{"id":47015,"date":"2023-12-22T11:09:37","date_gmt":"2023-12-22T05:39:37","guid":{"rendered":"https:\/\/www.aplustopper.com\/?p=47015"},"modified":"2023-12-22T17:06:52","modified_gmt":"2023-12-22T11:36:52","slug":"ml-aggarwal-class-9-solutions-for-icse-maths-chapter-14-chapter-test","status":"publish","type":"post","link":"https:\/\/www.aplustopper.com\/ml-aggarwal-class-9-solutions-for-icse-maths-chapter-14-chapter-test\/","title":{"rendered":"ML Aggarwal Class 9 Solutions for ICSE Maths Chapter 14 Theorems on Area Chapter Test"},"content":{"rendered":"
Question 1. (ii) Area of \u2206EFG = \\(\\frac{1}{2}\\) (area of || gm DEFC) (b) Given. PQRS is a || gm formed by drawing lines parallel to the diagonals of quadrilateral ABCD through its comers. Question P.Q. Question 2. (ii) ar (\u2206 ADQ) = \\(\\frac{1}{2}\\)ar (|| gm ABCD) ( Proved in part (i) above ) Question 3. Question P.Q. Question 4. Question 5. Question 6. Question 7. Question 8. ML Aggarwal Class 9 Solutions for ICSE Maths Chapter 14 Theorems on Area Chapter Test Question 1. (a) In the figure (1) given below, ABCD is a rectangle (not drawn to scale ) with side AB = 4 cm and AD = 6 cm. Find : (i) the area of parallelogram DEFC (ii)area of \u2206EFG […]<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_genesis_hide_title":false,"_genesis_hide_breadcrumbs":false,"_genesis_hide_singular_image":false,"_genesis_hide_footer_widgets":false,"_genesis_custom_body_class":"","_genesis_custom_post_class":"","_genesis_layout":"","footnotes":""},"categories":[3034],"tags":[],"yoast_head":"\n
\n(a) In the figure (1) given below, ABCD is a rectangle (not drawn to scale ) with side AB = 4 cm and AD = 6 cm. Find :
\n(i) the area of parallelogram DEFC
\n(ii)area of \u2206EFG
\n(b) In the figure (2) given below, PQRS is a parallelogram formed by drawing lines parallel to the diagonals of a quadrilateral ABCD through its corners. Prove that area of || gm PQRS = 2 \u00d7 area of quad. ABCD.
\n
\nAnswer:
\n(a)Given. ABCD is a rectangle AB = 4 cm and AD = 6 cm.
\nD C cm and AD = 6 cm.
\nRequired. (i) The area of || gm DEFC.
\n(ii) area of \u2206 EFG
\nAnswer:
\n(i) Since AB = 4cm and AD = 6 cm (given)
\n
\n\u2234 Area of rectangle ABCD = AB \u00d7 AD
\n= 4 cm \u00d7 6 cm = 24 cm2<\/sup>
\nNow, area of rectangle ABCD = area of || gm DEFC (\u2235 Both are on the same Base and between the same parallel lines)
\n\u21d2 Area of || gm DEFC = 24 cm2<\/sup><\/p>\n
\n(\u2234 Both are on the same base and between the same parallel lines)
\n\u2234 Area of AEFG = \\(\\frac{1}{2}\\)7 \u00d7 24 cm2<\/sup> =12 cm2<\/sup><\/p>\n
\nTo prove. Area of II gm PQRS = 2. area of quad. ABCD
\n
\nProof. ar(\u2206ACD) = \\(\\frac{1}{2}\\)ar(|| gm ACRS)
\n[\u2234 both are on same base AC and between the same || AC and SR ]
\n\u21d2 ar (|| gmACRS) = 2ar(\u2206ACD) …….(1)
\nSimilarly,
\nar ( \u2206 ABC) = \\(\\frac{1}{2}\\)ar (|| gm \u2206APQC)
\n\u21d2 ar (|| gm APQC) = 2ar ( \u2206 ABC) …..(2)
\nAdding (1) from (2),
\nar (|| gm ACRS) + ar (|| gm APQC) = 2ar(\u2206ACD) + 2ar (\u2206ABC)
\n\u21d2 (|| gm PQRS) = 2[ar (\u2206 ACD) + ar ( \u2206 ABC)]
\n\u21d2 ar (|| gm PQRS) = 2ar (quad. ABCD)
\nHence, area of || gm PQRS = 2.area of quad. ABCD. (Q.E.D.)<\/p>\n
\nIn the adjoining figure, ABCD and ABEF are parallelogram and P is any point on DC. If area of || gm ABCD = 90 cm2<\/sup>, find:
\n(i) area of || gm ABEF
\n(ii) area of \u2206ABP.
\n(iii) area of \u2206BEF.
\n
\nAnswer:
\nIn the given figure,
\nABCD and ABEF are parallelogram P is an point on DC
\nArea of ||gm ABCD = 90 cm2<\/sup>
\n||gm ABCD and ABEF are on the same base AB are between the same parallels
\n(i) \u2234 Area of ||gm ABEF = area of ||gm ABCD = 90 cm2<\/sup>
\n(ii) \u2235 \u2206ABP and ||gm ABCD are on the same base AB and between the same parallels
\n\u2234 Area \u2206ABP = \\(\\frac{1}{2}\\) area ||gm ABCD
\n= \\(\\frac{1}{2}\\) \u00d7 90 cm2<\/sup> = 45 cm2<\/sup>
\n(iii) \u2235\u2206BEF and ||gm ABEF are on the same base EF and between the same parallels
\n\u2235 Area \u2206BEF = \\(\\frac{1}{2}\\) area ||gm ABEF
\n= \\(\\frac{1}{2}\\) \u00d7 90 = 45 cm2<\/sup><\/p>\n
\nIn the parallelogram ABCD, P is a point on the side AB and Q is a point on the side BC. Prove that
\n(i) area of \u2206CPD = area of \u2206AQD
\n(ii)area of \u2206ADQ = area of \u2206APD + area of \u2206CPB.
\n
\nAnswer:
\nGiven. || gm ABCD in which P is a point on AB and Q is a point on BC.
\nTo prove. (i) ar (\u2206 CPD) = ar ( \u2206 AQD)
\n(ii) ar (\u2206 ADQ) = ar ( \u2206 APD) + ar (\u2206 CPB)
\nProof. \u2206 CPD and II gm ABCD are on the same base CD and between the same parallels lines AB and CD.
\n\u2234 ar (\u2206 CPD) = \\(\\frac{1}{2}\\) ar (|| gm ABCD) …..(1)
\n\u2206 ADQ and || gm ABCD are on the same base AD and between the same II lines AD and BC,
\nar ( \u2206 ADQ) = \\(\\frac{1}{2}\\) (|| gm ABCD) ………(2)
\nFrom (1) and (2),
\nar ( \u2206 CPD) = ar( \u2206 ADQ)
\nor ar ( \u2206 CPD) = ar (\u2206 ADQ) (Q.E.D.)<\/p>\n
\n\u21d2 2ar (\u2206 ADQ) = ar (|| gm ABCD)
\n\u21d2 ar (\u2206 ADQ) + ar( \u2206 ADQ) = ar (|| gm ABCD) ……..(3)
\nBut ar ( \u2206 ADQ) = ar (\u2206 CPD) …………(4)
\n(Proved in part (i) above)
\nFrom (3) and (4),
\nar ( \u2206 ADQ) + ar ( \u2206 CPD) = ar (|| gm ABCD)
\n\u21d2 ar ( \u2206 ADQ) + ar ( \u2206 CPD)
\n\u21d2 ar (\u2206 APD) + ar (\u2206 CPD) + ar (\u2206 CPB)
\n\u21d2 or (\u2206 ADQ) = ar( \u2206 APD) + ar (\u2206 CPB) (Q.E.D.)<\/p>\n
\nIn the adjoining figure, X and Y are points on the side LN of triangle LMN. Through X, a line is drawn parallel to LM to meet MN at Z. Prove that area of \u2206LZY = area of quad. MZYX.
\n
\nAnswer:
\nGiven : In the figure,
\nX and Y are points on side LN of ALMN. Through X, a line XZ || LM is drawn which meets MN at Z.
\nTo prove : area of \u2206LZY = area of quad. MZYX
\nConstruction : Join MX, ZY and LZ
\nProof: \u2206 LM || XZ
\nand \u2206LZX and \u2206MZX are on the same base XZ and between the same parallels
\n\u2234 area \u2206LZX = area \u2206MZX
\nAdding area \u2206XZY to both sides area \u2206LZX + area \u2206XZY
\n= area \u2206MZX + area \u2206XZY
\n\u21d2 area \u2206LZY = area quadrilateral MZYX<\/p>\n
\nIf D is a point on the base BC of a triangle ABC such that 2BD = DC, prove that area of \u2206ABD= \\(\\frac{1}{2}\\) area of \u2206 ABC.
\nAnswer:
\nGiven. \u2206 ABC in which base BC. D is a point on BC such that 2BD = DC.
\nTo prove, ar ( \u2206 ABD) = \\(\\frac{1}{3}\\) ar (ABC)
\n
\nConstruction. Let P is the mid-point of DC join AD = DC
\n\u21d2 BD = \\(\\frac{1}{2}\\)DC 2<\/sup>
\ni.e. BD = DP (P is mid-point of DC)
\n\u2234 D is mid-point of BP.
\nIn \u2206 ABP, AD is median of BP (D is mid-point of BP)
\n\u2234 ar (\u2206ABD) = ar ( \u2206 ADP) …..(1)
\nAgain in \u2206 ADC, AP is the median of DC. (P is mid-point of DC)
\nar (\u2206 ADP) = ar (\u2206 APC) ……………(2)
\nFrom (1) and (2),
\n\u2234 ar (\u2206 ABD) = ar (\u2206 ADP) = ar ( \u2206 APC)
\n\u2234 \u2206 ABC is divided into three equal triangles and each A will be of \\(\\frac{1}{3}\\) \u2206 ABC.
\n\u2234 ar (\u2206 ABD) = \\(\\frac{1}{3}\\)ar (\u2206 ABC) (Q.E.D.)<\/p>\n
\nPerpendiculars are drawn from a point within an equilateral triangle to the three sides. Prove that the sum of the three perpendiculars is equal to the altitude of the triangle.
\nAnswer:
\nABC is an equilateral triangle, i.e. AB = BC = CA. P is any point within an equilateral triangle to the three sides.
\n
\nPN, PM, and PL are perpendicular on side AB, AC and BC respectively. AD is any altitude from point A on side BC.
\nTo prove. AD = NP + LP + MP
\nConstruction. Join PA, PB and PC.
\nProof. Area of \u2206 ABC = \\(\\frac{1}{2}\\) \u00d7 Base \u00d7 Altitude
\nor(\u2206ABC)= \\(\\frac{1}{2}\\) \u00d7 BC \u00d7 AD ………(1)
\nNow, area of \u2206 APB = \\(\\frac{1}{2}\\) \u00d7 AB \u00d7 NP ………….(2)
\narea of \u2206 APC = \\(\\frac{1}{2}\\) \u00d7 AC \u00d7 MP
\narea of \u2206 BPC = \\(\\frac{1}{2}\\) \u00d7 BC \u00d7 LP …….(4)
\nAdding (2), (3) and (4)
\nar ( \u2206 APB) + ar (\u2206 APC) + ar (\u2206 BPC)
\n= \\(\\frac{1}{2}\\) \u00d7 AB \u00d7 NP + \\(\\frac{1}{2}\\) \u00d7 AC \u00d7 MP + \\(\\frac{1}{2}\\) \u00d7 BC \u00d7 LP
\nar(\u2206 ABC) = \\(\\frac{1}{2}\\) [AB \u00d7 NP + AC \u00d7 MP \u00d7 BC \u00d7 LP]
\n= \\(\\frac{1}{2}\\) [BC \u00d7 NP + BC MP \u00d7 BC \u00d7 LP] (\u2235 Ab = AC = CA)
\nar(\u2206ABC) = \\(\\frac{1}{2}\\) \u00d7 BC[NP + MP + LP ] ..(5)
\nFrom (4) and (5),
\n\\(\\frac{1}{2}\\) \u00d7 BC \u00d7 AD = \\(\\frac{1}{2}\\) \u00d7 BC \u00d7 (NP + LP + MP)
\n\u21d2 AD = NP + LP + MP
\n\u21d2 NP + LP + MP = AD
\ni.e.sum of three perpendiculars is equal to the altitude of the triangle.<\/p>\n
\nIf each diagonal of a quadrilateral’ divides it into two triangles of equal areas, then prove that the quadrilateral is a parallelogram.
\nAnswer:
\nGiven : In quadrilateral ABCD, diagonal AC bisects the quadrilateral ABCD in two triangle of equal area i.e.
\n
\nar (\u2206 ABC) = ar (\u2206 ADC)
\nTo prove : ABCD is a parallelogram.
\nProof : Join BD.
\nProof: \u2235 Diagonals of quad. ABCD divides the quad, into two triangles of equal area.
\n\u2234 ar( \u2206 ABC) = ar( \u2206 ABD)
\n= \\(\\frac{1}{2}\\) ar (ABCD)
\nBut, these are on the same base AB
\n\u2234 Their heights are equal
\n\u2234 DC || AB …(i)
\nSimilarly, we can prove that :
\nar (\u2206ABC) = ar (\u2206BDC)
\n\u2234 BC || AD …(ii)
\nFrom (i) and (ii)
\nABCD is a parallelogram.
\nHence proved.<\/p>\n
\nIn the given figure, ABCD is a parallelogram in which BC is produced to E such that CE = BC. AE intersects CD at F. If area of \u2206DFB = 3 cm2<\/sup>, find the area of parallelogram ABCD.
\n
\nSolution:
\nIn the figure, ABCD is a parallelogram BC is produced to E such that CE = BC
\n
\nJoin BD and AE
\nwhich intersects DC at F
\nJoin BF, AC and DE
\n\u2234 Area of \u2206DFB = 3 cm2<\/sup>
\nFind the area of ||gm ABCD
\nSolution:
\n\u2235 In \u2206ABE, C is mid-point of BE and CD || AB
\n\u2234 F is mid-point of AE and CD
\n\u2234 ABED is a ||gm (\u2235 Diagonals AE and CD bisect each other at F)
\n\u2235 BD is the diagonal of ||gm ABCD
\n\u2206BCD = \\(\\frac{1}{2}\\) ||gm ABCD
\n\u2235 F is mid-point of DC
\n\u2234 ADFB = \\(\\frac{1}{2}\\) \u2206BCD
\n\u21d2 ADFB = \\(\\frac{1}{2}\\) \u00d7 \\(\\frac{1}{2}\\) (||gm ABCD)
\n\u21d2 ADFB = \\(\\frac{1}{2}\\) (||gm ABCD)
\n\u2234 area ||gm ABCD = 4 area \u2206DFB
\n= 4 \u00d7 3 = 12 cm2<\/sup><\/p>\n
\nIn the given figure, ABCD is a square. E and F are mid-points of sides BC and CD respectively. If R is mid-point of EF, prove that: area of \u2206AER = area of \u2206AFR.
\n
\nSolution:
\nGiven : In square ABCD, BD is diagonals E and F are mid-point of BC and CD respectively. R is mid-point of EF.
\nTo prove : area (\u2206AER = area (\u2206AFR)
\nProof: In \u2206ABE and \u2206ADF
\nAB = AD (Sides of a square)
\n\u2220B = \u2220D (Each 90\u00b0)
\nBE = CE (E is mid-point of BC)
\n\u2234 \u2206ABE \u2245 \u2206ADF (SAS axiom)
\n\u2234 AE = AF (c.p.c.t.)
\nAgain in \u2206AER and \u2206AFR
\nAE = AF (Produced)
\nAR = AR (Common)
\nER = FR (R is mid-point of EF)
\n\u2234 \u2206AER = \u2206AFR (SSS axiom)
\n\u2234 area(\u2206AER) = area (\u2206AFR)<\/p>\n
\nIn the given figure, X and Y are mid-points of the sides AC and AB respectively of \u2206ABC. QP || BC and CYQ and BXP are straight lines. Prove that area of \u2206ABP = area of \u2206ACQ.
\n
\nSolution:
\nGiven : In the given figure,
\nX and Y are the mid-points of the sides AC and AB respectively of \u2206ABC
\nQP || BC
\nCYQ and BXP are straight lines
\nTo prove : area(\u2206ABP) = area(\u2206ACQ)
\nProof: v X and Y are the mid-points of sides AC and AB respectively
\n\u2234 YX || BC
\nBut QP || BC
\n\u2234 QP || BC || YX
\nIn \u2206BAP, Y is mid of AB and YX || QP
\n\u2234 X is mid-point of BP
\n\u2234YX = \\(\\frac{1}{2}\\)AP …(i)
\nSimilarly we can prove in \u2206AQC
\nYX = \\(\\frac{1}{2}\\) QA …(ii)
\nFrom (i) and (ii),
\nQA = AP
\nNow \u2206ABP and \u2206ACQ are on the equal base and between the same parallel lines
\narea(\u2206ABP) = area(\u2206ACQ)<\/p>\nML Aggarwal Class 9 Solutions for ICSE Maths<\/a><\/h4>\n","protected":false},"excerpt":{"rendered":"