Question
Two cubes each with side bare joined to form a cuboid (see the figure). What is the surface area of this cuboid? Is it $1 2 b ^{ 2 }$ ? Is the surface area of cuboid formed by joining three such cubes, $18 b^2 ?$ Why?
✓
Answer
By joining two cubes end-to-end each with side $b$, we get a cuboid whose length $=b+b=2 b$, breadth $=b$ and height $=b$
$\therefore$ Total surface area of this cuboid
$\begin{array}{l}=2(L B+B H+H L) \\ =2(2 b \times b+b \times b+b \times 2 b) \\ =2\left(2 b^2+b^2+2 b^2\right) \\ =2 \times 5 b^2=10 b^2\end{array}$
So, surface area of cuboid formed by joining two cubes is not equal to $12 b^2$, it is $10 b^2$.
Now, by joining three cubes end-to-end each with side $b$, we get
a cuboid whose length $=b+b+b=3 b$,
breadth $=b$ and height $=b$
$\therefore$ Total surface area of this cuboid
$\begin{array}{l}=2(L B+B H+H L) \\ =2(3 b \times b+b \times b+b \times 3 b) \\ =2 \times\left(3 b^2+b^2+3 b^2\right) \\ =2 \times 7 b^2=14 b^2\end{array}$
So, the surface area of cuboid formed by joining three such cuboids is, it is not equal to $18 b^2$, it is $14 b^2$.
$\therefore$ Total surface area of this cuboid
$\begin{array}{l}=2(L B+B H+H L) \\ =2(2 b \times b+b \times b+b \times 2 b) \\ =2\left(2 b^2+b^2+2 b^2\right) \\ =2 \times 5 b^2=10 b^2\end{array}$
So, surface area of cuboid formed by joining two cubes is not equal to $12 b^2$, it is $10 b^2$.
Now, by joining three cubes end-to-end each with side $b$, we get
a cuboid whose length $=b+b+b=3 b$,
breadth $=b$ and height $=b$
$\therefore$ Total surface area of this cuboid
$\begin{array}{l}=2(L B+B H+H L) \\ =2(3 b \times b+b \times b+b \times 3 b) \\ =2 \times\left(3 b^2+b^2+3 b^2\right) \\ =2 \times 7 b^2=14 b^2\end{array}$
So, the surface area of cuboid formed by joining three such cuboids is, it is not equal to $18 b^2$, it is $14 b^2$.
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Add the following:
(i) $7 \text{a}^2 \text{b} \text{c},-13 \text{a} \text{b} \text{c}^2, 13 \text{a}^2 \text{b} \text{c}, 2 \text{a} \text{b} \text{c}^2$
(ii) $21 \text{a}^2 \text{b} \text{c},-45 \text{a} \text{b} \text{c}^2, 18 \text{a}^2 \text{b} \text{c}, 21 \text{a} \text{b} \text{c}^2$
→(i) $7 \text{a}^2 \text{b} \text{c},-13 \text{a} \text{b} \text{c}^2, 13 \text{a}^2 \text{b} \text{c}, 2 \text{a} \text{b} \text{c}^2$
(ii) $21 \text{a}^2 \text{b} \text{c},-45 \text{a} \text{b} \text{c}^2, 18 \text{a}^2 \text{b} \text{c}, 21 \text{a} \text{b} \text{c}^2$