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Misc 2 The two equal sides of an isosceles triangle with fixed base b are decreasing at the rate of 3 cm per second. How fast is the area decreasing when the two equal sides are equal to the base ?Let x be the equal sides of isosceles triangle i.e. AB = AC = 𝒙 And, Base = BC = b Given that equal side of Triangle decreasing at 3 cm per second i.e. 𝑑π‘₯/𝑑𝑑= βˆ’ 3 cm/sec. We need to find how fast is the area decreasing when the two equal sides are equal to the base i.e. 𝒅𝑨/𝒅𝒕 when 𝒙 = b Finding Area Let’s draw perpendicular AD to BC i.e. AD βŠ₯ BC In Isosceles triangle, perpendicular from vertex to the side bisects the side i.e. D is the mid point of BC Thus, we can write BD = DC = 𝒃/𝟐 In βˆ† ADB Using Pythagoras theorem (𝐴𝐡)^2=(𝐴𝐷)^2+(𝐡𝐷)^2 (π‘₯)^2=(𝐴𝐷)^2+ (𝑏/2)^2 π‘₯2 – (𝑏/2)^2=(𝐴𝐷)^2 (𝐴𝐷)^2 = π‘₯2 – (𝑏/2)^2 𝑨𝑫=√(π’™πŸβˆ’(𝒃/𝟐)^𝟐 ) We know that Area of isosceles triangle = 1/2 Γ— Base Γ— Height A = 1/2 Γ— b Γ— √(π‘₯2βˆ’(𝑏/2)^2 ) A = 𝟏/𝟐 Γ— b Γ— √(π’™πŸβˆ’π’ƒ^𝟐/πŸ’) Finding 𝒅𝑨/𝒅𝒕 Differentiating w.r.t. t 𝑑𝐴/𝑑𝑑= 1/2 𝑏 . 𝑑(√(π‘₯^2 βˆ’ 𝑏^2/4))/𝑑𝑑 𝑑𝐴/𝑑𝑑= 1/2 𝑏 . 𝑑(√(π‘₯^2 βˆ’ 𝑏^2/4))/𝑑𝑑 ×𝑑π‘₯/𝑑π‘₯ 𝑑𝐴/𝑑𝑑= 1/2 𝑏 . 𝑑(√(π‘₯^2 βˆ’ 𝑏^2/4))/𝑑π‘₯ ×𝒅𝒙/𝒅𝒕 𝑑𝐴/𝑑𝑑= 1/2 𝑏 . 𝑑(√(π‘₯^2 βˆ’ 𝑏^2/4))/𝑑π‘₯ Γ— πŸ‘ 𝑑𝐴/𝑑𝑑= 1/2 𝑏 [1/(2√(π‘₯2 βˆ’ 𝑏^2/4)) Γ— 𝑑(π‘₯^2 βˆ’ 𝑏^2/4)/𝑑π‘₯]Γ— 3" " 𝑑𝐴/𝑑𝑑= 1/2 𝑏 [1/(2√(π‘₯2 βˆ’ 𝑏^2/4)) Γ—(2π‘₯βˆ’0)]Γ— 3" " 𝒅𝑨/𝒅𝒕= πŸ‘π’ƒπ’™/(𝟐√(π’™πŸ βˆ’ 𝒃^𝟐/πŸ’)) Finding 𝒅𝑨/𝒅𝒕 at 𝒙 = b β”œ 𝑑𝐴/𝑑𝑑─|_(π‘₯ = 𝑏)=(3𝑏^2)/(2√(𝑏^2 βˆ’ 𝑏^2/4)) = (6𝑏^2)/(4√((4𝑏^2 βˆ’ 𝑏^2)/4))= (6𝑏^2)/(4√((3𝑏^2)/4))= (6𝑏^2)/((4√3 𝑏)/2)= (6𝑏^2)/(2√3 𝑏)= 3𝑏/√3 =π‘βˆš3 Since dimension of area is cm2 and time is seconds ∴ 𝑑𝐴/𝑑𝑑 = π’ƒβˆšπŸ‘ cm2/s

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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, Social Science, Physics, Chemistry, Computer Science at Teachoo.