Question
Two cuboids with equal volumes will always have equal surface areas.
✓
Answer
We discard the statement by a counter example.
Let the dimensions of two cuboids be $1\ cm \times 1\ cm \times 2\ cm$ and $ 1\ cm \times \frac{1}{2}\ cm x 4\ cm,$
respectively,
Then, volume of first cuboid $= l \times b \times h = 1 \times 1 \times 2 = 2cm^3$
and volume of second cuboid $= l \times b \times h = 1 \times \frac{1}{2} \times 4 = 2cm^3$
Now, the surface area of first cuboid $= 2(lb + bh + hl)$
$= 2(1 \times 1 + 1 \times 2 + 2 \times 1)$
$= 2(1 + 2 + 2) = 10cm^2$
and surface area of the second cuboid $= 2(lb + bh + hl)$
$=2\Big(1\times\frac{1}{2}+\frac{1}{2}\times4+1\times4\Big)=2\Big(\frac{1}{2}+\frac{4}{2}+4\Big)$
$=2\Big(\frac{5}{2}+4\Big)=2\Big(\frac{13}{2}\Big)=13\text{cm}^2$
which are not equal.
So, the statement is false.
Let the dimensions of two cuboids be $1\ cm \times 1\ cm \times 2\ cm$ and $ 1\ cm \times \frac{1}{2}\ cm x 4\ cm,$
respectively,
Then, volume of first cuboid $= l \times b \times h = 1 \times 1 \times 2 = 2cm^3$
and volume of second cuboid $= l \times b \times h = 1 \times \frac{1}{2} \times 4 = 2cm^3$
Now, the surface area of first cuboid $= 2(lb + bh + hl)$
$= 2(1 \times 1 + 1 \times 2 + 2 \times 1)$
$= 2(1 + 2 + 2) = 10cm^2$
and surface area of the second cuboid $= 2(lb + bh + hl)$
$=2\Big(1\times\frac{1}{2}+\frac{1}{2}\times4+1\times4\Big)=2\Big(\frac{1}{2}+\frac{4}{2}+4\Big)$
$=2\Big(\frac{5}{2}+4\Big)=2\Big(\frac{13}{2}\Big)=13\text{cm}^2$
which are not equal.
So, the statement is false.
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