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Olympiad workbook

Exam: Olympiad workbook
Subject: IOQM
Chapter: Three Dimensional Geometry

Olympiad workbook · IOQM · Three Dimensional Geometry

Determine the sum of all possible surface areas of a cube two of...

Determine the sum of all possible surface areas of a cube two of whose vertices are

(1,2,0)

and

(3,3,2)

.

Determine the sum of all possible surface areas of a cube two of whose vertices are (1,2,0) and (3,3,2) .

Determine the sum of all possible surface areas of a cube two of whose vertices are (1,2,0) and (3,3,2) .

Determine the sum of all possible surface areas of a cube two of whose vertices are (1,2,0) and (3,3,2) .

Correct answer

99

Step-by-step solution

Case-I

: <img src="https://cdn-question-pool.Quantrex.app/NEET/Olympiad/rjKoXRRoQ3hJSc1-UBfAF5l0e_XDAIk86yjGNLr517A.original.fullsize.png"><br>

\begin{aligned} \text { SA } & =6 \times[\sqrt{4+1+4}]^2 \\ & =54 \text { sq.unit }{ }^2 \\ \text { Case-II: } & \end{aligned}

<img src="https://cdn-question-pool.Quantrex.app/NEET/Olympiad/dBSOHpHepqoxSDmyIaS1ap0V92XgdxkoOaZud8bIs8k.original.fullsize.png"><br>

2 a^2+9 a=\frac{3}{\sqrt{2}} \quad S A=6

\left(\frac{9}{2}\right)=27

sq unit Case-III

:<img src="https://cdn-question-pool.Quantrex.app/NEET/Olympiad/5rMXaUWzbmwZ4MW9wC03sx-b2H42SpVFCZYFaeLjEpU.original.fullsize.png"><br>

\begin{aligned} & \sqrt{3} a=3 \\ & \Rightarrow a=\sqrt{3} \\ & S A=6 a^2=6 \times 3=18 \text { sq.unit }{ }^2 \\ & \therefore \text { Possible } S A=18+27+54=99\end{aligned}

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