{"id":31856,"date":"2018-10-08T11:58:56","date_gmt":"2018-10-08T11:58:56","guid":{"rendered":"https:\/\/www.aplustopper.com\/?p=31856"},"modified":"2020-11-25T14:58:36","modified_gmt":"2020-11-25T09:28:36","slug":"ml-aggarwal-class-10-solutions-for-icse-maths-chapter-14-chapter-test","status":"publish","type":"post","link":"https:\/\/www.aplustopper.com\/ml-aggarwal-class-10-solutions-for-icse-maths-chapter-14-chapter-test\/","title":{"rendered":"ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 14 Similarity Chapter Test"},"content":{"rendered":"
These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths<\/a>. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 14 Similarity Chapter Test.<\/p>\n ML Aggarwal Solutions<\/a>ICSE Solutions<\/a>Selina ICSE Solutions<\/a><\/p>\n Question 1.<\/strong><\/span> Question 2.<\/strong><\/span> Question 3.<\/strong><\/span> Question 4.<\/strong><\/span> Question 5.<\/strong><\/span> Question 6.<\/strong><\/span> Question 7.<\/strong><\/span> Question 8.<\/strong><\/span> Question 9.<\/strong><\/span> Question 10.<\/strong><\/span> Question 11.<\/strong><\/span> Question 12.<\/strong><\/span> Question 13.<\/strong><\/span> Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 14 Similarity Chapter Test are helpful to complete your math homework.<\/p>\n
\nIn the given figure, \u22201 = \u22202 and \u22203 = \u22204. Show that PT x QR = PR x ST.<\/strong>
\nSolution:<\/strong><\/span>
\nGiven : In the given figure,
\n\u22201 = \u22201 and \u22203 = \u22204
\n
\nTo prove : PT x QR = PR x ST
\nProof: \u22201 = \u22202
\nAdding \u22206 to both sides
\n\u22201 + \u22206 = \u22202 + \u22206
\n\u2220SPT = \u2220QPR
\nIn \u2206PQR and \u2206PST
\n<\/p>\n
\nIn the adjoining figure, AB = AC. If PM \u22a5 AB and PN \u22a5 AC, show that PM x PC = PN x PB.<\/strong>
\n
\nSolution:<\/strong><\/span>
\nGiven : In the given figure,
\nAB = AC, PM \u22a5 AB and PN \u22a5 AC
\nTo prove : PM x PC = PN x PB
\nProof: In \u2206ABC, AB = AC
\n\u2220B = \u2220C
\nNow in \u2206CPN and \u2206BPM,
\n<\/p>\n
\n(a) In the figure (1) given below. \u2220AED = \u2220ABC. Find the values of x and y.<\/strong>
\n(b) In the fig. (2) given below, CD = \\(\\\\ \\frac { 1 }{ 2 } \\) AC, B is mid-point of AC and E is mid-point of DF. If BF || AG, prove that :<\/strong>
\n(i) CE || AG<\/strong>
\n(ii) 3 ED = GD.<\/strong>
\n
\nSolution:<\/strong><\/span>
\n(a) Given : In following figure, \u2220AED = \u2220ABC
\nRequired : The values of x and y.
\nNow, in \u2206ABC and \u2206ADE
\n\u2220AED = \u2220ABC (given)
\n\u2220A = \u2220A (common)
\n\u2234 \u2206ABC ~ \u2206ADE
\n(By A.A. axiom of similarity)
\n
\n
\n<\/p>\n
\nIn the given figure, 2 AD = BD, E is mid-point of BD and F is mid-point of AC and EC || BH. Prove that:<\/strong>
\n(i) DF || BH<\/strong>
\n(ii) AH = 3 AF.<\/strong>
\n
\nSolution:<\/strong><\/span>
\nGiven : E is the mid-point of BD and F is mid-point of AC also 2 AD = BD and EC || BH
\nTo Prove : (i) DF || BH
\n(ii) AH = 3 AF
\n
\n<\/p>\n
\nIn a \u2206ABC, D and E are points on the sides AB and AC respectively such that DE || BC. If AD = 2.4 cm, AE = 3.2 cm, DE = 2 cm and BC = 5 cm, find BD and CE.<\/strong>
\nSolution:<\/strong><\/span>
\nGiven : In \u2206ABC, D and E are the points on the sides AB and AC respectively
\nDE || BC
\nAD = 2.4 cm, AE = 3.2 cm, DE = 2 cm, BC = 5 cm
\n
\n<\/p>\n
\nIn a \u2206ABC, D and E are points on the sides AB and AC respectively such that AD = 5.7cm, BD = 9.5cm, AE = 3.3cm and AC = 8.8cm. Is DE || BC? Justify your answer.<\/strong>
\nSolution:<\/strong><\/span>
\nIn \u2206ABC, D and E are points on the sides AB and AC respectively
\nAD = 5.7 cm, BD = 9.5 cm, AE = 3.3 cm and AC = 8.8 cm
\n
\n<\/p>\n
\nIn a \u2206ABC, DE is parallel to the base BC, with D on AB and E on AC. If \\(\\frac { AD }{ DB } =\\frac { 2 }{ 3 } ,\\frac { BC }{ DE } \\)<\/strong>
\nSolution:<\/strong><\/span>
\nIn \u2206ABC, DE || BC
\nD is on AB and E is on AC
\n
\n<\/p>\n
\nIf the area of two similar triangles are 360 cm\u00b2 and 250 cm\u00b2 and if one side of the first triangle is 8 cm, find the length of the corresponding side of the second triangle.<\/strong>
\nSolution:<\/strong><\/span>
\nLet \u2206ABC and \u2206DEF are similar and area of
\n\u2206ABC = 360 cm\u00b2
\n
\n<\/p>\n
\nIn the adjoining figure, D is a point on BC such that \u2220ABD = \u2220CAD. If AB = 5 cm, AC = 3 cm and AD = 4 cm, find<\/strong>
\n(i) BC<\/strong>
\n(ii) DC<\/strong>
\n(iii) area of \u2206ACD : area of \u2206BCA.<\/strong>
\n
\nSolution:<\/strong><\/span>
\nIn \u2206ABC and \u2206ACD
\n\u2220C = \u2220C (Common)
\n\u2220ABC = \u2220CAD (given)
\n\u2234 \u2206ABC ~ \u2206ACD
\n
\n<\/p>\n
\nIn the adjoining figure the diagonals of a parallelogram intersect at O. OE is drawn parallel to CB to meet AB at E, find area of DAOE : area of ||gm ABCD.<\/strong>
\n
\nSolution:<\/strong><\/span>
\n(a) In the figure
\nDiagonals of parallelogram ABCD are AC and BD which intersect each other at O.
\nOE is drawn parallel to CB to meet AB in E.
\n
\n<\/p>\n
\nIn the given figure, ABCD is a trapezium in which AB || DC. If 2AB = 3DC, find the ratio of the areas of \u2206AOB and \u2206COD.<\/strong>
\n
\nSolution:<\/strong><\/span>
\nIn the given figure, ABCD is trapezium in
\nwhich AB || DC, 2AB = 3DC
\n<\/p>\n
\nIn the adjoining figure, ABCD is a parallelogram. E is mid-point of BC. DE meets the diagonal AC at O and meet AB (produced) at F. Prove that .<\/strong>
\n
\n(i) DO : OE = 2 : 1<\/strong>
\n(ii) area of \u2206OEC : area of \u2206OAD = 1 : 4.<\/strong>
\nSolution:<\/strong><\/span>
\nGiven : In || gm ABCD,
\nE is mid point of BC and DE meets the diagonal AC at O and meet AB produced at F.
\nTo prove : (i) DO : OE = 2 : 1
\n(ii) area of \u2206OEC : area of \u2206OAD = 1 : 4
\nProof: In \u2206AOD and \u2206EDC
\n\u2220AOD = \u2220EOC (vertically opposite angle)
\n\u2220OAD = \u2220OCB (alt. angles)
\n\u2206AOD ~ \u2206EOC (AA postulate)
\n<\/p>\n
\nA model of a ship is made to a scale of 1 : 250. Calculate :<\/strong>
\n(i) the length of the ship, if the length of model is 1.6 m.<\/strong>
\n(ii) the area of the deck of the ship, if the area of the deck of model is 2.4 m\u00b2.<\/strong>
\n(iii) the volume of the model, if the volume of the ship is 1 km\u00b3.<\/strong>
\nSolution:<\/strong><\/span>
\nScale factor (k) of the model of the ship = \\(\\\\ \\frac { 1 }{ 250 } \\)
\n(i) Length of model = 1.6 m
\n<\/p>\n