ar (AOB) (ii) ar (DCB) = ar (ACB) (iii) DA || CB or ABCD is a parallelogram. [Hint : From D and B, draw perpendiculars to AC.] 1 × CA × DL ...(iv) 2 From (ii) and (iv), we get ar (DOC)= ar (AOB) (ii) Now ar (DOC)= ar (AOB) Adding ar (BOC) to both sides. ar (DOC) + ar (BOC) = ar (AOB) + ar (BOC) Hence, ar (DCB) = ar (ABC) (iii) ar (DCB) = ar (ABC) As the ∆DCB and ∆ABC lie on the same base and they are equal in area. Therefore, DA || CB As DA || CB and AB = CD. Hence ABCD is a parallelogram. Proved. Q.7. D and E are points on sides AB and AC respectively of ∆ABC such that ar (DBC) = ar (EBC). Prove that DE || BC. Ans. Given : A quadrilateral ABCD in which diagonal AC and BD intersect at O such that OB = OD. Also AB = CD. To prove : (i) ar (DOC) = ar (AOB) (ii) ar (DCB) = ar (ACB) (iii) DA || CB. Construction : From D and B, draw DL and DM Ans. Given : D and E are points on sides AB and AC respectively of ∆ABC, such that ar (∆BCE) = ar (∆BCD). To prove : D E | | B C. Proof : ar (∆BCE) = ar (∆BCD) Also, ∆BCE and ∆BCD have he same base BC. Since, triangles having the same base and between the same parallel lines are equal in area. 1 Hence, area (DEF) = ar (ABC) 4 (iii) ar (∆BDEF) = ar (∆FBD) + ar (∆DEF) = ar (∆DEF) + ar (∆DEF) = 2ar (∆DEF) + (From (iii)] = 2ar (∆ABC) ⇒ ar ∆ABC = or ar (AOB) = ∴ ∆BCE and ∆BCD must be between the same parallel lines. Hence, D E || BC. (Proved) Q.8. XY is a line parallel to side BC of a triangle ABC. If BE || AC and CF || AB meet XY at E and F respectively, show that ar (ABE) = ar (ACF) Ans. Given : XY is a line parallel to side BE of a ∆ABC. BE || AC and CF || AB To prove : ar (ABE) = ar (ACF) Proof : ∆ABE and parallelogram BCYE are on the same base BE and between the same parallels BE and AC. 1 ar (BCYE) ...(i) 2 [If a triangle and ||gm are on the same base and between the same parallels then area of triangle is half of the area of ||gm] Similarly, ∴ ar (ABE) = ar (ACF) = 1 ar (BCFX) 2 ...(ii) [Same Reason] But parallelogram BCYE and BCFX are on the same base BE and between the same parallels BC and EF. ∴ ar (BCYE) = ar (BCFX) ...(iii) From (i), (ii) and (iii), we get Hence, ar (ABE) = ar (ACF) Proved. Q.9. The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q and then parallelogram PBQR is completed (see Fig.). Show that ar (ABCD) = ar (PBQR). [Hint : Join AC and PQ. Now compare ar (ACQ) and ar (APQ).] Ans. Join AC and PQ. Now, ar (ACQ) = ar (AQP) AQ || CP and base AQ is same. Subtracting ar (AQB) from the both sides, ar (ACQ) – ar (AQB) = ar (AQB) – ar (AQB) ar (ABC) = ar (BPQ) 1 ar (ABC) = ar (ABCD) 2 1 and ar (BPQ) = ar (PBQR) 2 Hence, ar (ABCD) = ar (PBQR). Q.10. Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O. Prove that ar (AOD) = ar (BOC). Ans. Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O. As ∆ABC and ∆ABD are on the same base and between the same parallels. ∴ ar (ABD) = ar (ABC) ⇒ ar (ABD) – ar (AOB) = ar (ABC) – ar (AOB) [Subtracting ar (AOB) from the both sides) ⇒ ar (AOD) = ar (BOC). Q.11. In Fig., ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that (i) ar (ACB) = ar (ACF) (ii) ar (AEDF) = ar (ABCDE) Ans. Given ABCDE is a pentagon and BF || AC. (i) ∆ACB and ∆ACF being on the same base AC and between the same parallels AC and BF. ∴ ar (∆ACB) = ar (∆ACF) ...(i) (ii) ar (AEDC) = ar (AEDC) + ar (∆ACF) = ar (AEDC) + ar (∆ACB) = ar (ABCDE) Q.12. A villager Itwaari has a plot of land of the shape of a quadrilateral. The Gram Panchayat of the village decided to take over some portion of his plot from one of the corners to construct a Health Centre. Itwaari agrees to the above proposal with the condition that he should be given equal amount of land in lieu of his land adjoining his plot so as to form a triangular plot. Explain how this proposal will be implemented. Ans. We are given a quadrilateral plot ABCD. Since the quadrilateral has to be constructed into a triangle of equal area, so we proceed as follows : Draw CE || BD which inersects AB produced at E. Join D and E, intersecting BC at K. Now AED is the required triangular plot left after giving to Gram Panchayat the triangular plot CDK, we know that ar (BDE) = ar (BCD) [Triangles on the same base and between the same parallels] Adding ar (ABD) to both sides, we get ar (ABD) + ar (BDE) = ar (ABD) + ar (BCD) ∆AED) = ar (quad. ABCD) ar (∆ Proved. Q.13. ABCD is a trapezium with AB || DC. A line parallel to AC intersects AB at X and BC at Y. Prove that ar (ADX) = ar (ACY). [Hint : Join CX.] Ans. Given, ABCD is a trapezium and AB || CD. Also, XY || AC. Now, join CX. Since, ∆ADX and ∆ACX lie on the same base AX and between the same parallel lines AB and DC. ∴ ar (∆ADX) = ar (∆ACX) ...(i) Again, ∆ACX and ∆ACY lie on the same base AC and between the same parallel lines AC and XY. ∴ ar (∆ACX) = ar (∆ACY) ...(ii) Hence, from Eqs.(i) and (ii), we get ar (∆ADX) = ar (∆ACY) 14. In Fig., AP || BQ || CR. Prove that ar (AQC) = ar (PBR). Ans. Given : As given in the figure AP || BQ || CR. To prove : ar (AQC) = ar (PBR) Proof : ∆ABQ and ∆PBQ are on the same base BQ and between the same parallels AP and BQ. ∴ ar (ABQ) = ar (PBQ) ...(i) Again ∆BQC and ∆QBR are on the same base BQ and between the same parallels BQ and CR, ∴ ar (BQC) = ar (QBR) ...(ii) Adding (i) and (ii), we get ar (ABQ) + ar(BQC) = ar (PBQ) + ar (QBR) ar (AQC) = ar (PBR). Proved. Q.15. Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that ar (AOD) = ar (BOC). Prove that ABCD is a trapezium. Ans. Given : A quadrilateral ABCD in which diagonals AC and BD intersect at O in such a way that ar (AOD) = ar (BOC) To prove : ABCD is a trapezium. Proof : ar (AOD) = ar (BOC) (Given) Adding ar (COD) to both sides, we get ar (AOD) + ar (COD) = ar (BOC) + ar (COD) ∴ ar (ACD) = ar (BCD) we see that ∆ACD and ∆BCD are on the same base CD and their areas are equal, therefore, AB || CD. Now two opposite sides of quadrilateral ABCD are parallel, therefore ABCD is a trapezium. Q.16. In Fig., ar (DRC) = ar (DPC) and ar (BDP) = ar (ARC). Show that both the quadrilaterals ABCD and Δ CPR are trapeziums. Ans. Given : ar (∆DPC) = ar (∆DRC) ...(i) and ar (∆BDP) = ar (∆ARC) ...(ii) On subtracting Eq. (i) from Eq. (ii), we get ar (∆BDP) – ar (∆DPC) = ar (∆ARC) – ar (∆DRC) ⇒ ar (∆BDC) = ar (∆ADC) Since, these two triangles are on the same base ∴ DC || AB Hence, ABCD is a trapezium. Also, ar (∆DRC) = ar (∆DPC) Since, both triangles have the same base DC. ∴ RP || DC Hence, PRCD is a trapezium. DC. EXERCISE 9.4 (Optional) Q.1. Parallelogram ABCD and rectangle ABEF are on the same base AB and have equal areas. Show that the perimeter of the parallelogram is greater than that of the rectangle. Ans. Given : ABC is a triangle and D and E are two points on BC, such that Ans. Given : Parallelogram ABCD and rectangle ABEF are on the same base AB and ar (ABCD) = ar (ABEF). To prove : Perimeter of parallelogram > perimeter of rectangle. or AB + BC + CD + DA > AB + BE + EF + AF. Proof : In ∆AFD, ∠F = 90° ∴ DA > FA ...(i) Also in ∆BCE, ∠CEF = 90° ∴ BC > BE ...(ii) Also AB = EF ...(iii) (Opposite sides of a rectangle) CD = AB ...(iv) (Opp. sides of a ||gm) From (i), (ii), (iii) and (iv), we get DA + BC + AB + CD > FA + BE + EF + AB Hence, AB + BC + CD + DA > AB + BE + EF + FA. Q.2. In Fig., D and E are two points on BC such that BD = DE = EC. Show that ar (ABD) = ar (ADE) = ar (AEC). Can you now answer the question that you have left in the ‘Introduction’ of this chapter, whether the field of Budhia has been actually divided into three parts of equal area? [Remark: Note that by taking BD = DE = EC, the triangle ABC is divided into three triangles ABD, ADE and AEC of equal areas. In the same way, by dividing BC into n equal parts and joining the points of division so obtained to the opposite vertex of BC, you can divide ∠ABC into n triangles of equal areas.] BD = DE = EC Let AO be the perpendicular to BC. ∴ ar (∆ABD) = 1 × BD × AO 2 ar (∆ADE) = 1 × DE ×AO 2 1 × EC × AO 2 Since, BD = DE = EC (Given) ∴ ar (∆ABD) = ar (∆ADE) = ar (∆AEC) Yes, altitudes of all triangles are same. Budhia has used the result of this question in dividing her land in three equal parts. Q.3. In Fig., ABCD, DCFE and ABFE are parallelograms. Show that ar (ADE) = ar (BCF). and ar (∆AEC) = Ans. Given : There are three parallelograms ABCD, DCEF and ABFE. To prove : ar (ADE) = ar (BCF). Proof : In ∆ADE and ∆BCF, AE = BF AD = BC DE = CF (Opposite sides of a parallelogram) So, ∆ADE ≅ ∆BCF (SSS Rule) We know that congruent triangles are equal in areas. Hence ar (ADE) = ar (BCF). Proved. Q.4. In Fig., ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ. If AQ intersect DC at P, show that ar (BPC) = ar (DPQ). [Hint : Join AC.] Thus ar (DPQ) = ar (PCQ) ...(i) Also in PBQ, BC = AD = CQ ∴ B C = CQ ∴ ar (PCQ) = ar (BPC) ...(ii) [Median of D divides it into two triangles of equal area] From (i) and (ii), we get ar (BPC) = ar (DPQ) Proved. Q.5. In Fig., ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, show that 1 ar (ABC) 4 (i) ar (BDE) = Ans. Given : ABCD is a parallelogram. BC is produced to a point Q such that AD = CQ, AQ intersects DC at P. To prove : ar (BPC) = ar (DPQ) 1 ar (BAE) 2 (iii) ar (ABC) = 2 ar (BEC) (iv) ar (BFE) = ar (AFD) (v) ar (BFE) = 2 ar (FED) (ii) ar (BDE) = 1 ar (AFC) 8 [Hint : Join EC and AD. Show that BE || AC and DE || AB, etc.] Ans. Given : Two equilatral triangles ABC and BDE such that D is the mid-point of BC. AE intersects BC at F. To prove : (vi) ar (FED) = Construction : Join A and C. AD = CQ (Given) and AD | CQ ∴ ADQC is a parallelogram. [A pair of opposite sides are equal and parallel] ∴ ar (∆PDQ) = ar (PCQ) Diagonals divide a parallelogram in 4 triangles of equal area. (i) ar (BDE) = 1 ar (ABC) 4 1 ar (BAE) 2 (iii) ar (ABC) = 2 ar (BEC) (iv) ar (BFE) = ar (AFD) (v) ar (BFE) = 2 ar (FED) (ii) ar (BDE) = 1 ar (AFC) 8 Construction : Join E and C and also join A and D. Draw EL ⊥ BC. Proof : (i) Let side of ∆BDE = x. (vi) ar (FED) = 3 2 x 4 Now side of ∆ABC = 2x ∴ at (BDE) = ...(i) 3 ( 2 x) 2 = 3x 2 4 On dividing (i) by (ii), we get Thus, ar (ABC) = 4 ar (BDE) Hence, ∴ ar (ABC)= ...(ii) 1 ar (ABC) 4 (ii) ∠ACB = ∠EBC (each = 60°) But these are alternate angles because transversal BC intersects BE and AC. ∴ B E || AC Now ΔBEC and ΔBAE are on the same base BE and between the same parallels AC and BE. ∴ ar (BEC) = ar (BAE) ...(ii) Again in ΔBEC, Δ is the mid-point of BC. 1 ar (ABC) 2 Hence ar (ABC) = 2 ar( BEC). (iv) ∠BDE = ∠DBA = 60° But these are alternate angles because BD intersects two lines AB and ED. ∴ AB | DE Now ∆ADE and ∆BED are on the same base ED and between the same parallels AB and DE. ∴ ar (ADE)= ar (BED) Subtracting ar (FED) from both sides. ar (ADE) – ar (FED) = ar (BED) – ar (FED) ar (AFD) = ar (BFE) (v) In ∆ABC, AD2 = AB2 – BD2 ∴ ar (BDE) = 1 ar (BEC) 2 From (i) and (ii), we get ∴ ar (BDE) = ar (BEC) = = ⇒ a2 − a 2 3a 2 = 4 4 3 a 2 AD = [Let AB = a then BD = In ∆BED EL2 = DE2 – DL2 2 ...(ii) 1 ar (BDE) = ar (BEC) Proved. 2 (iii) As median divides a ∆ into two triangles of equal area, therefore ar (BDE) = ar (CDE) ar (BCE) = 2 ar (BED) From part (i), ⎡1 ⎤ But (BED) = 2 ⎢ ar (ABC)⎥ ⎣4 ⎦ a ] 2 2 = ⎛a⎞ ⎛a⎞ ⎜ ⎟ −⎜ ⎟ ⎝2⎠ ⎝4⎠ = a 2 a 2 3a 2 − = 4 16 16 a 3a [BL = DL = ] 4 4 [ABC is an equilateral ∆ and AD is median] So, AD will be ⊥ on BC, i.e. ∠ADF = 90°. ∆ABD ≅ ∆ACD ⇒ EL = ∴ ar (AFD)= (Proved above) = and ar (EFD) = = 1 × FD × AD 2 1 3 a × FD × 2 2 ...(i) 1 × FD × EL 2 1 3 a × FD × 2 2 ...(ii) From (i) and (ii), we have ar (AFD)= 2 ar (EFD) Combining this result with part (iv). We have ar (BFE) = ar (AFD) = 2 ar (EFD) (iv) From part (i) 1 ar (ABC) 4 ar (BDE) = ∴ ar (BEF) + ar (FED) = Also ar (CPD) = and ar (BPC) = L.H.S. = 1 × 2ar (ADC) 4 ⇒ 2ar (FED) + ar (FED) = 1 (ar(AFC) – ar (AFD)) [Using part (v)] 2 ⇒ 3ar (FED) = ⇒ 4ar FED = 1 1 ar (AFC) – × 2ar (FED) 2 2 1 ar (AFC) 2 1 ar (AFC) 8 Q.6. Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that ar (APB) × ar (CPD) = ar (APD) × ar (BPC). [Hint : From A and C, draw perpendiculars to BD.] ⇒ ar (FED) = = R.H.S. = 1 × PD × CM 2 1 × BP × CM 2 ar (APB) × ar (CPD) 1 (BP × AN) × 2 1 (PD × CM) 2 ar (APD) × ar (CPD) 1 1 (PD × AN) × (BP × CM) 2 2 Hence, L.H.S. - R.H.S. Proved. Q.7. P and Q are respectively the mid-points of sides AB and BC of a triangle ABC and R is the midpoint of AP, show that = (i) ar (PRQ) = 1 ar (ARC) 2 3 ar (ABC) 8 (iii) ar (PBQ) = ar (ARC) Ans. Given : A triangle ABC, P and Q are midpoints of AB and BC, R is the mid-point of AP. Proof : Since CP is a median of ∆ABC. (ii) ar (RQC) = 1 (ABC) 2 [We know that the median divides a triangle into two triangles of equal area] ... (i) ⇒ ar (APC) = ar (PBC) = ar Ans. Given : A quadrilateral in which diagonals AC and BD intersect each other at P. To prove : ar (APB) × ar (CPD) = ar (APD) × ar (BPC) Proof : ar (APB) = 1 × BP × AN 2 and 1 × PD × AN 2 ar (APD)= Similarly, CR is a median of ∆APC ∴ ar (ARC) = ar (PRC) = 1 ar (APC) ...(ii) 2 QR is a median of ∆APQ. ∴ ar (ARQ) = ar (PRQ) = 1 ar (APQ) ...(iii) 2 PQ is a median of ∆PBC 1 ∴ ar (PQC) = ar (PQB) = ar (PBC) 2 RQ is a median of ∆RBC. ar (RQC) = ar (RQB) = ...(iv) 1 ar (RBC) ...(v) 2 (i) ar (PQA) = ar (PQC) (Triangles on the same base PQ and between the same parallels PQ and AC) ⇒ ar (ARQ) + ar (PRQ) = 1 ar (ABC) ...(ii) 4 From equations (i) and (ii), we get ar (PBQ) = ar (ARC). Proved. Q.8. In Fig., ABC is a right triangle right angled at A. BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment AX ⊥ DE meets BC at Y. Show that: = 1 ar (PBC) 2 [From (iv)] 1 ar (APC) [From (i), (iii)] 2 ⇒ 2(ar (PRQ) = ar (ARC) [From (ii)] ⇒ 2ar (PRQ) = ⇒ ar (PRQ) = 1 ar (ARC) 2 Proved. [From equation (v)] (ii) We have, 1 ar (RQC) = ar (RBC) 2 = ⇒ ar (RQC) = = 1 1 ar (PBC) + ar (PRC) 2 2 1 1 ar (ABC) + ar (APC) 2 4 [From (i) and (ii)] 1 1 ar (ABC) + ar (ABC) 8 4 ⇒ ar (RQC) = 2ar (ABC) + ar (ABC) 8 ⇒ ar (RQC) = 3 ar (ABC) 8 Proved. (iii) We have ar (PBQ)= = ar (ARC)= 1 ar (PBC) 2 (Proved above in eqn. (iv)] 1 ar (ABC) 4 1 ar (APC) 2 ...(i) (i) ∆MBC ≅ ABD (ii) ar (BYXD) = 2 ar (MBC) (iii) ar (BYXD) = ar (ABMN) (iv) ∆FCB ≅ ACE (v) ar (CYXE) = 2 ar (FCB) (vi) ar (CYXE) = ar (ACFG) (vii) ar (BCED) = ar (ABMN) + ar (ACFG) Note : Result (vii) is the famous Theorem of Pythagoras. You shall learn a simpler proof of this theorem in Class X. Ans. Given : ∆ABC is right angled at A. On sides BC, CA and AB there are respective squares BCED, ACFG and ABMN on sides BC, CA and AB. Line segment AX ⊥ DE which meets BC at Y. To prove : (i) ∆MBC ≅ ∆ABD (ii) ar (BYXD) = 2 ar (MBC) (iii) ar (BYXD) = ar (ABMN) (iv) ∆FCB ≅ ∆ACE (v) ar (CYXE) = 2 ar (FCB) (vi) ar (CYXE) = ar (ACFG) (vii) ar (BCED) = ar (ABMN) + ar (ACFG) Proof : (i) In ∆MBC and ∆ABD MB = AB (Sides of a square) BC = BD (Sides of a square) ∠MBC = ∠ABD [MBC = 90° + ∠ABC and ∠ABD = 90° + ∠ABC] ∴ ∆MBC ≅ ∆ABC (SAS Congruence Rule) (ii) ∴ ar (MBC) = ar (ABC) ...(i) [Areas of congruent triangles are equal] Now rectangle BYXD and ΔABD are on the same base BD and between the same parallels, So, ar (BYXD) = 2ar (ABD) ...(ii) From (i) and (ii), we get ar (BYXD) = 2 ar (MBC) ...(iii) (Proved) (iii) Again square ABMN and ΔMBC are on the same base MB and between the same parallels MB and NA, So 2ar (MBC) = ar (ABMN) ...(iv) From (iii) and (iv), we get ar (BYXD) = ar (ABMN) ...(A) (iv) In ∆FCB and ∆ACE, BC = CE (Sides of a ractangle) CF = AC (sides of a square) ∠BCF = ∠ACE [90° + ∠ACB = ∠BCF = ∠ACE] ∴ ∆FCB ≅ ∆ACE (SAS congruence rule) (v) ∴ ar (FCB) = ar (ACE) ...(v) (Congruent triangles, are equal in area) ∆ACE and rectangle CYXE are on the same base CE and between the same parallels AX and CE. ∴ ar (CYXE) = 2 ar (ACE) ...(vi) From (v) and (vi), we get ar (CYXE) = 2 ar (FCB) ...(vii) (vi) ∆BCF and square ACFG are on the same base AC and between the same parallels CF and BAG. ∴ ar (ACFG) = 2 ar (CBF) ...(viii) From (v), (vi), (vii) and (viii), we get ar (CYXE) = ar (ACFG) ...(ix) (Proved) (vii) Adding (viii) and (ix), we get ar (BYXD) + ar (CYXE) = ar (ABMN) + ar (ACFG) Hence ar (BCED) = ar (ABMN) + ar (ACFG) (Proved) Additional Questions Q.1. If the ratio of altitude and the area of the parallelogram is 2 : 11, then find the length of the base of parallelogram. Ans. Let the length of the altitude by x and base of the parallelogram be b. Then area of the parallelogram = b × x. And, area of triangle = 1 × base × altitude 2 ⎛1 ⎞ = ⎜ × 24 × h ⎟ cm2 2 ⎝ ⎠ Also, area of triangle = Area of rectangle Now, ⇒ altitude area x b× x = 2 11 = 2 11 ⇒ 1 b = 2 11 ⇒ b = 11 Units. 2 Q.2. The area of a triangle is equal to the area of a rectangle whose length and breadth are 18 cm and 12 cm respectively. If the base of the triangle is 24 cm, then, find its altitude. Ans. Area of rectangle = length × breadth = (18 × 12) cm2 Therefore, 1 × 24 × h = 18 × 12 2 2 × 18 × 12 24 ⇒ h = 18 cm Q.3.In figure ABCD and EFGH are two parallelogram and G is the mid-point of CD. Then, ⇒ h = Ans. False. We know that, if a triangle and a parallelogram are on the same base and btween the same parallels, then area of triangle is equal to half of the area of parallelogram. ∴ ar (DPC) = 1 (ABCD) = ar (EFGD) 2 Q.4.In which of the following (fig.). You find two polygons on the same base and between the same parallels? Ans. (d) in figure (d) we have two polygons PQRA and PQOA on the same base PQ and between the same parallels PQ and AR. Similarly we have BQRS and BORS on the same base RS and between the same parallels RS and OB. Q.5. In Fig., ABCD is a paralelogram and EFCD is a rectangle. Also, AL ⊥ DC. Prove that (i) ar (ABCD) = ar (EFCD) (ii) ar (ABCD) = DC × AL Ans. (i) As a rectangle is also a parallelogram, therefore, ar (ABCD) = ar (EFCD) (Theorem 9.1) (ii) From above result, ar (ABCD) = DC × FC ...(i) (Area of the rectangle = length × breadth) As AL ⊥ DC, therefore, AFCL is also a rectangle So, AL = FC ...(ii) Therefore, ar (ABCD) = DC × AL [From (i) and (ii)] Q.6. P is any point on the median AD of ΔABC. Show that ar (APB) = ar (ACP) Ans. Since AD is the median of ∆ABC, therefore, ar (∆ABD) = ar (∆ACD) ...(i) Also, PD is the median of ∆PBC, therefore, ar (∆PBD) = ar (∆PCD) ...(ii) Subtracting (ii) from (i), we get ar (∆ABD) – ar (∆PBD) = ar (∆ACD) – ar (∆PCD) ⇒ ar (∆APB) = ar (∆APC) Proved. Q.7. XY is a line parallel to side BC of a triangle ABC passing through A. If BE || AC and CF || AB meet XY at E and F respectively, show that : ar (ABE) = ar (ACF), Ans. Given : XY is a line parallel to side BE of ∆ABC, BE || AC and CF || AB. To prove : ar (ABE) = ar (ACF) 1 ar (BCYE) ...(i) 2 [ΔABE and || gm BCYE are on the same base BE and BE || AC] Similarly, Proof : ar (ABE) = 1 ar (ACF) = ar (BCFX) ...(ii) 2 But ar (BCYE) = ar (BCFX) ...(iii) [Parallelograms on the same base and between the same parallels are equal in area] ∴ from (i), (ii) and (iii) we get ar (ABE) = ar (ACF) Proved. Q.8. X and Y are points on the side LN of the triangle LMN such that LX = XY = YN. Through X, a line is drawn parallel to LM to meet MN at Z (See Fig.). Prove that ar (LZY) = ar (MZYX) (iii) ar (BEF) Proof : (i) ar (ABEF) = ar (||gm ABCD) = 90cm2 [Since ||gm lie on the same base and between the same parallels) ⇒ ar (ABEF) = 90cm2 1 ar (||gm ABCD) 2 [A diagonal of a ||gm divides it into two equal parts) (ii) ar (∆ABD) (iii) ar (∆BEF) Ans. Given : LX = XY = YN and LM || XZ To prove : ar (LZY) = ar (MZYX) Proof : LM || XZ and ∆LXA and ∆XMZ are on the same base (XZ) and between the same parallels. ∴ ar (∆LXZ) = ar (∆XMZ) ...(i) Adding ar (∆XYZ) to both the sides of equation (i), we get ar (∆LXZ) + ar (∆XYZ) = ar (∆XMZ) + ar (∆XYZ) ⇒ ar (∆LZY) = ar (MZYX) Q.9. The area of the parallelogram ABCD is 90 cm2 (See fig). Find (i) ar (ABEF) (ii) ar (ABD) (iii) ar (BEF) Ans. Given : ar (||gm ABCD) = 90cm2 To find : (i) ar (ABEF) (ii) ar (ABD) = = 1 × 90 = 45 cm2 2 = 1 ar ( ||gm ABEF) 2 = 1 × 90 = 45 cm2 2 Q.10. O is any point on the diagonal PR of a parallelogram PQRS (Fig.). Prove that ar (PSO) = ar (PQO). Ans. Given : PQRS is a || gm and O is any point on PR. To prove : as (PSO) = ar (PQO) Construction : Join Q to S. Proof : ST = TQ ∴ PT is a median of ∆PQS [Diagonal of a ||gm bisect each other] ∴ ar (∆PTS) = ar (∆PTQ) ...(i) [Median of a ∆ divides a triangle into two triangles of equal area.] Also, OT is a median of ∆SOQ ∴ ar (∆OST) = ar (∆OQT) ...(ii) On adding (i) and (ii), we get ar (∆PTS) + ar (∆OPST) = ar (∆PTQ) + ar (∆OQT) ⇒ ar (PSO) = ar (PQO) Proved. Multiple Choice Questions Q.1. Two parallelogram are no n equal bases and between the same parallels. The ratio of their areas is: (a) 1:2 (b) 1:1 (c) 2:1 (d) 3:1 Ans. (b) 1:1 Q.2. The median of a triangle divides it into two : (a) right triangles (b) isosceles triangles (c) triangles of equal area (d) congruent triangles. Ans. (c) triangles of equal area Q.3. The mid-point of the sides of a triangle along with any of the vertices as the fourth point make a parallelogram of area equal to (a) 1 ar (ABC) 2 (c) 1 ar ABC 4 (b) 1 ar (ABC) 3 (d) ar (ABC) 1 ar (ABC) 2 Q.4. ABCD is a quadrilateral whose diagonal AC divides it into two parts, equal in area, then ABCD : (a) is a rectanlge (b) is always a rhombus (c) is a parallelogram (d) need not be any of (A), (B) or (C) Ans. (d) need not be any of (A), (B) or (C) Q.5. If a triangle and a parallelogram are on the same base and between same parallels, then Ans. (a) the ratio of the area of the triangle to the area of parallelogram is : (a) 1 : 3 (b) 1 : 2 (c) 3 :1 (d) 1 : 4 Ans. (b) 1 : 2 Q.6. If area of the parallelogram ABCD is 92.4 cm2. If E and F are the mid– point of AB and CD respectively then area of parallelogram AEFD is: (a) 92.4 cm2 (b)184.8 cm2 2 (c) 46.7 cm (d) 46.2 cm2 2 Ans. (d) 46.2 cm Q.7. AD is median of ΔABC. If area of ΔABC is 50cm2, the areaof ΔABD is is: (a) 100 cm2 (b) 25 cm2 2 (c) 50 cm (d) 75 cm2 2 Ans. (b) 25 cm Q.8. If the length of the diagonal of a square is 8cm, thenits area is : (a) 64cm2 (b) 32 cm2 2 (c) 16 cm (d) 48 cm2 2 Ans. (b) 32 cm Q.9. In ΔABC,D, E and F are the mid– points of BC, ΔABC)= 56cm2, CA and AB respectively. if ar (Δ the ar (AEDF) is: (a) 92cm2 (b) 28 cm2 2 (c) 46 cm (d) 42 cm2 2 Ans. (b) 28 cm Q.10. Two figures are called congruence if they have: (a) same area (b) same shape (c) same size (d) same shape and size Ans. (d) same shape and size