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Example 25 If length of three sides of a trapezium other than base are equal to 10 cm, then find the area of the trapezium when it is maximum.Let ABCD be given trapezium Given length of sides other than base is 10 āˆ“ AD = DC = CB = 10cm. Draw a perpendicular DP & CQ on AB. Let AP = š’™ cm By symmetry QB = š’™ cm Let A be the area of trapezium ABCD A = 1/2 (Sum of parallel sides) Ɨ (Height) A = šŸ/šŸ (DC + AB) Ɨ DP Now, Since DP & CQ is perpendicular to AB, And CD was parallel to AB Thus, DPCQ forms a rectangle. āˆ“ PQ = DC = 10 cm Thus, AB = AP + PQ + QB = x + 10 + x = 2x + 10 Finding DP In Ī” ADP By Pythagoras theorem DP2 + x2 = 102 DP2 + x2 = 100 DP2 = 100 ā€“ š‘„2 DP = āˆš(šŸšŸŽšŸŽ āˆ’š’™šŸ) From (1) A = šŸ/šŸ (DC + AB) DP A = 1/2 (10+2š‘„+10) (āˆš(100āˆ’š‘„2)) A = 1/2 (2š‘„+20) (āˆš(100āˆ’š‘„2)) A = (2(š‘„ +10) (āˆš(100 āˆ’ š‘„2)))/2 A = (š’™+šŸšŸŽ) āˆš(šŸšŸŽšŸŽāˆ’š’™šŸ) We need to find area of trapezium when it is maximum i.e. We need to Maximize Area A = (š‘„+10) (āˆš(100āˆ’š‘„2)) Since A has a square root It will be difficult to differentiate Let Z = A2 = (š‘„+10)^2 (100āˆ’š‘„2) Since A is positive, A is maximum if A2 is maximum So, we maximize Z = A2 Differentiating Z Z =(š‘„+10)^2 " " (100āˆ’š‘„2) Differentiating w.r.t. x Zā€™ = š‘‘((š‘„ + 10)^2 " " (100 āˆ’ š‘„2))/š‘‘š‘˜ Zā€™ = [(š‘„ + 10)^2 ]^ā€² (100 āˆ’ š‘„^2 )+(š‘„ + 10)^2 " " (100 āˆ’ š‘„^2 )^ā€² Zā€™ = 2(š‘„ + 10)(100 āˆ’ š‘„^2 )āˆ’2š‘„(š‘„ + 10)^2 Zā€™ = 2(š‘„ + 10)[100 āˆ’ š‘„^2āˆ’š‘„(š‘„+10)] Zā€™ = 2(š‘„ + 10)[100 āˆ’ š‘„^2āˆ’š‘„^2āˆ’10š‘„] Zā€™ = 2(š‘„ + 10)[āˆ’2š‘„^2āˆ’10š‘„+100] Zā€™ = āˆ’4(š‘„ + 10)[š‘„^2+5š‘„+50] Putting š‘‘š‘/š‘‘š‘„=0 āˆ’4(š‘„ + 10)[š‘„^2+5š‘„+50] =0 (š‘„ + 10)[š‘„^2+5š‘„+50] =0 (š‘„ + 10) [š‘„2+10š‘„āˆ’5š‘„āˆ’50]=0 (š‘„ + 10) [š‘„(š‘„+10)āˆ’5(š‘„+10)]=0 (š‘„ + 10)(š‘„āˆ’5)(š‘„+10)=0 (š‘„ + 10)(š‘„āˆ’5)(š‘„+10)=0 So, š‘„=5 & š‘„=āˆ’10 Since š‘„ represents distance & distance cannot be negative So, š’™ = 5 only By First Derivative Test Hence, š‘„ = 5 is point of Maxima āˆ“ Z is Maximum at š‘„ = 5 That means, Area A is maximum when x = 5 Finding maximum area of trapezium A = (š‘„+10) āˆš(100āˆ’š‘„2) = (5+10) āˆš(100āˆ’(5)2) = (15) āˆš(100āˆ’25) = 15 āˆš75 = 75āˆš3 Hence, maximum area is 75āˆššŸ‘ cm2

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo