Question 6 - End-of-Chapter Exercises - Chapter 4 Class 9 - Exploring Algebraic Identities (Ganita Manjari I)

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Question 6 (i) Find possible expressions for the length, breadth, and heights of each of the following cuboids whose volumes are given by the following expressions in cubic units. (i) 6𝑎^2−24𝑏^2 Because Volume = Length × Breadth × Height We factor the expression into 3 brackets to give us Length, Breadth and Height Now, our expression is 6𝑎^2−24𝑏^2 Taking 6 common =6(a^2−4b^2 ) =𝟔(𝐚^𝟐−(𝟐𝐛)^𝟐 ) Using 𝑎^2−𝑏^2=(𝑎+𝑏)(𝑎−𝑏) Where 𝑎 = 𝑎, b = 2𝑏 =𝟔(𝒂+𝟐𝒃)(𝒂−𝟐𝒃) Now, we can consider any bracket as length Thus, Length = 𝟔 Breadth = 𝒂+𝟐𝒃 Height = 𝒂−𝟐𝒃 =𝟔(𝒂+𝟐𝒃)(𝒂−𝟐𝒃) Now, we can consider any bracket as length Thus, Length = 𝟔 Breadth = 𝒂+𝟐𝒃 Height = 𝒂−𝟐𝒃 Using 𝑎^2−𝑏^2=(𝑎+𝑏)(𝑎−𝑏) Where 𝑎 = 𝑎, b = 2𝑏 Question 6 (ii) Find possible expressions for the length, breadth, and heights of each of the following cuboids whose volumes are given by the following expressions in cubic units. (ii) 3𝑝𝑠^2−15𝑝𝑠+12𝑝 Because Volume = Length × Breadth × Height We factor the expression into 3 brackets to give us Length, Breadth and Height Now, our expression is 3𝑝𝑠^2−15𝑝𝑠+12𝑝 Taking 3 common = 3(𝑝𝑠^2−5𝑝𝑠+4𝑝) We can also take p common = 3 × 𝑝(𝑠^2−5𝑠+4) = 𝟑𝒑(𝒔^𝟐−𝟓𝒔+𝟒) Factorising the bracket by splitting the middle term = 3𝑝(𝑠^2−𝟒𝒔−𝒔+4) = 3𝑝(𝑠(𝑠−4)−1(𝑠−4)) = 𝟑𝒑(𝒔−𝟏)(𝒔−𝟒) Now, we can consider any bracket as length Thus, Length = 𝟑𝒑 Breadth = 𝒔−𝟏 Height = 𝒔−𝟒 Splitting the middle term method We need to find two numbers whose Sum = –5 Product = 1 × 4 = 4 Since sum is negative but product is positive. Thus, both numbers are negative