# Example 37

Last updated at March 11, 2017 by Teachoo

Last updated at March 11, 2017 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 = 𝑥 cm. In ΔAPD & Δ BQC ∠ APD = ∠ BQC AD = BC DP = CQ ∴ ΔAPD ≅ Δ BQC ∴ QB = AP ⇒ QB = 𝑥 cm. Let A be the area of trapezium ABCD A = 12 (Sum of parallel sides) × (Height) A = 12 (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 Δ ADP By Pythagoras theorem DP2 + x2 = 102 DP2 + x2 = 100 DP2 = 100 – 𝑥2 DP = 100 −𝑥2 From (1) A = 12 (DC + AB) DP A = 12 10+2𝑥+10 100−𝑥2 A = 12 2𝑥+20 100−𝑥2 A = 2𝑥 +10 100 − 𝑥22 A = 𝑥+10100−𝑥2 We need to find area of trapezium it is maximum i.e. we need to Maximize Area A = 𝑥+10 100−𝑥2 Diff w.r.t 𝑥 𝑑𝐴𝑑𝑥= 𝑑𝑥 + 10 100 − 𝑥2𝑑𝑥 𝑑𝐴𝑑𝑥 = 𝑑𝑥 − 10𝑑𝑥 . 100−𝑥2 + 𝑑100 − 𝑥2𝑑𝑥 . 𝑥+10 𝑑𝐴𝑑𝑥 = 1+0 100−𝑥2 + 12100 − 𝑥2 . 𝑑100 − 𝑥2𝑑𝑥 . 𝑥+10 𝑑𝐴𝑑𝑥 = 100−𝑥2 + 12100 − 𝑥2 . 0−2𝑥𝑥+10 𝑑𝐴𝑑𝑥 = 100−𝑥2 – 2𝑥𝑥 + 102100 − 𝑥2 𝑑𝐴𝑑𝑥 = 100 − 𝑥22− 𝑥𝑥 + 10100 − 𝑥2 𝑑𝐴𝑑𝑥 = 100 − 𝑥2 − 𝑥𝑥 + 10100 − 𝑥2 𝑑𝐴𝑑𝑥 = −2𝑥2− 10𝑥 + 100100 − 𝑥2 Putting 𝑑𝐴𝑑𝑥=0 −2𝑥2− 10𝑥 + 100100 − 𝑥2=0 −2𝑥2− 10𝑥 + 100=0 −2𝑥2+5𝑥−50=0 𝑥2+5𝑥−50=0 𝑥2+10𝑥−5𝑥−50=0 𝑥𝑥+10−5𝑥+10=0 𝑥−5𝑥+10=0 So, 𝑥=5 & 𝑥=−10 Since 𝑥 represents distance & distance cannot be negative So, 𝑥 = 5 only. Hence value of 𝑥 = 5 Hence, 𝑥 = 5 is point of Maxima ∴ A is Maximum at 𝑥 = 5 Finding maximum area of trapezium A = 𝑥+10 100−𝑥2 = 5+10 100−52 = 15100−25 = 15 75 = 753 Hence, maximum area is 75𝟑 cm2

Ex 6.3,5
Important

Ex 6.3,7 Important

Ex 6.3,12 Important

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

Ex 6.5,18 Important

Ex 6.5,20 Important

Ex 6.5,23 Important

Ex 6.5,26 Important

Ex 6.5,28 Important

Example 46 Important

Example 47 Important

Misc 6 Important

Misc 11 Important

Misc 13 Important

Misc 17 Important

Misc 22 Important

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 7 years. He provides courses for Mathematics and Science from Class 6 to 12. You can learn personally from here https://www.teachoo.com/premium/maths-and-science-classes/.