Area of Parallelogram and Triangles

AREA OF PARALLELOGRAM AND TRIANGLES  IMPORTANT POINTS  Parallelograms on the same base and between the same parallels are equal in area. Area of a parallelogram is the product of its any side and the corresponding altitude. Parallelogram on the same base 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 Two congruent figures having same area. ar (||gm ABCD) = Base × Corresponding altitude = AB × BD = 5.2 × 4 sq.cm      Ex.2 In parallelogram ABCD, AB = 10 cm. The altitudes corresponding to the sides AB and AD are respectively 7 cm and 8 cm. Find AD. Sol. We have, Area of a ||gm = Base × Height.  ar (||gm ABCD) = AB × DM = (10 × 7) cm2 .... (i) Also, ar (||gmABCD) = AD × BN = (AD × 8) cm2 ....(ii)   EXAMPLES  Ex.1 ABCD is a quadrilateral and BD is one of its diagonals as shown in fig. Show that ABCD is a parallelogram and find its area. Sol. Since diagonal BD intersects transversals AB and DC at B and D respectively such that From (i) and (ii), we get 10 × 7 = AD × 8 ABD = CDB [Each equal to 90°] i.e., alternate interior angles are equal.  AB || DC Also, AB = DC [Each equal to 5.2 cms (Given)] Thus, one pair of opposite sides AB and DC of quadrilateral ABCD are equal and parallel. Hence, ABCD is a parallelogram.  Ex.3 AD = 10 7 cm = 8.75 cm. 8 In the adjoining figure, ABCD is a ||gm whose diagonals AC and BD intersect at O. A line segment through O meets AB at P and 1 DC at Q. Prove that are (APQD) = ar (||gm 2 ABCD). Now Q D C O A P B Sol. A Diagonal AC of ||gm ABCD divides it into two triangles of equal area.  ar(ACD) = 1 ar(||gm ABCD) ….(i) 2 In OAP and OCQ, we have L D B OA = OC [diagonals of a ||gm bisect each other] D is the mid-point of BC. AOP = COQ [vert. opp. ]  BD = DC PAO = QCO [alt. int. ]  BD × AL = DC × AL [Multiplying both sides by AL]  OAP OCQ  ar(OAP) = ar(OCQ)   ar(OAP) + ar(quad. AOQD) ALITER Since s ABD and ADC have equal bases and the same altitude AL. Therefore, ar ( ABC) = ar ( ADC).  ar(quad. APQD) = ar(ACD)  ar(APQD) = 1 ar(||gm ABCD) [using (i)] 2 1 1 (BD × AL) = (DC × AL) 2 2  ar (ABD) = ar (ADC) = ar(OCQ) + ar(quad. AOQD) = C Ex.5 1 ar(||gm ABCD) 2 In figure, AD is a median of ABC and DE is a median of DAC. Show that ar(AED) = Sol. 1 ar(ABC) 4 AD is a median of ABC  ar (ABD) = ar (ACD) ar (ACD) = 1 ar (ABC) 2 .... (1) DE is a median of DAC ar (AED) = Ex.4 Sol. Show that a median of a triangle divides it into two triangles of equal area. ar (AED) = To Prove ar (ABD) = ar (ADC) Proof Since AD is the median of  ABC. Therefore, .... (2) From (1) and (2), Given : A  ABC in which AD is the median. Construction : Draw AL BC. 1 ar (ACD) 2 Ex.6 1 1  1  ar(ABC) = ar (ABC) 2 2  4 The diagonals of ABCD, AC and BD intersect in O. Prove that if BO = OD, the triangles ABC and ADC are equal in area. Sol. Given : A quadrilateral ABCD in which its diagonals AC and BD intersect at O such that BO = OD. In DAC, S and R are midpoints of AD and DC respectively. To Prove : ar ( ABC) = ar ( ADC)  SR || AC and SR = Proof : In  ABD, we have BO = OD. [Given] Thus, PQ || SR and PQ = SR.  PQRS is a ||gm. C D Now, median AR divides ACD into two  of equal area. O  ar (ARD) = B A  AO is the median  ar (DSR) =  ar (AOB) = ar (AOD) ....(i) ar (DSR) = In CBD, O is the mid-point of BD.  CO is a median 1 ar (ARD) 2 ….(2) 1 ar (ACD) 4 Similarly, ar (BQP) =  ar (COB) = ar (COD) ....(ii) 1 ar (ABC) 4  ar (DSR) + ar (BQP) Adding (i) and (ii), we get = ar (AOB) + ar (COB) = ar(AOD) + ar(COD)   Let P, Q, R, S be respectively the midpoints of the sides AB, BC, CD and DA of quad. ABCD. Show that PQRS is a parallelogram 1 such that ar(||gm PQRS) = ar(quad. ABCD). 2 C 1 [ar (ACD) + ar (ABC)] 4  ar (DSR) + ar (BQP)  ar (ABC) = ar (ADC). = 1 ar [quad. ABCD) …..(3) 4 Similarly, ar (CRQ) + ar (ASP) = 1 ar (quad. ABCD) ….(4) 4 Adding (3) and (4), we get ar (DSR) + ar (BQP) + ar (CRQ) R + ar (ASP) Q D = S A ….(1) From (1) and (2), we get  Median divides a  into  two s of equal area   Sol. 1 ar (ACD) 2 Median RS divides ARD into two  of equal area.  O is the mid-point of BD Ex.7 1 AC 2 B P But, ar (DSR) + ar (BQP) + ar (CRQ) + ar (ASP) + ar (||gm PQRS) Join AC and AR. In ABC, P and Q are midpoints of AB and BC respectively.  PQ || AC and PQ = 1 ar (quad. ABCD) ….(5) 2 1 AC 2 = ar (quad. ABCD) ….(6) Subtracting (5) from (6), we get ar (||gm PQRS) = 1 ar (quad. ABCD) 2 Ex.8 Sol. D The medians BE and CF of a triangle ABC intersect at G. Prove that area of GBC = area of quadrilateral AFGE. G C H Join EF. Since the line segment Joining the mid-points of two sides of a triangle is parallel to the third side. So, EF || BC. F A E B Since HGF and ||gm HDCF are on the same base HF and between the same parallel lines.  ar(HGF) = 1 ar (||gm HDCF) 2 ....(i) Similarly, HEF and ||gm HABF are on the same base HF and between the same parallels. ar (HEF) = Clearly, sBEF and CEF are on the same base EF and between the same parallel lines. So, 1 ar (||gm HABF) 2 ....(ii) Adding (iii) and (iv),we get ar (HGF) + ar (HEF) = ar (BEF) = ar (CEF) 1 [ar(||gm HDCF) 2 + ar( ||gm HABF)] ar(BEF) –ar(GEF)  ar (||gm EFGH) = = ar(CEF) – ar (GEF)  ar (BFG) = ar (CEG) .... (i) We know that a median of a triangle divides it into two triangels of equal area. Therefore, 1 ar (||gm ABCD). 2 Ex.10 Two segments AC and BD bisect each other at O. Prove that ABCD is a parallelogram. Sol. Given: AC and BD are two segments bisecting each other at O. To Prove: ABCD is a parallelogram. ar (BEC) = ar (ABE) Construction : Join AB, BC, CD and DA. ar(BGC) + ar (CEG) = ar(quad. AFGE) + ar (BFG) ar(BGC) + ar(BFG) = ar (quad. AFGE) + ar (BFG) [Using (i)]  ar (BGC) = ar (quad. AFGE) Ex.9 Sol. E, F, G, H are respectively, the mid-points of the sides AB, BC, CD and DA of parallelogram ABCD. Show that the area of quadrilateral EFGH is half the area of the parallelogram ABCD. Given: A quadrilateral ABCD in which E, F, G, H are respectively the mid-points of the sides AB, BC, CD and DA. AO = CO [Given] BO = DO [Given] and, AOB = COD [Vertically opp. s] So, by SAS criterion of congruence AOB COD To Prove : (i) ar (||gm EFGH) = Proof : In s AOB and COD, we have 1 ar (||gm ABCD) 2 Construction : Join AC and HF  AB = CD [ Corresponding parts of congruent triangles are equal] and, 1 = 2. Thus, AB and DC intersect AC at A and C respectively such that 1 = 2 i.e. alternate interior angles are equal.  AB || DC Ex.12 A point O inside a rectangle ABCD is joined to the vertices. Prove that the sum of the areas of a pair of opposite triangles so formed is equal to the sum of the other pair of triangles. Given : A rectangle ABCD and O is a point inside it, OA, OB, OC and OD have been joined. Sol. Thus, in quadrilateral ABCD, we have AB = DC and AB || DC i.e. a pair of opposite sides are equal and parallel. Hence, ABCD is a parallelogram. To Prove : ar (AOD) + ar (BOC) = ar (AOB) + ar (COD). Hence, ABCD is a parallelogram. Construction : Draw EOF|| AB and LOM || AD. Ex.11 ABCD is a parallelogram. L and M are points on AB and DC respectively and AL = CM. Prove that LM and BD bisect each other. Proof : We have, ar (AOD) + ar (BOC) We have, AL = CM Sol.  AB – BL = CD – DM   – BL = – DM [ ABCD is a ||gm  AB = DC]  1 1 (AD × OE) + (BC × OF) 2 2 1 1  (AD × OE) + (AD × OF) [AD = BC] 2 2 1  (AD × (OE + OF)) 2 =   BL = DM .... (i)   Now, AB || DC and transversals BD and LM intersect them. D E A   3 = 4 and 1 = 2 ....(ii) Thus, in s OBL and ODM, we have 1 = 2 [From (ii)] BL = MD [From (i)] 3 = 4 [From (ii)]  1 (AD × EF) 2  1 (AD × AB) 2  1 ar (rect ABCD) 2 =   Corresponding parts of  congruent triangles are equal     BD and LM bisect each other. B [EF = AB] and,   1 (AB × OM) 2 = 1 AB × (OL + OM) 2 = 1 (AB × LM) 2  OB = OD and OL = OM  O is the mid-point of BD and LM both. F 1 1 1 (AB × OL) + (CD ×OM) 2 2 2 = (AB × OL) + OBL ~  ODM  0 L C ar(AOB)+ ar ( COD) So, by ASA criterion of congruence  M [ AB  CD ] [ LM  AD ] 1 1 = (AB × AD) = ar (rect ABCD) 2 2 ar (AOD) + ar (BOC) = ar(AOB)+ ar (COD) Ex.13 ABCD is a rhombus and P, Q, R, S are the mid-points of AB, BC, CD, DA respectively. Prove that PQRS is a rectangle.  PB = BQ ....(i)  P and Q are the mid  po ints   of AB and BC respectively    Sol. Given: A rhombus ABCD in which P, Q, R, S are the mid-points of sides AB, BC, CD and DA respectively. PQ, QR, RS and SP are joined. Now, in PBQ we have, PB = BQ  1 = 2 ....(ii)  Angles opposite to    equal sides are equal To Prove: PQRS is rectangle. Construction : Join AC. Again, ABCD is a rhombus  AB = BC = CD = AD  AB = BC, CD = AD  1 1 1 1 AB = BC, CD = AD 2 2 2 2  AP = CQ, CR = AS Proof : In order to prove that PQRS is a rectangle, it is sufficient to show that it is a parallelogram whose one angle is a right angle. First we shall prove that PQRS is parallelogram. In ABC, P and Q are the mid-points of AB and BC respectively. 1  PQ || AC and PQ = AC 2 ....(i) In ADC, R and S are the mid-points of CD and AD respectively. 1  RS || AC and RS = AC 2 ....(ii) From (i) and (ii), we have PQ || RS and PQ = RS Thus, PQRS is a quadrilateral such that one pair of opposite sides PQ and SR is equal and parallel. So, PQRS is a parallelogram. Now, we shall prove that one angle of parallelogram PQRS is a right angle. Now, in s APS and CQR, we have AP = CQ [From (iii)] AS = CR [From (iii)] and PS = QR [ PQRS is ||gm  PS = QR] So, by SSS criterion of congruence APS ~  CQR  3 = 4 ....(iv)   Corresponding parts of    congruent triangles are equal Now, 3 + SPQ + 2 = 180º and, 1 + PQR + 4 = 180º  3 + SPQ + 2 = 1 + PQR + 4  SPQ = PQR ...(v)   1   2, from (ii )    and 3  4, from (iv )  Now, transversal PQ cuts parallel lines SP and RQ at P and Q respectively.  SPQ + PQR = 180º ABCD is a rhombus  SPQ + SPQ = 180º  AB = BC  SPQ = 90º [ All sides of a rhombus are equal]  1 1 AB = BC 2 2 ....(iii) [Using (v)] Thus, PQRS is a parallelogram such that    SPQ = 90º. Hence, PQRS is a rectangle. Ex.14 Triangles ABC and DBC are on the same base BC with A, D on opposite sides of line BC, such that ar ( ABC) = ar ( DBC). Show that BC bisects AD. Sol. Adding (i) and (ii), we get ar (AOB) + ar (COD) = Since s ABC and DBC are equal in area and have a common side BC. Therefore the altitudes corresponding to BC are equal i.e. 1 ar (||gm ABCD) 2 (ii) To Prove: ar(AOB) + ar(COD) = ar (BOC) + ar (AOD). AE = DF. Construction: Draw EOF ||AB and GOH || AD. Now, in s AEO and DFO, we have  1 = 2 [Vertically opp. angles]  AEO = DFO [Each equal to 90°] and, AE = DF A Proof: Since GH || DE and EF || DC  OG || DE and OE || GD B E 1 O F 2  EOGD is a parallelogram C Similarly, EAHO, HBFO and FOGC are parallelograms. Now, OD is a diagonal of parallelogram EOGD D  ar (EOD) = ar (DOG) So, by AAS criterion of congruence, OA is a diagonal of parallelogram EAHO  AEO  DFO  AO = DO  ar (EOA) = ar (AOH)  BC bisects AD. OB is a diagonal of parallelogram HBFO  ar (BOF) = ar (BOH) Ex.15 ABCD is a parallelogram and O is any point in its interior. Prove that : (i) ar (OB)+ ar (COD) = 1 ar (||gm ABCD) 2 Adding (iii), (iv) and (v), we get ar (EOD) + ar (EOA) + ar (BOF) + ar(FOC) Given: A parallelogram ABCD and O is a point in its interior. = ar (DOG) + ar (AOH) + ar (BOH) + ar (COG) (i) Since AOB and parallelogram ABFE are on the same base AB and between the same parallel lines AB and EF. 1 ar (||gm ABFE) 2 ....(i) Similarly, 1 ar (||gm DEFC) ar (COD) = 2  ar (AOD) + ar (BOC) = ar (AOB) + ar (COD) Ex.16 A quadrilateral ABCD is such that diagonal BD divides its area in two equal parts. Prove that BD bisects AC. Sol. ....(ii) ....(v)  ar (FOC) = ar (COG) = ar (BOC) + ar (AOD)  ar (AOB) = ....(iv) OC is a diagonal of parallelogram FOGC (ii) ar (AOB) + ar (COD) Sol. .... (iii) Given: A quadrilateral ABCD in which diagonal BD bisects it, i.e. ar (ABD) = ar (BDC) Construction: Join AC. Since segment joining the mid-points of two sides of a triangle is half of the third side. Suppose AC and BD intersect at O. Draw AL  BD and CM  BD. 1 DC 2  EG = To Prove : AO = OC. ....(i) Now, ABCD is a trapezium in which AB || DC. Proof: We have, ar ( ABD) = ar( BDC) Thus, s ABD and ABC are on the same base AB and have equal area. Therefore, their corresponding altitudes are equal. i.e., AL = CM But, EF || DC  EF || AB  GF || AB In ABC, G is the mid-point of AC (proved above) and EF || AB.  F is the mid-point of BC 1 AB ....(ii) 2   Segment joining the mid  po int s of    two sides of a  is half of the third sides From (i) and (ii), we have  GF = Now, in s ALO and CMO, we have 1 = 2  [Vertically opposite angles] ALO = CMO and,AL = CM [Each equal to 90°] [Proved above] EG + GF = So, by AAS criterion of congruence ALO CMO  EF =  AO = OC BD bisects AC. Ex.17 In Fig. ABCD is a trapezium in AB is parallel to side DC and E point of side AD. If F is a point BC such that the segment EF is side DC. which side is the midon the side parallel to 1 1 (DC) + (AB) 2 2 1 (AB + DC) 2 Ex.18 In ABC, D is the mid-point of AB. P is any point of BC. CQ || PD meets AB in Q. Show 1 that ar (BPQ) = ar (ABC). 2 Sol. To Prove: ar (BPQ) = 1 ar (ABC) 2 Construction: Join CD. A Q Prove that EF = Sol. 1 (AB + DC). 2 Given : A trapezium ABCD in which AB || DC, E is the mid-point of AD and F is a point on BC such that EF || DC. To Prove: EF = 1 (AB + DC) 2 Proof: In ADC, E is the mid-point of AD and EG || DC (Given)  G is the mid-point of AC D B P C Proof: Since D is the mid-point of AB. So, in ABC, CD is the median. ar (BCD) = 1 ar (ABC) 2 .... (i) Since sPDQ and PDC are on the same base PD and between the same parallel lines PD and QC.  ar(PDQ) = ar (PDC) ....(ii) Now, from (i)  3 ar ( AGB) = ar ( ABC) ar (BCD) = 1 ar(ABC) 2  ar ( AGB) =  ar (BPD) + ar (PDC) = 1 ar (ABC) 2  ar (BPD) + ar (PDQ) = 1 ar (ABC) 2 [Using (ii)]  ar (BPQ) = 1 ar (ABC) 2 1 ar ( ABC) 3 Hence, ar ( AGB) = ar ( AGC) = ar ( BGC) 1 = ar ( ABC). 3 Ex.20 In a parallelogram ABCD, E, F are any two point on the sides AB and BC respectively. Show that ar (ADF) = ar (DCE). Sol. Construction: Draw EG || AD and FH || AB. Ex.19 If the medians of a ABC intersect at G, show that ar(AGB) = ar( AGC) 1 = ar( BGC) = ar( ABC). 3 Sol. Given: A ABC such that its medians AD, BE and CF intersect at G. To Prove : ar ( AGB) = ar ( BGC) 1 = ar (CGA) = ar ( ABC) 3 Proof : We know that the median of a triangle divides it into two triangles of equal area. Proof: Since FH || AB (by construction). Therefore, ABFH is a parallelogram. Now, AF is a diagonal of ||gm ABFH  ar (AFH) = .... (i) In  GBC, GD is the median  ar ( GBD) = ar ( GCD)  ar(DFH) = 1 ar(||gm DCFH) 2 ....(ii) ar ( ABD) – ar ( GBD) = ar ( ACD) – ar ( GCD) ....(iii) Similarly, ar (AFH) + ar(DFH) = A 1 1 ar (||gm ABFH) + ar(||gm DCFH) 2 2  ar (AFH) + ar (DFH) = 1 [ar (||gm ABFH) + (ar (||gm DCFH)] 2 ar(AFH) + ar(DFH) = E F  ar(ADF) = G B D C ar ( AGB) = ar ( BGC) 1 ar (||gm ABCD) 2 1 ar(||gm ABCD) 2 ....(iii) In ||gm AEGD, DE is a diagonal. ....(iv)  ar(DEG) = From (iii) and (iv), we get ar( AGB) = ar( BGC) = ar( AGC) ....(ii) From (i) and (ii), we have Subtracting (ii) from (i), we get  ar ( AGB) = ar ( AGC) .... (i) In ||gm DCFH, DF is a diagonal. In  ABC, AD is the median  ar ( ABD) = ar ( ACD) 1 ar(||gm ABFH) 2 ....(v) But, ar ( AGB) + ar ( BGC) + ar ( AGC) = ar ( ABC) 1 ar (||gm AEGD) 2 ....(iv) In ||gm CBEG, CE is a diagonal.  ar(CEG) = 1 ar (||gm CBEG) 2 ....(v) From (iv) and (v),we have 2 = 3 ar (DEG) + ar (CEG) = OL = PL   Opposite sides of equal  angles in triangle are equal   1 1 ar ((||gm AEGD) + ar ((||gm CBEG) 2 2  ar (DEG) + ar (CEG) Since OQ is the bisector of Q 1 1 = [ar (||gm AEGD + ar (||gm CBEG)] 2 2  4 = 5 1  ar (DCE) = ar (||gm ABCD) 2   4 = 6 ....(v) ....(vi) From (iv) and (v), we get From (iii) and (vi), we get 5 = 6 ar (ADF) = ar (DCE). Thus, in OQM, we have Ex.21 In Fig. PQRS is a parallelogram, PQ and QO are respectively, the angle bisectors of P and Q. Line LOM is drawn parallel to PQ. Prove that : Sol. ....(iv) Also, PQ || LM and transversal OQ intersects them 1  ar(DEG)+ ar(CEG)= [ar(||gm ABCD) 2 (i) PL = QM ...(iii) 5 = 6  OM = QM ....(vi) [ Opp. sides of equal angles are equal] (ii) LO = OM. Since PQRS is a parallelogram. But,PL = QM  PS || QR PL || QM So, from (iii), (vi) and (vii), we get Thus, we have OL = OM. S L 2 1 3 M 6 [As proved above] Ex.22 The diagonals of a quadrilateral ABCD are perpendicular. Show that the quadrilateral, formed by joining the mid-points of its sides, is a rectangle. R O ....(vii) 5  PQML is parallelogram. Given: A quadrilateral whose diagonals AC and BD are perpendicular to each other, P, Q, R, S are the mid-points of sides AB, BC, CD and DA respectively. PQ, QR, RS and SP are joined. PL = QM [ Opp. sides of a || gm are equal] To Prove: PQRS is a rectangle. P 4 Sol. Q PL || QM and LM || PQ [Given] This proves (i). Now, OP is the bisector of P  1 = 2 ....(i) Now, PQ || LM and transversal OP intersects them  1 = 3 From (i) and (ii), we get 2 = 3 Thus, in OPL, we have ....(ii) Proof: In ABC, P and Q are the mid-points of AB and BC respectively.  PQ || AC and PQ = 1 AC 2 .... (i) In ADC, R and S are the mid-points of CD and AD respectively.  RS || AC and RS = 1 AC 2 Sol. ....(ii) Given: ABCD is a quadrilateral in which P, Q, R and S are the mid-points of sides AB, BC, CD and DA respectively. To Prove: PQRS is a parallelogram. Construction: Join A and C. From (i) and (ii), we have D PQ || RS and PQ = RS Thus, in quadrilateral PQRS, a pair of opposite sides are equal and parallel. So, PQRS is a parallelogram. R C Q S Suppose the diagonals AC and BD of quadrilateral ABCD intersect at O. Now in ABD, P is the mid-point of AB and S is the mid-point of AD. B P A Proof: In ABC, P and Q are the mid-points of sides AB and BC respectively.  PS || BD PN || MO  PQ || AC and PQ = Also, from (i) , PQ || AC 1 AC 2 .... (i) In ADC, R and S are the mid-points of CD and AD respectively.  PM || NO Thus, in quadrilateral PMON, we have PN || MO and PM || NO  RS || AC and RS =  PMON is a parallelogram. 1 AC 2 ....(ii) From (i) and (ii), we have  MPN = MON PQ = RS and PQ || RS [ Opposite angles of a ||gm are equal] Thus, in quadrilateral PQRS one pair of opposite sides are equal and parallel.  MPN = BOA [ BOA = MON] Hence, PQRS is a parallelogram.  MPN = 90º [ AC  BD BOA = 90º]  QPS = 90º [ MPN = QPS] Thus, PQRS is a parallelogram whose one angle QPS = 90º Hence PQRS is a rectangle. Ex.23 In a parallelogram ABCD diagonals AC and BD intersect at O and AC = 6.8cm and BD = 13.6 cm. Find the measures of OC and CD. Sol. Since the diagonals of a parallelogram bisect each other. Therefore, O is the mid-point of AC and BD.  OC = 1 1 AC = × 6.8 cm = 3.4 cm 2 2 and, OD = 1 1 BD = × 5.6 cm = 2.8 cm 2 2 Ex.24 Prove that the figure formed by joining the mid-points of the pairs of consecutive sides of a quadrilateral is a parallelogram. Ex.25 The side AB of a parallelogram ABCD is produced to any point P. A line through A parallel to CF meets CB produced in Q and the parallelogram PBQR completed. Show that ar (||gm ABCD) = ar (||gm BPRQ). Sol. Construction: Join AC and PQ. D A C P B Q R To prove: ar (||gm ABCD) = ar(||gm BPRQ) Proof: Since AC and PQ are diagonals of parallelograms ABCD and BPQR respectively.  ar (ABC) = 1 ar (||gm ABCD) 2 and, ar (PBQ) = 1 ar (||gm BPRQ) 2 ....(i)   1 = 2.  BX = AB [ Sides opposite to equal  ar(ACQ) – ar (ABQ) angles in a  are equal] = ar (AQP)–ar(ABQ) 2BX = 2AB [Multiplying both sides by 2] [Subtracting ar (ABQ) from both sides]  ar (ABC) = ar (BPQ)  BC = 2AB [X is the mid-point of BC  1 1 ar (||gm ABCD ) = ar (||gm BPRQ) 2 2  ar(||gm ABCD) = ar(||gm BPRQ). Ex.26 In a parallelogram ABCD, the bisector of A also bisects BC at X. Prove that AD = 2AB. 1 A 2 .... (i) Since ABCD is a parallelogram.     AD = BC] [ ABCD is a ||gm  AD = BC] Ex.27 In Fig. BC || XY, BX || CA and AB || YC. Prove that: ar (ABX) = ar (ACY). Sol. Since AX is the bisector of A.  AD = 2AB [Using (i) and (ii)]  1 = ....(ii) Thus, in ABX, we have  ar(ACQ) = ar (AQP) Sol. 1  2 From (i) and (ii), we have 1 = 2. ....(ii) Now, s ACQ and AQP are on the same base AQ and between the same parallels AQ and CP    Join XC and BY. Since s BXC and BCY are on the same base BC and between the sum parallels BC and XY  ar(BXC) = ar(BCY) B Therefore, AD || BC and AB intersects them. .... (i) C  A + B = 180º [ Sum of interior angles is 180º] D X Q P Y C A X 2 1 Also, s BXC and ABX are on the same base BX and between the same parallels BX and AC.  ar (BXC) = ar (ABX) ....(ii)  B = 180º – A Clearly, sBCY and ACY are on the same base CY and between the same parallels AB and CY. In ABX, we have  ar(BCY) = ar (ACY) A B 1 + 2 + B = 180º  1 A + 2 + 180º – A = 180º 2  2 – 1  2 ....(iii) From (i), (ii) and (iii), we get ar(ABX) = ar (ACY). Ex.28 In Fig. AD and BE are medians of ABC and BE || DF. DE is a diagonal of ||gm DCEF A  ar (DCE) = ar (DEF) E ....(ii) FE is a diagonal of ||gm AFDE F  ar (AFE) = ar (DEF) D B ar (BDF) = ar (DCE) = ar (AFE) = ar (DEF) In BEC, DF is a line through the mid-point D of BC and parallel to BE intersecting CE at F. Therefore, F is the mid-point of CE. Because the line drawn through the mid-point of one side of a triangle and parallel to another side bisects the third side. But, ar (BDF) + ar (DCE) + ar (AFE) + ar (DEF) = ar (ABC)  4 ar (DEF) = ar (ABC)  ar (DEF) = Now, F is the mid-point of CE  ar (||gm BDEF) = 2 ×  CF =  E is the mid  po int  1 1   1  AC   2 2   of AC  EC  2 AC     CF = 1 AC 4 Ex.29 D, E, F are the mid-points of the sides BC, CA and AB respectively of ABC, prove that BDEF is a parallelogram whose area is half that of ABC. Also, show that Sol. 1 ar (ABC). 4 Now, ar (||gm BDEF) = 2 ar (DEF) 1  CF = CE 2 ar (DEF) = ....(iii) From (i), (ii) and (iii), we have 1 AC 4 Prove that CF = Sol. C 1 are (ABC). 4 = 1 ar (ABC) 2 Ex.30 ABC and DEF are two triangles such that AB, BC are respectively equal and parallel to DE, EF; show that AC is equal and parallel to DF. Sol. Since D and E are the mid-points of sides BC and AC respectively. 1 ar(ABC) 4 Given:Two triangles ABC and DEF such that AB = DE and AB || DE. Also BC = EF and BC || EF To Prove: AC = DF and AC || DF D A A F E C B B D Therefore, DE || BA C F Proof : Consider the quadrilateral ABED. DE || BF Similarly, FE || BD. So, BDEF is a parallelogram. Similarly, DCEF and AFDE are parallelograms. Now, DF is a diagonal of ||gm BDEF.  ar (BDF) = ar (DEF) E .... (i) We have, AB = DE and AB || DE  One pair of opposite sides are equal and parallel  ABED is a parallelogram.  AD = BE and AD || BE ....(i) Now, consider quadrilateral BCFE. We have, BC = EF and BC || EF  One pair of opposite sides are equal and parallel Ex.32 In ABC, AD is the median through A and E is the mid-point of AD. BE produced meets 1 AC in F. Prove that AF = AC. 3 Sol.  BCFE is a parallelogram.  CF = BE and CF || BE Through D, draw DK || BF. In ADK, E is the mid-point of AD and EF || DK. ....(ii) A From (i) and (ii), we have AD = CF and AD || CF F  ACFD is a parallelogram E K AC = DF and AC || DF Ex.31 Parallelogram ABCD & rectangle ABEF have the same base AB and also have equal areas. Show that perimeter of the parallelogram is greater than that of the rectangle. Sol. Given: A ||gm ABCD and a rectangle ABEF with the same base AB and equal areas. To Prove: Perimeter of ||gm > Perimeter of rectangle ABEF ABCD i.e. AB + BC + CD + AD > AB + BE + EF + AF. F D D B C  F is the mid-point of AK  AF = FK In BCF, D is the mid-point of BC and DK || BF  K is the mid-point of FC  FK = KC AF = FK = KC Now, AC = AF + FK + KC  AC = AF + AF + AF [Using (iii)] B  AC = 3 (AF) Proof: Since opposite sides of a parallelogram and a rectangle are equal.  AB = DC and, AB = EF [ ABCD is a ||gm] [ ABEF is a rectangle]  DC = EF .... (i)  AB + DC = AB + EF ....(ii) Since, of all the segments that can be drawn to a given line from a point not lying on it, the perpendicular segment is the shortest. BE < BC and AF < AD BC > BE and AD > AF  BC + AD > BE + AF ....(ii) From (i) and (ii), we have E C A .... (i) ....(iii) Adding (ii) and (iii), we get AB + DC + BC + AD > AB + EF + BE + AF AB + BC + CD + DA > AB + BE + EF + FA.  AF = 1 AC 3 ....(iii) IMPORTANT POINTS TO BE REMEMBERED 1. Two figures are said to be on the same base and between the same parallels, if they have a common side (base) and the vertices (or the vertex) opposite to the common base of each figure lie on a line parallel to the base. 2. Two congruent figures have equal areas but the converse need not be true. 3. A diagonal of a parallelogram divides it into two triangles of equal area. 4. Parallelograms on the same base and between the same parallels are equal in area. 5. The area of a parallelogram is the product of its base and the corresponding altitude. 6. Parallelograms on equal bases and between the same parallels are equal in area. 7. 8. 9. If a triangle and a parallelogram are on the same base and between the same parallels, then the area of the triangle is equal to the half of the parallelogram. Triangles on the same bases and between the same parallels are equal in area. 10. The area of a trapezium is half the product of its height and the sum of parallel sides. The area of a triangle is half the product of any of its sides and the corresponding altitude. 11. Triangles having equal areas and having one side of one of the triangles, equal to one side of the other, have their corresponding altitudes equal. 12. If each diagonal of a quadrilateral separates it into two triangles of equal area, then the quadrilateral is a parallelogram. 13. The area of a rhombus is half the product of the lengths of its diagonals. 14. Diagonals of a parallelogram divide it into four triangles of equal area. 15. A median of a triangle divides it into two triangles of equal area.