# Example 37

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Example 37 If length of three sides of a trapezium other than base are equal to 10cm, then find the area of the trapezium when it is maximum. Let ABCD be given trapezium & Given length of three side other than base is 10 AD = DC = CB = 10cm. Draw a perpendicular DP & CQ ON AB. Let AP = cm. In APD & BQC APD = BQC AD = BC DP = CQ APD BQC QB = AP QB = cm. Let A be the area of trapezium ABCD A = 1 2 (Sum of parallel sides) (Height) A = 1 2 (DC + AB) DP Now, Since DP & CQ is perpendicular to AB, DPCQ forms a rectangle. PQ = DC = 10 cm Thus, AB = AP + PQ + QB = x + 10 + x = 2x + 10 & DC = 10 cm Finding DP In ADP By Pythagoras theorem DP2 + x2 = 102 DP2 + x2 = 100 DP2 = 100 2 DP = 100 2 From (1) A = 1 2 (DC + AB) DP A = 1 2 10+2 +10 100 2 A = 1 2 2 +20 100 2 A = 2 +10 100 2 2 A = +10 100 2 We need to find area of trapezium it is maximum i.e. we need to Maximize Area A = +10 100 2 Diff w.r.t = + 10 100 2 = 10 . 100 2 + 100 2 . +10 = 1+0 100 2 + 1 2 100 2 . 100 2 . +10 = 100 2 + 1 2 100 2 . 0 2 +10 = 100 2 2 + 10 2 100 2 = 100 2 2 + 10 100 2 = 100 2 + 10 100 2 = 2 2 10 + 100 100 2 Putting =0 2 2 10 + 100 100 2 =0 2 2 10 + 100=0 2 2 +5 50 =0 2+5 50=0 2+10 5 50=0 +10 5 +10 =0 5 +10 =0 So, =5 & = 10 Since represents distance & distance cannot be negative So, = 5 only. Hence value of = 5 Hence, = 5 is point of Maxima A is Maximum at = 5 Finding maximum area of trapezium A = +10 100 2 = 5+10 100 5 2 = 15 100 25 = 15 75 = 75 3 Hence, maximum area is 75 cm2

Chapter 6 Class 12 Application of Derivatives

Ex 6.3,5
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Ex 6.3,7 Important

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Ex 6.3,15 Important

Ex 6.3,26 Important

Example 35 Important

Example 37 Important You are here

Example 38 Important

Example 40 Important

Ex 6.5,1 Important

Ex 6.5,5 Important

Ex 6.5,7 Important

Ex 6.5,11 Important

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Ex 6.5,28 Important

Example 46 Important

Example 47 Important

Misc 6 Important

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Class 12

Important Question for exams Class 12

- Chapter 1 Class 12 Relation and Functions
- Chapter 2 Class 12 Inverse Trigonometric Functions
- Chapter 3 Class 12 Matrices
- Chapter 4 Class 12 Determinants
- Chapter 5 Class 12 Continuity and Differentiability
- Chapter 6 Class 12 Application of Derivatives
- Chapter 7 Class 12 Integrals
- Chapter 8 Class 12 Application of Integrals
- Chapter 9 Class 12 Differential Equations
- Chapter 10 Class 12 Vector Algebra
- Chapter 11 Class 12 Three Dimensional Geometry
- Chapter 12 Class 12 Linear Programming
- Chapter 13 Class 12 Probability

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.