There is a bridge whose length of three sides of a trapezium other than base are equal to 10 cm.

Based on the above information answer the following:

This question is inspired from Example 37 - Chapter 6 Class 12 (AOD) - Maths

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Question 1

What is the value of DP?

(A) √( 100 - x 2 )  

(B) √( x 2 - 100 )

(C) 100 - x 2   

(D) x 2 − 100

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Question 2

What is the area of trapezium A(x)?

(A) (x - 10 )√( 100 - x 2 )  

(B) ( x + 10) √( 100 - x 2 )

(C) ( x - 10 ) (100 - x 2 )

(D) ( x + 10)(100 - x 2 )

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Question 3

If A'(x) = 0, then what are the values of x?

(A) 5,-10 

(B) - 5, 10

(C) - 5,-10 

(D) 5,10

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Question 4

What is the value of maximum Area?

(A) 75 √2  cm 2  

(B) 75 √3  cm 2

(C) 75 √5  cm 2   

(D) 75 √7  cm 2  

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  1. Chapter 6 Class 12 Application of Derivatives (Term 1)
  2. Serial order wise

Transcript

Question There is a bridge whose length of three sides of a trapezium other than base are equal to 10 cm. Based on the above information answer the following: Question 1 What is the value of DP? (A) โˆš(100โˆ’๐‘ฅ^2 ) (B) โˆš(๐‘ฅ^2โˆ’100) (C) 100โˆ’๐‘ฅ^2 (D)ใ€– ๐‘ฅใ€—^2 โˆ’ 100 In ฮ” ADP By Pythagoras theorem DP2 + AP2 = AD2 DP2 + x2 = 102 DP2 + x2 = 100 DP2 = 100 โ€“ ๐‘ฅ2 DP = โˆš(๐Ÿ๐ŸŽ๐ŸŽ โˆ’๐’™๐Ÿ) So, the correct answer is (A) Question 2 What is the area of trapezium A(x)? (A) (๐‘ฅ โˆ’10)โˆš(100โˆ’๐‘ฅ^2 ) (B) (๐‘ฅ+10)โˆš(100โˆ’๐‘ฅ^2 ) (C) (๐‘ฅโˆ’10)(100โˆ’๐‘ฅ^2) (D) (๐‘ฅ+10)(100โˆ’๐‘ฅ^2 ") " Let A be the area of trapezium ABCD A = 1/2 (Sum of parallel sides) ร— (Height) A = ๐Ÿ/๐Ÿ (DC + AB) ร— DP A = 1/2 (10+2๐‘ฅ+10) (โˆš(100โˆ’๐‘ฅ2)) A = 1/2 (2๐‘ฅ+20) (โˆš(100โˆ’๐‘ฅ2)) A = (๐’™+๐Ÿ๐ŸŽ) (โˆš(๐Ÿ๐ŸŽ๐ŸŽโˆ’๐’™๐Ÿ)) So, the correct answer is (B) Question 3 If A'(x) = 0, then what are the values of x? (A) 5,โˆ’10 (B) โˆ’5, 10 (C) โˆ’5,โˆ’10 (D) 5,10 A = (๐’™+๐Ÿ๐ŸŽ) (โˆš(๐Ÿ๐ŸŽ๐ŸŽโˆ’๐’™๐Ÿ)) Since A has a square root It will be difficult to differentiate Let Z = A2 = (๐‘ฅ+10)^2 (100โˆ’๐‘ฅ2) Where A'(x) = 0, there Zโ€™(x) = 0 So, the correct answer is (A) Differentiating Z Z =(๐‘ฅ+10)^2 " " (100โˆ’๐‘ฅ2) Differentiating w.r.t. x Zโ€™ = ๐‘‘((๐‘ฅ + 10)^2 " " (100 โˆ’ ๐‘ฅ2))/๐‘‘๐‘˜ Using product rule As (๐‘ข๐‘ฃ)โ€ฒ = uโ€™v + vโ€™u Zโ€™ = [(๐‘ฅ + 10)^2 ]^โ€ฒ (100 โˆ’ ๐‘ฅ^2 )+(๐‘ฅ + 10)^2 " " (100 โˆ’ ๐‘ฅ^2 )^โ€ฒ Zโ€™ = 2(๐‘ฅ + 10)(100 โˆ’ ๐‘ฅ^2 )โˆ’2๐‘ฅ(๐‘ฅ + 10)^2 Zโ€™ = 2(๐‘ฅ + 10)[100 โˆ’ ๐‘ฅ^2โˆ’๐‘ฅ(๐‘ฅ+10)] Zโ€™ = 2(๐‘ฅ + 10)[100 โˆ’ ๐‘ฅ^2โˆ’๐‘ฅ^2โˆ’10๐‘ฅ] Zโ€™ = 2(๐‘ฅ + 10)[โˆ’2๐‘ฅ^2โˆ’10๐‘ฅ+100] Zโ€™ = โˆ’๐Ÿ’(๐’™ + ๐Ÿ๐ŸŽ)[๐’™^๐Ÿ+๐Ÿ“๐’™+๐Ÿ“๐ŸŽ] Putting ๐’…๐’/๐’…๐’™=๐ŸŽ โˆ’4(๐‘ฅ + 10)[๐‘ฅ^2+5๐‘ฅ+50] =0 (๐‘ฅ + 10)[๐‘ฅ^2+5๐‘ฅ+50] =0 (๐‘ฅ + 10) [๐‘ฅ2+10๐‘ฅโˆ’5๐‘ฅโˆ’50]=0 (๐‘ฅ + 10) [๐‘ฅ(๐‘ฅ+10)โˆ’5(๐‘ฅ+10)]=0 (๐’™ + ๐Ÿ๐ŸŽ)(๐’™โˆ’๐Ÿ“)(๐’™+๐Ÿ๐ŸŽ)=๐ŸŽ So, ๐‘ฅ=๐Ÿ“ & ๐’™=โˆ’๐Ÿ๐ŸŽ So, the correct answer is (A) Question 4 What is the value of maximum Area? (A) 75 โˆš2 cm^2 (B) 75 โˆš3 cm^2 (C) 75 โˆš5 cm^2 (D) 75 โˆš7 cm^2 We know that Zโ€™(x) = 0 at x = 5, โˆ’10 Since x is length, it cannot be negative โˆด x = 5 Finding sign of Zโ€™โ€™ for x = 5 Now, Zโ€™ = โˆ’4(๐‘ฅ + 10)[๐‘ฅ^2+5๐‘ฅ+50] Zโ€™ = โˆ’4[๐‘ฅ(๐‘ฅ^2+5๐‘ฅ+50)+10(๐‘ฅ^2+5๐‘ฅ+50)] Zโ€™ = โˆ’4[๐‘ฅ^3+5๐‘ฅ^2+50๐‘ฅ+10๐‘ฅ^2+50๐‘ฅ+500] Zโ€™ = โˆ’๐Ÿ’[๐’™^๐Ÿ‘+๐Ÿ๐Ÿ“๐’™^๐Ÿ+๐Ÿ๐ŸŽ๐ŸŽ๐’™+๐Ÿ“๐ŸŽ๐ŸŽ] Differentiating w.r.t x Zโ€™โ€™ = ๐‘‘(โˆ’4[๐’™^๐Ÿ‘ + ๐Ÿ๐Ÿ“๐’™^๐Ÿ + ๐Ÿ๐ŸŽ๐ŸŽ๐’™ + ๐Ÿ“๐ŸŽ๐ŸŽ])/๐‘‘๐‘˜ Zโ€™โ€™ =โˆ’4[3๐‘ฅ^2+15 ร— 2๐‘ฅ+100] Zโ€™โ€™ =โˆ’4[3๐‘ฅ^2+30๐‘ฅ+100] Putting x = 5 Zโ€™โ€™ (5) = โˆ’4[3(5^2) +30(5) +100] = โˆ’4 ร— 375 = โˆ’1500 < 0 Hence, ๐‘ฅ = 5 is point of Maxima โˆด Z is Maximum at ๐‘ฅ = 5 That means, Area A is maximum when x = 5 Finding maximum area of trapezium A = (๐‘ฅ+10) โˆš(100โˆ’๐‘ฅ2) = (5+10) โˆš(100โˆ’(5)2) = (15) โˆš(100โˆ’25) = 15 โˆš75 = 75โˆš๐Ÿ‘ cm2 So, the correct answer is (C)

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.