Class IX Chapter 5 Areas of Parallelograms and Triangles Maths Exercise 5.. Which of the following figures lie on the same base and between the same parallels. In such a case, write the common base and two parallels. (i) PCD and trapezium ABCD or on the same base CD and between the same parallels AB and DC. (ii) Parallelogram ABCD and APQD are on the same base AD and between the same parallels AD and BQ. (iii) Parallelogram ABCD and PQR are between the same parallels AD and BC but they (iv) are not on the same base. QRT and parallelogram PQRS are on the same base QR and between the same parallels QR and PS (v) Parallelogram PQRS and trapezium SMNR on the same base SR but they are not between the same parallels. (vi) Parallelograms PQRS, AQRD, BCQR and between the same parallels also, parallelograms PQRS, BPSC and APSD are between the same parallels..
Class IX Chapter 5 Areas of Parallelograms and Triangles Maths Exercise 5.. In fig below, ABCD is a parallelogram, AE DC and CF AD. If AB = 6 cm, AE = 8 cm and CF = 0 cm, find AD. Given that, In a parallelogram ABCD, CD AB 6cm [Opposite sides of a parallelogram are equal] We know that, Area of parallelogram = base corresponding attitude Area of parallelogram ABCD CD AE AD CF 6cm8cm AD0cm 68 AD cm 8cm 0 Thus, the length of AD is 8cm. In Q. No, if AD = 6 cm, CF = 0 cm, and AE = 8cm, find AB. We know that, Area of parallelogram ABCD = ADCF () Again area of parallelogram ABCD DC AE () Compare equation () and equation () AD CF DC AE 60 D B 60 D 75 cm 8 AB DC 75 cm [Opposite sides of a parallelogram are equal]
Class IX Chapter 5 Areas of Parallelograms and Triangles Maths 3. Let ABCD be a parallelogram of area 4 cm. If E and F are the mid-points of sides AB and CD respectively, then find the area of parallelogram AEFD. Given, Area of parallelogram ABCD 4cm Construction: draw AP DC Proof: Area of parallelogram And area of parallelogram And AFED DF AP EBCF FC AP...... DF FC... 3 [F is the midpoint of DC] Compare equation (), () and (3) Area of parallelogram AEFD Area of parallelogram EBCF Area of parallelogram ABCD Area of parallelogram AEFD 4 6cm 4. If ABCD is a parallelogram, then prove that ar ( ABD) = ar ( BCD) = ar ( ABC) = ar ( ACD) = ar ( gm ABCD) Given: ABCDis a parallelogram To prove: area ABD ar ABC are ACD gm ar ABCD Proof: we know that diagonals of a parallelogram divides it into two equilaterals. Since, AC is the diagonal.
Class IX Chapter 5 Areas of Parallelograms and Triangles Maths Then, ar ABC ar ACD ar gm ABCD... Since, BD is the diagonal Then, ar ABD ar BCD ar gm ABCD... Compare equation () and () ar ABC ar ACD ar ABD ar BCD ar ABCD gm Exercise 5.3. In the below figure, compute the area of quadrilateral ABCD. Given that DC 7cm AD 9cm and BC 8cm In BCD we have CD BD BC 7 BD 8 BD 89 64 BD 5 In ABD, we have AB AD BD 5 AB 9 AB AB 5 8 44 quad, ar ABCD ar ABD ar BCD ar quad, ABCD 9 87 54 68
Class IX Chapter 5 Areas of Parallelograms and Triangles Maths cm ar (quad, ABCD 9 85 54 60cm 4cm. In the below figure, PQRS is a square and T and U are respectively, the mid-points of PS and QR. Find the area of OTS if PQ = 8 cm. From the figure T and U are the midpoints of PS and QR respectively. TU PQ TO PQ Thus, in PQS, T is the midpoint of PS and TO PQ TO PQ 4cm Also, TS PS 4cm ar OTS TOTS 4 4 cm 8 cm 3. Compute the area of trapezium PQRS is Fig. below. We have
Class IX Chapter 5 Areas of Parallelograms and Triangles Maths trap rect are ar PQRS ar PSRT a QRT ar PQRS PT RT QT RT 8 RT 8 RT RT In trap QRT, we have QR QT RT RT QR QT RT 7 8 5 RT 5 ar trap PQRS 5cm 80cm Hence, 4. In the below fig. AOB = 90, AC = BC, OA = cm and OC = 6.5 cm. Find the area of AOB. Since, the midpoint of the hypotenuse of a right triangle is equidistant from the vertices CA CB OC CA CB 65 cm AB 3cm In a right angle triangle OAB, we have AB OB OA 3 OB OB OB 5 3 69 44 5 ar AOB OAOB 5 30 cm
Class IX Chapter 5 Areas of Parallelograms and Triangles Maths 5. In the below fig. ABCD is a trapezium in which AB = 7 cm, AD = BC = 5 cm, DC = x cm, and distance between AB and DC is 4cm. Find the value of x and area of trapezium ABCD. Draw AL DC, BM DC Then, AL BM 4cm and LM 7cm In ADL, we have AD AL DL 5 6 DL DL 3cm Similarly MC BC BM 5 6 3cm x CD CM ML CD 3 73 3cm ar trap ABCD AB CD AL 7 3 4cm 40cm 6. In the below fig. OCDE is a rectangle inscribed in a quadrant of a circle of radius 0 cm. If OE = 5, find the area of the rectangle. Given OD 0cm and OE 5cm By using Pythagoras theorem OD OE DE DE OD OF 0 5 4 5cm rect 5 4 5 ar DCDE OE DE cm 40cm 5 5 5
Class IX Chapter 5 Areas of Parallelograms and Triangles Maths 7. In the below fig. ABCD is a trapezium in which AB DC. Prove that ar ( AOD) = ar( BOC). Given: ABCDis a trapezium with To prove: ar AOD ar BOC Proof: Since AB DC ADC and BDC are on the same base DC and between same parallels AB and DC Then, ar ADC ar( BDC ar AOD ar DOC ar BOC ar DOC ar AOD ar BOC 8. In the given below fig. ABCD, ABFE and CDEF are parallelograms. Prove that ar ( ADE) = ar ( BCF) Given that, ABCDis a parallelogram AD BC CDEF is a parallelogram DE CF ABFE is a parallelogram AE BF Thus, in s ADE and BCF, we have AD BC, DE CF and AE BF So, by SSS criterion of congruence, we have ADE ABCF ar ADE ar BCF
Class IX Chapter 5 Areas of Parallelograms and Triangles Maths 9. Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that: ar( APB) ar( CPD) = ar( APD) ar ( BPC) Construction: Draw BQ AC and DR AC Proof: L.H.S ar APB ar CPD AP BQ PC DR PC BQ AP DR ar BPC ar APD RHS LHS RHS Hence proved. 0. In the below Fig, ABC and ABD are two triangles on the base AB. If line segment CD is bisected by AB at O, show that ar ( ABC) = ar ( ABD) Given that CD is bisected at O by AB To prove: ar ABC ar ABD Construction: Draw CP AB and DQ AB
Class IX Chapter 5 Areas of Parallelograms and Triangles Maths Proof:- ar ABC AB CP i ar ABC AB DQ... ii In CPO and DQO... CPQ DQO Each 90 Given that CO DO COP DOQ [vertically opposite angles are equal] Then, CPO DQO [By AAS condition] CP DQ... 3 [CP.C.T] Compare equation (), () and (3) Area ABC area of ABD. If P is any point in the interior of a parallelogram ABCD, then prove that area of the triangle APB is less than half the area of parallelogram. Draw DN AB and PM AB. Now, Area ABCD AB DN, ar APB AB PM gm Now, PM DN AB PM AB DN AB PM AB DN area APB ar Parragram ABCD
Class IX Chapter 5 Areas of Parallelograms and Triangles Maths. If AD is a median of a triangle ABC, then prove that triangles ADB and ADC are equal in area. If G is the mid-point of median AD, prove that ar ( BGC) = ar ( AGC). Draw AM BC Since, AD is the median of BD DC ABC BD AM DC AM BD AM DC AM... ar ABD ar ACD i In BGC, GD is the median... ar BGD area OGD ii In ACD, CG is the median area AGC area CGD... iii From (i) and (ii), we have Area BGD ar AGC But, ar BGC ar BGD ar BGC ar AGC 3. A point D is taken on the side BC of a ABC such that BD = DC. Prove that ar( ABD) = ar ( ADC).
Class IX Chapter 5 Areas of Parallelograms and Triangles Maths Given that, In ABC, BD DC To prove: ar ABD ar ADC Construction: Take a point E on BD such that BE ED Proof: Since, BE ED and BD DC Then, BE ED DC le We know that median of divides it into two equal In is a median ABD, AE Then, area ABD ar AED... i In AEC, AD is a median Then area AED area ADC.. ii Compare equation (i) and (ii) Area ABD ar ADC. les 4. ABCD is a parallelogram whose diagonals intersect at O. If P is any point on BO, prove that: (i) ar ( ADO) = ar ( CDO) (ii) ar ( ABP) = ar ( CBP) Given that ABCDis a parallelogram To prove: (i) ar ADO ar CDO (ii) ar ABP ar CBP Proof: We know that, diagonals of a parallelogram bisect each other AO OC and BO OD (i) In DAC, since DO is a median Then area ADO area CDO (ii) In BAC, Since BO is a median Then; area BAO area BCO... In a PAC, Since PO is a median Then, area PAO area PCO... Subtract equation () from equation ()
Class IX Chapter 5 Areas of Parallelograms and Triangles Maths area BAO ar PAO ar BCO area PCO Area ABP Area of CBP 5. ABCD is a parallelogram in which BC is produced to E such that CE = BC. AE intersects CD at F. (i) Prove that ar ( ADF) = ar ( ECF) (ii) If the area of DFB = 3 cm, find the area of gm ABCD. In triangles ADF and ECF, we have ADF ECF [Alternative interior angles, Since AD EC Since AD BC CE And DFA CFA [vertically opposite angles] So, by AAS congruence criterion, we have ADF ECF area ADF area ECF and DF CF. Now, DF CF BF is a median in BCD 3 6 area BCD ar BDF area BCD cm cm Hence, cm ar ABCD ar BCD 6cm gm AD BE ] 6. ABCD is a parallelogram whose diagonals AC and BD intersect at O. A line through O intersects AB at P and DC at Q. Prove that ar ( POA) = ar ( QOC).
Class IX Chapter 5 Areas of Parallelograms and Triangles Maths In triangles POA and QOC, we have AOP COQ [vertically opposite angles] OA OC [Diagonals of a parallelogram bisect each other] PAC QCA [ AB DC; alternative angles] So, by ASA congruence criterion, we have POA QOC Area POA area QOC. 7. ABCD is a parallelogram. E is a point on BA such that BE = EA and F is a point on DC such that DF = FC. Prove that AE CF is a parallelogram whose area is one third of the area of parallelogram ABCD. Construction: Draw FG AB Proof: We have BE EA and DF FC AB AE EA and DC FC FC AB 3EA and DC 3FC AE AB and FC DC... 3 3 But AB DC Then, AE = DC Then, AE FC. [opposite sides of gm ]
Class IX Chapter 5 Areas of Parallelograms and Triangles Maths Thus, AE = FC and AE FC. Then, AECF is a parallelogram gm Now ar AECF AE FG ar AECF AB FG from () 3 gm gm 3 ar AECF AB FG... Compare equation () and (3) gm and area ABCD AB FG... 3 gm gm 3 ar AECF area ABCD area AECF area ABCD 3 gm gm 8. In a ABC, P and Q are respectively the mid-points of AB and BC and R is the mid-point of AP. Prove that : (i) ar ( PBQ) = ar ( ARC) (ii) (iii) ar ( PRQ) = ar ( ARC) ar ( RQC) = 3 ar ( ABC). 8 (i) le We know that each median of a divides it into two triangles of equal area Since, OR is a median of CAP ar CRA ar CAP... i Also, CP is a median of CAB... ar CAP ar CPB ii From (i) and (ii) we get area ARC ar CPB... iii PQ is the median of PBC... area CPB area PBQ iv
Class IX Chapter 5 Areas of Parallelograms and Triangles Maths (ii) From (iii) and (iv) we get... area ARC area PBQ v Since QP and QR medians of s QAB and QAP respectively. ar QAP area QBP... vi And area QAP ar QRP... vii From (vi) and (vii) we have PRQ ar PBQ... viii From (v) and (viii) we get Area PRQ area ARC (iii) Since, Ris a median of area ARC ar CAP Area CAP ar ABC area ABC 4 Since RQ is a median of RBC ar RQC ar RBC ar ABC ar ARC ar ABC ABC 4 3 ABC 8 9. ABCD is a parallelogram, G is the point on AB such that AG = GB, E is a point of DC such that CE = DE and F is the point of BC such that BF = FC. Prove that: (i) ar ADEG ar GBCE 6 ar EFC ar EBF (ii) ar EGB ar ABCD (iii) (iv) ar EBG ar EFC (v) Find what portion of the area of parallelogram is the area of EFG.
Class IX Chapter 5 Areas of Parallelograms and Triangles Maths Given, ABCDis a parallelogram AG GB, CE DE and BF FC To prove: (i) ar ADEG ar GBCE (ii) ar EGB are ABCD 6 (iii) ar EFC area EBF 3 (iv) area EBG area EFC (v) Find what portion of the area of parallelogram is the area of EFG. Construction: draw EP AB and EQ BC Proof : we have, AG GB and CE DE and BF FC AB GB GB and CD DE DE and BC FC FC AB GB GB and CD DE DE and BC FC FC. AB 3GB and CD 3DE and BC 3FC GB AB and DE CD and FC BC... i 3 3 3 (i) ar ADEG AG DE EP ar ADEG AB CD EP 3 3 [By using ()] ar ADEG ar ADEG AB AB EP 3 3 AB EP AB CD And ar GBCE GB CE EP...
Class IX Chapter 5 Areas of Parallelograms and Triangles Maths ar GBCE AB CD EP 3 3 [By using ()] ar GBCE AB AB EP AB CD 3 3 ar GBCE AB EP... Compare equation () and (3) ar EGB GB EP AB EB 6 9 m ar ABCD. 6 EFC FC EQ... 4 And area EBF BF EQ ar EBF FC EQ BF FC given (ii) (iii) Area... 5 ar EBF FC EQ Compare equation 4 and 5 Area EFC area EBF (iv) From (i) part 5 m ar EGB ar ABCD... 6 6 From (iii) part ar EFC ar EBF ar EFC ar EBC 3 ar EFC CE EP 3 CD EP 3 3 gm ar ABCD 6 3 ar EFC ar EGB [By using] 3
Class IX Chapter 5 Areas of Parallelograms and Triangles Maths 3 ar EGB ar EFC. (v) Area EFG ar Trap BGEC ar BGF GB EC EP Now, area (trap BGEC) AB CD EP 3 3 ABEP 5m ar ABCD Area 5 m EFC area ABCD [From iv part] 9 And area BGF BF GR BC GR 3 BC GR 3 ar GBC 3 GB EP 3 AB EP 3 3 AB EP 9 gm ar ABCD [From ()] 9 gm gm gm ar EFG ar ABCD ar ABCD ar ABCD 9 9 5 gm ar ABCD. 8 0. In Fig. below, CD AE and CY BA. (i) Name a triangle equal in area of CBX (ii) Prove that ar ( ZDE) = ar ( CZA) (iii) Prove that ar (BCZY) = ar ( EDZ)
Class IX Chapter 5 Areas of Parallelograms and Triangles Maths Since, BCA and BYA are on the same base BA and between same parallels BA and CY Then area BCA ar BYA ar CBX ar BXA ar BXA ar AXY ar CBX ar AXY... Since, ACE and ADE are on the same base AE and between same parallels CD and AE Then, ar ACE ar ADE ar CLA ar AZE ar AZE ar DZE ar CZA DZE... Since CBY and CAY are on the same base CY and between same parallels BA and CY Then ar CBY ar CAY Adding ar CYG on both sides, we get ar CBX ar CYZ ar CAY ar CYZ ar BCZX ar CZA... 3 Compare equation () and (3) ar BCZY ar DZE. In below fig., PSDA is a parallelogram in which PQ = QR = RS and AP BQ CR. Prove that ar ( PQE) = ar ( CFD). Given that PSDA is a parallelogram Since, AP BQ CR DS and AD PS PQ CD...i In BED, C is the midpoint of BD and CF F is the midpoint of ED BE
Class IX Chapter 5 Areas of Parallelograms and Triangles Maths EF PE Similarly EF PE PE FD... In SPQE and CFD, we have PE FD EDQ FDC, And PQ CD [Alternative angles] So by SAS congruence criterion, we have PQE DCF.. In the below fig. ABCD is a trapezium in which AB DC and DC = 40 cm and AB = 60 cm. If X and Y are respectively, the mid-points of AD and BC, prove that: (i) XY = 50 cm (ii) DCYX is a trapezium (iii) ar (trap. DCYX) = 9 ar (trap. (XYBA)) (i) (ii) Join DY and produce it to meet AB produced at P In 's BYP and CYD we have BYP CYD [Vertical opposite angles] DCY PBY DC AP And BY CY So, by ASA congruence criterion, we have BYP CYD DY YP and DC BP y is the midpoint of DP Also, x is the midpoint of AD XY AP and XY AD xy AB BD xy BA DC xy 60 40 We have XY AP
Class IX Chapter 5 Areas of Parallelograms and Triangles Maths XY AB and AB DC [As proved above] XY DC DCY is a trapezium (iii) Since x and y are the midpoint of DA and CB respectively Trapezium DCXY and ABYX are of the same height say hm Now ar Trap DCXY DC XY h 50 40 hcm 45 hcm 3 ar trap ABXY AB XY h 60 50 hm ar trap ABYX AB XY h 60 50 hcm 55cm ar trapyx 45h 9 ar trap ABYX 55h 9 ar trap ABXY ar trap DCYX 3. In Fig. below, ABC and BDE are two equilateral triangles such that D is the mid-point of BC. AE intersects BC in F. Prove that 4 (i) ar BDE ar ABC (ii) area BDE ar BAE (iii) ar BEF ar AFD. (iv) area ABC area BEC 8 (v) ar FED ar AFC (vi) ar BFE ar EFD
Class IX Chapter 5 Areas of Parallelograms and Triangles Maths Given that, ABC and BDE are two equilateral triangles. Let AB BC CA x. Then (i) (ii) We have 3 ar ABC x 4 x BD DE BE 3 x 3 ar ABC x 4 4 4 3 x ar BDE 4 It is given that triangles ABC and BED are equilateral triangles ACB DBE 60 (Since alternative angles are equal) BE AC Triangles BAF and BEC are on the same base BE and between the same parallel BE and AC ar BAE area BEC ar BAE ar BDE [ ED is a median of EBC; ar BEC ar BDE ] area BDE ar BAE (iii) Since ABC and BDE are equilateral triangles ABC 60 and BDE 60 ABC BDE AB DE (Since alternative angles are equal) Triangles BED and AED are on the same base ED and between the same parallels AB and DE. ar BED area AED. ar BED area EFD area AED area EFD ar BEF ar AFD (iv) Since ED is the median of BEC area BEC ar BDE ar BEC ar ABC [from (i)] 4
Class IX Chapter 5 Areas of Parallelograms and Triangles Maths (v) ar BEC area ABC area ABC area BEC Let h be the height of vertex E, corresponding to the side BD on triangle BDE Let H be the height of the vertex A corresponding to the side BC in triangle ABC From part (i) ar BDE ar ABC 4 BD h ar ABC 4 BD h BC H 4 h H... From part..(iii) BFE ar AFD Area FD H FD H FD h ar EFD (vi) area AFC area AFD area ADC ar BFE ar ABC [using part (iii); and AD is the median ABC ] ar BFE 4 ar BDE using part (i)] ar BFE ar FED... 3 Area BDE ar BFE ar FED R ar FED ar FED FED 3 ar... 4 From (), (3) and (4) we get AFC area FED 3ar FED Area 8 ar FED 8 Hence, area FED area AFC
Class IX Chapter 5 Areas of Parallelograms and Triangles Maths 4. D is the mid-point of side BC of ABC and E is the mid-point of BD. if O is the mid-point of AE, prove that ar ( BOE) = ar ( ABC). 8 Given that D is the midpoint of side BC of ABC. E is the midpoint of BD and O is the midpoint of AE Since AD and AE are the medians of ABC and ABD respectively ar ABD ar ABC i ar ABE ar ABD ii OB is a median of ABE ar BOE ar ABE From i, (ii) and (iii) we have ar BOE ar ABC 8...... 5. In the below fig. X and Y are the mid-points of AC and AB respectively, QP BC and CYQ and BXP are straight lines. Prove that ar ( ABP) = ar ( ACQ). Since x and y are the midpoint AC and AB respectively XY BC Clearly, triangles BYC and BXC are on the same base BC and between the same parallels XY and BC
Class IX Chapter 5 Areas of Parallelograms and Triangles Maths area BYC area BXC area BYC ar BOC ar BXC ar BOC ar BOY ar COX ar BOY ar XOY ar COX ar XOY ar BXY ar CXY We observe that the quadrilateral XYAP and XYAQ are on the same base XY and between the same parallel XY and PQ.... area quad XYAP ar quad XYPA ii Adding (i) and (ii), we get ar ABP ar ACQ ar BXY ar quad XYAP ar CXY ar quad XYQA 6. In the below fig. ABCD and AEFD are two parallelograms. Prove that (i) PE = FQ (ii) ar ( APE) : ar ( PFA) = ar (QFD) : ar ( PFD) (iii) ar PEA ar( QFD) Given that, ABCD and AEFD are two parallelograms To prove: (i) PE FQ (ii) ar APE ar QFD ar PFA ar PFD (iii) ar PEA ar QFD Proof: (i) In EPA and FQD PEA QFD [ Corresponding angles] EPA FQD [Corresponding angles] PA QD opp sides of gm Then, EPA FQD [By. AAS condition]
Class IX Chapter 5 Areas of Parallelograms and Triangles Maths EP FQ c. p. c. t (ii) Since, PEA and QFD stand on the same base PE and FQ lie between the same parallels EQ and AD ( ) () ar PEA ar QFD... AD ar PFA ar PFD Divide the equation (i) by equation () area of PEA arqfd area of PFA ar PFD (iii) From (i) part EPA FQD Then, ar EDA ar FQD 7. In the below figure, ABCD is parallelogram. O is any point on AC. PQ AB and LM AD. Prove that ar ( gm DLOP) = ar ( gm BMOQ) Since, a diagonal of a parallelogram divides it into two triangles of equal area areaadc area ABC gm area APO area DLOP area OLC... area AOM ar gmdlop area OQC i Since, AO and OC are diagonals of parallelograms AMOP and OQCL respectively. areaapo area AMO... ii And, area OLC Area OQC... iii Subtracting (ii) and (iii) from (i), we get Area gm gm DLOP area BMOQ 8. In a ABC, if L and M are points on AB and AC respectively such that LM BC. Prove that: (i) ar LCM ar LBM (ii) ar ( LBC) ar MBC (iii) ar ABM ar ACL (iv) ar LOB ar MOC (i) Clearly Triangles LMB and LMC are on the same base LM and between the same parallels LM and BC.
Class IX Chapter 5 Areas of Parallelograms and Triangles Maths (ii)... ar LMB ar LMC i We observe that triangles LBC and MBC area on the same base BC and between the same parallels LM and BC arc LBC ar MBC... ii (iii) We have ar LMB ar LMC [from ()] ar ALM ar LMB ar ALM ar LMC ar ABM ar ACL (iv) We have ar CBC ar MBC [from ()] ar LBC ar BOC ambc ar BOC ar LOB ar MOC 9. In the below fig. D and E are two points on BC such that BD = DE = EC. Show that ar ( ABD) = ar ( ADE) = ar ( AEC). Draw a line through A parallel to BC Given that, BD DE EC We observe that the triangles ABD and AEC are on the equal bases and between the same parallels C and BC. Therefore, Their areas are equal. Hence, ar ABD ar ADE ar ACDE
Class IX Chapter 5 Areas of Parallelograms and Triangles Maths 30. If below 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: (i) MBC ABD (ii) ar BYXD ar MBC (iii) (iv) (v) (vi) (vii) ar (BYXD) = ar ( ABMN) FCB ACE ar (CYXE) = ar ( FCB) ar (CYXE) = ar (ACFG) ar (BCED) = ar (ABMN) + ar (ACFG) (i) In MBC and ABD, we have MB AB BC BD And MBC ABD [ MBC and ABC are obtained by adding ABC to a right angle] So, by SAS congruence criterion, We have MBC ABD... ar MBC ar ABD (ii) Clearly, ABC and BYXD AX and BD Area ABD Area rect BYXD ar rect BYXD ar ABD are on the same base BD and between the same parallels... are rect BYXD ar MBC ar ABD ar MBC... from ( i)
Class IX Chapter 5 Areas of Parallelograms and Triangles Maths (iii) Since triangle M BC and square MBAN are on the same Base MB and between the same parallels MB and NC ar MBC ar MBAN... 3 From () and (3) we have ar rect BYXD. ar sq MBAN (iv) In triangles FCB and ACE we have (v) FC AC CB CF And FCB ACE [ FCB and ACE are obtained by adding So, by SAS congruence criterion, we have FCB ACE We have FCB ACE ar FCB ar ECA ACB to a right angle] Clearly, ACE and rectangle CYXE are on the same base CE and between the same parallels CE and AX ar ACE ar CYXE... 4 (vi) Clearly, FCB and rectangle FCAG are on the same base FC and between the same parallels FC and BG ar FCB ar FCAG... 5 From (4) and (5), we get Area CYXE ar ACFG (vii) Applying Pythagoras theorem in ACB, we have BC AB AC BC BD AB MB AC FC area BCED area ABMN ar ACFG