{"id":25560,"date":"2021-06-23T15:22:54","date_gmt":"2021-06-23T09:52:54","guid":{"rendered":"https:\/\/ncertsolutions.guru\/?p=25560"},"modified":"2021-06-23T15:22:54","modified_gmt":"2021-06-23T09:52:54","slug":"ncert-solutions-for-class-9-maths-chapter-9-ex-9-3","status":"publish","type":"post","link":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-9-maths-chapter-9-ex-9-3\/","title":{"rendered":"NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.3"},"content":{"rendered":"

These NCERT Solutions for Class 9 Maths<\/a> Chapter 9 Areas of Parallelograms and Triangles Ex 9.3 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\n

NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Exercise 9.3<\/h2>\n

Question 1.
\nIn fig. 9.23, E is any point on median AD of a \u2206ABC Show that ar(\u2206ABE) = ar(\u2206ACE).
\n\"NCERT
\nSolution:
\nGiven: In \u2206ABC, AD is median. Any point E lie of median AD.
\nTo prove that: ar(\u2206ABE) = ar(\u2206ACE)
\nProof: In \u2206ABC, AD is median.
\n\u2234 ar(\u2206ABD) = ar(\u2206ACD) ……(i)
\n(A median of a triangle divides it into two triangles of equal areas)
\nNow, In \u2206EBC, ED is median.
\n\u2234 ar(\u2206EBD) = ar(\u2206CED) ……(ii)
\n(A median of a triangle divides it into two triangles of equal areas)
\nSubtract equation (ii) from equation (i)
\nar(\u2206ABD) – ar(\u2206EBD) = ar(\u2206ACB) – ar(\u2206CED)
\n\u2234 ar(\u2206ABE) = ar(\u2206ACE)<\/p>\n

\"NCERT<\/p>\n

Question 2.
\nIn a triangle ABC, E is the mid point of median AD. Show that ar(BED) = \\(\\frac {1}{4}\\) ar(ABC).
\nSolution:
\nGiven: ABC is a triangle in which AD is median and E is the mid point of median AD.
\n\"NCERT
\nTo prove that: ar(BED) = \\(\\frac {1}{4}\\) ar(ABC)
\nConstruction: Join BE.
\nProof: In \u2206ABD, E is the mid point of AD. Therefore, BF is median.
\nWe know that a median of a triangle divide triangles of equal areas.
\n\u2234 ar(\u2206BFD) = ar(\u2206ABE)
\nor, ar(\u2206BED) = \\(\\frac {1}{4}\\) ar(\u2206ABD) …..(i)
\nNow, In \u2206ABC,
\nar (\u2206ABD) = ar (\u2206ACD)
\n(because AD is median of \u2206ABC)
\nor, ar (\u2206ABD) = \\(\\frac {1}{4}\\) ar (\u2206ABC) …….(ii)
\nPut Value of ar (\u2206ABD) in equation (i) we have
\nar (\u2206BED) = \\(\\frac{1}{2} \\times \\frac{1}{2}\\) ar (\u2206ABC)
\nor, ar (\u2206BED) = \\(\\frac {1}{4}\\) ar (\u2206ABC)<\/p>\n

\"NCERT<\/p>\n

Question 3.
\nShow that the diagonals of a parallelogram divide it into four triangles of equal area.
\nSolution:
\nGiven: ABCD is a parallelogram in which diagonal AC and BD bisects each other at O.
\n\"NCERT
\nTo prove that: ar (\u2206AOB) = ar (\u2206BOC) = ar (\u2206COD) = ar (\u2206DOA)
\nProof: In \u2206ABC, O is mid point of side AC.
\nTherefore, BO is median of \u2206ABC.
\n\u2234 ar (\u2206AOB) = ar (\u2206BOC) ……(i)
\n(Median divide a A in two equal parts)
\nNow, In \u2206ADC, DO is median
\n\u2234 ar (\u2206AOD) = ar (\u2206COD) …….(ii)
\n(Median divide a \u2206 in two equal parts)
\nAgain in \u2206ADB, AO is a median.
\n\u2234 ar (\u2206AOD) = ar (\u2206AOB) ………(iii)
\nFrom equation (i), (ii) and (iii)
\nar (\u2206AOB) = ar (\u2206BOC) = ar (\u2206COD) = ar (\u2206DOA)
\nTherefore, the diagonals of a parallelogram divide it into four triangles of equal area.<\/p>\n

\"NCERT<\/p>\n

Question 4.
\nIn fig. 9.24, 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).
\n\"NCERT
\nSolution:
\nIn \u2206ACD, O is mid point of CD.
\n\u2234 AO is median
\nWe know that a median of a triangle divides it into triangles of equal areas.
\n\u2234 ar (\u2206AOQ) = ar (\u2206AOD) …..(i)
\nSimilarly, In \u2206BCD, BO is median
\n\u2234 ar (\u2206BOC) = ar (\u2206BOD)
\nAdding equation (i) and (ii)
\nar (\u2206AOC) + ar (\u2206BOC) = ar (\u2206AOD) + ar (\u2206BOD)
\nSo, ar (\u2206ABC) = ar (\u2206ABD)<\/p>\n

\"NCERT<\/p>\n

Question 5.
\nD, E and F are respectively the mid points of the sides BC, CA and AB of a \u2206ABC. Show that
\n(i) BDEF is a parallelogram
\n(ii) ar (DEF) = \\(\\frac {1}{4}\\) ar (ABC)
\n(iii) ar (BDEF) = \\(\\frac {1}{2}\\) ar (ABC)
\nSolution:
\nGiven: ABC is a triangle in which D, E and F are the mid points of side BC, AC and AB respectively.
\nTo prove that:
\n(i) BDEF is a parallelogram
\n(ii) ar(DEF) = \\(\\frac {1}{4}\\) ar (ABC)
\n(iii) ar(BDEF) = \\(\\frac {1}{2}\\) ar (ABC)
\n\"NCERT
\nProof: (i) In \u2206ABC,
\nE and F are the mid point of side AC and AB respectively.
\nWe know that the line joining the mid points of two sides of a triangle is parallel to third side and half of it.
\n\u2234 EF || BC …….(i)
\nand EF = \\(\\frac {1}{2}\\) BC
\nor, EF = BD …(ii) (\u2235 D is mid point of BC)
\nFrom (i) and (ii)
\nBDEF is a parallelogram.<\/p>\n

(ii) We have,
\nBDEF is a parallelogram.
\n\u2234 ar (\u2206BDF) = ar (\u2206DEF) ……(i)
\n(Diagonals of a || gm divide it in two equal parts)
\nSimilarly, CDFE is parallelogram.
\n\u2234 ar (\u2206CDE) = ar (\u2206DEF) …..(ii)
\nand ar (\u2206AEF) = ar (\u2206DEF) ……(iii)
\nFrom equation (i), (ii) and (iii)
\nar (\u2206BDF) = ar (\u2206CDE) = ar (\u2206AEF) = ar (\u2206DEF)
\nNow,
\nar (\u2206BDF) + ar (\u2206CDE) + ar (\u2206AFE) + ar (\u2206DEF) = ar (\u2206ABC)
\nor, ar (\u2206DEF) + ar (\u2206DEF) + ar (\u2206DEF) + ar (\u2206DEF) = ar (\u2206ABC)
\n[\u2235 ar (\u2206BDF) + ar (\u2206CDE) = ar (\u2206AFF)]
\nor, 4 ar (\u2206DEF) = ar (\u2206ABC)
\nor, ar (\u2206DEF) = \\(\\frac {1}{4}\\) ar (\u2206ABC)<\/p>\n

(iii) ar (|| gm BDEF) = 2 ar (\u2206DEF)
\nor, ar (|| gm BDEF) = 2 . \\(\\frac {1}{4}\\) ar (\u2206ABC)
\n[\u2235 ar (\u2206DEF) = \\(\\frac {1}{4}\\) ar (\u2206ABC)]
\nSo, ar (|| gm BEDF) = \\(\\frac {1}{4}\\) ar (\u2206ABC)<\/p>\n

\"NCERT<\/p>\n

Question 6.
\nIn fig. 9.25, diagonals AC and BD of quadrilateral ABCD intersect at O, such that OB = OD. If AB = CD then show that
\n(i) ar (DOC) = ar (AOB)
\n(ii) ar (DCB) = ar (ACB)
\n(iii) DA || CB or ABCD is a parallelogram.
\n\"NCERT
\nSolution:
\nGiven: ABCD is a quadrilateral in which diagonals AC and BD intersect each other at O.
\nTo prove that:
\n(i) ar(\u2206DOC) = ar(\u2206AOB)
\n(ii) ar (\u2206DCB) = ar(\u2206ACB)
\n(iii) DA || CB or ABCD is a parallelogram.
\nConstruction: Draw DN \u22a5 AC and BM \u22a5 AC.
\n\"NCERT
\nProof: (i) In \u2206DON and \u2206BOM
\n\u2220DNO = \u2220BMO (Each 90\u00b0)
\n\u2220DNO = \u2220BOM (Vertically opposite angle)
\nDO = BO (Given)
\nBy A-A-S congruency condition
\n\u2206DON \u2245 \u2206BOM
\nSo, ar (\u2206DON) = ar (\u2206BOM) ……(i)
\n(Congruent \u2206s have equal area)
\nDN = BM (By C.P.C.T)
\nNow, In \u2206DNC and \u2206BMA
\n\u2220DNC = \u2220BMA (Each 90\u00b0)
\nCD = BA (Given)
\nDN = BM (Prove above)
\nBy R.H.S Congruency Condition
\n\u2206DNC = \u2206BMA
\nSo, ar (\u2206DNC) = ar (\u2206BMA) …….(ii)
\n(Congruent \u2206s have equal area)
\nAdding equatiqn (i) and (ii)
\nar (\u2206DON) + ar (\u2206DNC) = ar (\u2206BOM) + ar (\u2206BMA)
\nar (\u2206DOC) = ar (\u2206AQB)<\/p>\n

(ii) We have,
\nar (\u2206DOC) = ar (\u2206AOB) (Prove above)
\n\u2234 ar (\u2206DOC) + ar (\u2206OCB) = ar (\u2206AOB) + ar (\u2206QCB) [add ar (\u2206OCB) both side]
\nor, ar (\u2206DCB) = ar (\u2206ACB)<\/p>\n

(iii) We have
\nar (\u2206DCB) = ar (\u2206ACB)
\nWe know that if two mangles on same base and equal in area, then it must be lie between same parallels.
\nTherefore, DA || CB …..(A)
\n\u2220DCO = \u2220BAO (\u2206DNC \u2245 \u2206BMA)
\nwhich is the pair of alternate interior angles.
\nCD || BA …….(B)
\nFrom (A) and (B),
\nABCD is a parallelogram.<\/p>\n

\"NCERT<\/p>\n

Question 7.
\nD and B are points on sides AB and AC respectively of \u2206ABC such that ar (DBC) = ar (EBC).
\nProve that DE || BC.
\n\"NCERT
\nSolution:
\nSince \u2206s BCE and ABCD are equal in area and have a same base BC.
\nTherefore, altitude from E of \u2206BCE = altitude from D of \u2206BCD.
\nor, \u2206s BCE and BCD are between the same parallel lines.
\nDE || BC.<\/p>\n

Question 8.
\nXY 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)
\nSolution:
\nGiven: In \u2206ABC, XY || BC, BE || AC and CF || AB.
\nwhere E and F lie on XY.
\nTo prove that: ar (\u2206ABE) = ar (\u2206ACF)
\n\"NCERT
\nProof: We have
\nXY || BC
\nor EY || BC ……(i)
\nand BE || AC
\nor BE || CY …….(ii)
\nFrom, equation (i) and (ii)
\nEBCY is a parallelogram.
\nAgain, parallelogram EBCY and \u2206AEB lie on same base EB and between same parallels BE and AC.
\nWe know that 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.
\nSo, ar(\u2206ABE) = \\(\\frac {1}{2}\\) ar(|| gm FBCY) …….(A)
\nNow, XY || BC
\nor, XF || BC ……(iii)
\nand, CF || AB
\nor, CF || BX ……(iv)
\nFrom equation (iii) and (iv)
\nBCFX is a parallelogram.
\nAgain, parallelogram BCFX and \u2206ACF lie on sane base CF and between same parallels CF and AB.
\n\u2234 ar(\u2206ACF) = \\(\\frac {1}{2}\\) ar (|| gm BCFX) ……(B)
\nNow, parallelogram EBCY and parallelogram BCFX lie on same base BC and between same parallels BC and EF.
\nWe know that, parallelograms on the same base and lie between the same parallels having equal area.
\n\u2234 ar(|| gm EBCY) = ar (|| gm BCFX) ……(C)
\nTherefore from equation (A), (B) and (C)
\nwe have ar(\u2206ABE) = ar(\u2206ACF)<\/p>\n

\"NCERT<\/p>\n

Question 9.
\nThe side AB of a parallelogram ABCD is produced to any points. A line through A and parallel to CD meet CB produced at Q arid then parallelogram PBQR is completed (see Fig 9.26). Show that ar(ABCD) = ar(PBQR).
\n\"NCERT
\nSolution:
\nGiven: ABCD is a parallelogram in which P is any point on AB produced and AQ || CP.
\nTo prove that: ar(|| gm ABCD) = ar(|| gm BPRQ)
\nConstruction: Join AC and PQ.
\nProof: Since AC and PQ are diagonals of parallelograms ABCD and BPQR respectively.
\n\"NCERT
\n\u2234 ar(\u2206ABC) = \\(\\frac {1}{2}\\) ar(|| gm ABCD) ……(i)
\nand ar(\u2206PBQ) = \\(\\frac {1}{2}\\) ar(|| gm BPRQ) ……(ii)
\nNow, \u2206s ACQ and AQP are on die same base AQ and between same parallels AQ and CP.
\n\u2234 ar(\u2206ACQ) = ar(\u2206AQP)
\nor, ar(\u2206ACQ) – (\u2206ABQ) = ar(\u2206AQP) – ar(\u2206ABQ)
\n[Subtracting ar(\u2206ABQ) from both sides]
\nor, ar(\u2206ABC) = ar(\u2206BPQ)
\nor, \\(\\frac {1}{2}\\) ar(|| gm ABCD) = \\(\\frac {1}{2}\\) ar (|| gm BPRQ) [From equ. (i) and (ii)]
\nor, ar(|| gm ABCD) = ar (|| gm BPRQ)<\/p>\n

Question 10.
\nDiagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O. prove that ar(AOD) = ar(BOC).
\n\"NCERT
\nSolution:
\nGiven: ABCD is a trapezium, in which AB || DC, and diagonal AC and BD intersect each other at O.
\nTo prove that: ar(\u2206AOD) = ar(\u2206BOC)
\nProof: Since \u2206DAC and \u2206DBC lie on same base BC and between same parallels AB and DC.
\n\u2234 ar(\u2206DAC) = ar(\u2206DBC)
\n[\u2206s on the same base and between same parallels are equal in area]
\nor, ar (\u2206DAC) – ar (\u2206DOC) = ar (\u2206DBC) – ar (\u2206DOC) [Subtract ar (\u2206DOC) both side]
\nar (\u2206AOD) = ar (\u2206BOC)<\/p>\n

\"NCERT<\/p>\n

Question 11.
\nIn fig. 9.27, ABCDE is a pantagon. A line through B parallel to AC meet DC produced at F.
\nShow that
\n(i) ar (ACB) = ar (ACF)
\n(ii) ar (AEDF) = ar (ABCDE)
\n\"NCERT
\nSolution:
\nGiven; ABCDE is a pentagon, and AC || BF.
\nTo prove that:
\n(i) ar (ACB) = ar (ACF)
\n(a) ar (AEDE) = ar (ABCDE)
\nProof:
\n(i) Since \u2206s ACB and ACF lie on the same base AC and between same parallels AC and BF.
\n\u2234 ar(\u2206ACB) = ar(\u2206ACF)
\n[\u2206s on same base and betwen same paralles are equal in area]<\/p>\n

(ii) We have,
\nar (\u2206ACB) = ar(\u2206ACF) [Prove above]
\nar(\u2206ACB) + ar(ACDE) = ar(\u2206ACF) + ar(ACDE) [Add ar (ACDE) both side]
\n\u2234 ar (AEDF) = ar (ABCDE)<\/p>\n

Question 12.
\nA villager Itwaari has a plot oi land of the shape of a quadrilateral. Tht Gram Panchayat of the village dicided b take over some portion of his plot from ore of the comers to construct a Health Centre. Itwaari agrees to the above proposal wih the condition that he should be given eqial amount of land in lieu of his land adjoiniag his plot so as to form a triangular pbt. Explain how this proposal will be implemented.
\n\"NCERT
\nSolution:
\nLet a villager Itwaari has a plot of land of the shape of a quadrilateral ABCD.
\nThe Gram Panchayat decided to take comer portion of the land \u2206ACD. Itwaari agreed. So the remaining part of the land belong to Itwaari is
\nar (ABCD) – ar (\u2206ACD) = ar (\u2206ABC)
\nNow, Itwaari demands the adjoining plot. So, if DE is drawn parallel to AC and join A to E. In this way we add ar (\u2206ACE) in the remaining plot of Itwaari. In this way proposal will be implemented.
\nAs, ar (\u2206ABC) + ar (\u2206ACD) = ar (\u2206ABC) + ar (\u2206ACE)
\nHere ar(\u2206ACD) = ar(\u2206ACE)
\n(Triangles made between two parallels and same base AC)
\nar(ABCD) = ar(\u2206ABE)<\/p>\n

\"NCERT<\/p>\n

Question 13.
\nABCD is a trapezium with AB || DC. A line parallel to AC intersect AB at X and BC at Y. Prove that ar (ADX) = ar (ALY).
\nSolution:
\nGiven: ABCD is a trapezium, in which AB || DC and AC || XY. .
\nTo prove that: ar(\u2206ADX) = ar(\u2206ACY)
\n\"NCERT
\nConstruction: Join CX.
\nProof: Since \u2206s ADX and ACX lie on same base AX and between same parallels AB and DC.
\n\u2234 ar (\u2206ADX) = ar (\u2206ADX) ……(i)
\n[\u2206s on same base and between same parallels are equal in area]
\nNow, \u2206ACY and \u2206ACX lie on same base AC and betwen same parallels AC and XY.
\nar (\u2206ACY) = ar (\u2206ACX) …….(ii)
\nFrom equation (i) and (ii) we get
\nar (\u2206ADX) = ar (\u2206ACY)<\/p>\n

Question 14.
\nIn fig 9.28, AP || BQ || CR. Prove that ar (AQC) = ar (PBR).
\n\"NCERT
\nSolution:
\nGiven: AP || BQ || CR
\nTo prove that: ar (\u2206AQC) = ar (\u2206PBR)
\nProof: Since \u2206s ABQ and PBQ lie on same base and betwen same parallels AP and BQ.
\n\u2234 ar (\u2206ABQ) = ar (\u2206PBQ) ……(i)
\n[\u2206s on same base and between same parallels are equal in area]
\nAgain, \u2206s BQC and BQR lie on same base BQ and between same parallels BQ and CR.
\nar (\u2206BQC) = ar (\u2206BQR) ……(ii)
\nAdding equation (i) and (ii)
\nar(\u2206ABQ) + ar(\u2206BQC) = ar(\u2206PBQ) + ar(\u2206BQR)
\nor, ar (\u2206AQC) = ar (\u2206PBR).<\/p>\n

\"NCERT<\/p>\n

Question 15.
\nDiagonals 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.
\n\"NCERT
\nSolution:
\nGiven: ABCD is a quadrilateral in which diagonal AC ab BD intersect each other at O, and ar (\u2206AOD) = ar (\u2206BOC).
\nTo prove that: ABCD is a trapezium.
\nProof: We have
\nar(\u2206AOD) = ar(\u2206BOC) (Given)
\nor, ar(\u2206AOD) + ar(\u2206DOC) = ar(\u2206BOC) + ar(\u2206DOC) [Add ar (\u2206DOC) both side]
\nor, ar(\u2206ADC) = ar(\u2206BDC)
\nWe know that if two triangles on same base and have equal area, then it must lie in between same parallels.
\nHere, \u2206s ADC and BDC lie on same base DC, and
\nar (\u2206ADC) = ar (\u2206BDC)
\n\u2234 DC must be parallel to AB.
\n\u2234 ABCD is a trapezium.<\/p>\n

\"NCERT<\/p>\n

Question 16.
\nIn fig.9.29, ar (DRC) = ar (DPC) and ar (BDP) = ar (ARC). Show that both the quadrilaterals ABCD and DCPR are trapeziums.
\n\"NCERT
\nSolution:
\nWe have,
\nar(\u2206BDP) = ar(\u2206ARC) (Given)
\nar(\u2206BDP) – ar(\u2206DPC) = ar(\u2206ARC) – ar(\u2206DRC)
\nar (\u2206DRC) = ar (\u2206DBC) (Given)
\nor, ar (\u2206BDC) = ar (\u2206ADC)
\nNow, \u2206s BDC and ADC lie on same base and having equal area therefore they must lie betwen two parallel lines.
\nThen, AB || DC
\nTherefore, ABCD is a trapezium.
\nAgain, ar(\u2206DRC) = ar(\u2206DPC) (Given)
\nHere, \u2206s DRC and DPC lie on same base DC and having equal area.
\n\u2234 DC || RP
\nTherefore, DCPR is a trapezium.<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.3 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Exercise 9.3 Question 1. In fig. 9.23, E is any point on median AD of …<\/p>\n

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NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Exercise 9.3 Question 1. 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