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) √(100−𝑥^2 ) (B) √(𝑥^2−100) (C) 100−𝑥^2 (D)〖 𝑥〗^2 − 100
In Δ ADP
By Pythagoras theorem
DP2 + AP2 = AD2
DP2 + x2 = 102
DP2 + x2 = 100
DP2 = 100 – 𝑥2
DP = √(𝟏𝟎𝟎 −𝒙𝟐)
So, the correct answer is (A)
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 ") "
Let A be the area of trapezium ABCD
A = 1/2 (Sum of parallel sides) × (Height)
A = 𝟏/𝟐 (DC + AB) × DP
A = 1/2 (10+2𝑥+10) (√(100−𝑥2))
A = 1/2 (2𝑥+20) (√(100−𝑥2))
A = (𝒙+𝟏𝟎) (√(𝟏𝟎𝟎−𝒙𝟐))
So, the correct answer is (B)
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
A = (𝒙+𝟏𝟎) (√(𝟏𝟎𝟎−𝒙𝟐))
Since A has a square root
It will be difficult to differentiate
Let Z = A2
= (𝑥+10)^2 (100−𝑥2)
Where A'(x) = 0, there Z’(x) = 0
So, the correct answer is (A)
Differentiating Z
Z =(𝑥+10)^2 " " (100−𝑥2)
Differentiating w.r.t. x
Z’ = 𝑑((𝑥 + 10)^2 " " (100 − 𝑥2))/𝑑𝑘
Using product rule
As (𝑢𝑣)′ = u’v + v’u
Z’ = [(𝑥 + 10)^2 ]^′ (100 − 𝑥^2 )+(𝑥 + 10)^2 " " (100 − 𝑥^2 )^′
Z’ = 2(𝑥 + 10)(100 − 𝑥^2 )−2𝑥(𝑥 + 10)^2
Z’ = 2(𝑥 + 10)[100 − 𝑥^2−𝑥(𝑥+10)]
Z’ = 2(𝑥 + 10)[100 − 𝑥^2−𝑥^2−10𝑥]
Z’ = 2(𝑥 + 10)[−2𝑥^2−10𝑥+100]
Z’ = −𝟒(𝒙 + 𝟏𝟎)[𝒙^𝟐+𝟓𝒙+𝟓𝟎]
Putting 𝒅𝒁/𝒅𝒙=𝟎
−4(𝑥 + 10)[𝑥^2+5𝑥+50] =0
(𝑥 + 10)[𝑥^2+5𝑥+50] =0
(𝑥 + 10) [𝑥2+10𝑥−5𝑥−50]=0
(𝑥 + 10) [𝑥(𝑥+10)−5(𝑥+10)]=0
(𝒙 + 𝟏𝟎)(𝒙−𝟓)(𝒙+𝟏𝟎)=𝟎
So, 𝑥=𝟓 & 𝒙=−𝟏𝟎
So, the correct answer is (A)
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
We know that
Z’(x) = 0 at x = 5, −10
Since x is length, it cannot be negative
∴ x = 5
Finding sign of Z’’ for x = 5
Now,
Z’ = −4(𝑥 + 10)[𝑥^2+5𝑥+50]
Z’ = −4[𝑥(𝑥^2+5𝑥+50)+10(𝑥^2+5𝑥+50)]
Z’ = −4[𝑥^3+5𝑥^2+50𝑥+10𝑥^2+50𝑥+500]
Z’ = −𝟒[𝒙^𝟑+𝟏𝟓𝒙^𝟐+𝟏𝟎𝟎𝒙+𝟓𝟎𝟎]
Differentiating w.r.t x
Z’’ = 𝑑(−4[𝒙^𝟑 + 𝟏𝟓𝒙^𝟐 + 𝟏𝟎𝟎𝒙 + 𝟓𝟎𝟎])/𝑑𝑘
Z’’ =−4[3𝑥^2+15 × 2𝑥+100]
Z’’ =−4[3𝑥^2+30𝑥+100]
Putting x = 5
Z’’ (5) = −4[3(5^2) +30(5) +100]
= −4 × 375
= −1500
< 0
Hence, 𝑥 = 5 is point of Maxima
∴ Z is Maximum at 𝑥 = 5
That means,
Area A is maximum when x = 5
Finding maximum area of trapezium
A = (𝑥+10) √(100−𝑥2)
= (5+10) √(100−(5)2)
= (15) √(100−25)
= 15 √75
= 75√𝟑 cm2
So, the correct answer is (C)
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, Social Science, Physics, Chemistry, Computer Science at Teachoo.
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