Volume

Asad Abbas

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Joined
Mar 30, 2011
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10
Q1.The area of a wooden box is 24cm2,and its volume is 96cm3.What is the height of the box?

Q2.The volume of a cuboid is 64cm3,while the height of the box is 4 cm.What is the area of the box?What is the length of each side of the cuboid

I know it sounds silly when you read it but I haven't been taught how to do the above 2 questions.Please help me.
 
Small wonder you have not been taught such things. They could make more sense.

1-1) Boxes don't have area unless you tell us which surface you are describing.
1-2) We may need to know the shape of the box, but with the hint from 1-1, maybe we don't need that.

Assuming 26 cm^3 = Surface area of the base of the box.
H = height of the box

(26 cm^2) * H = 96 cm^3

2-1) Again, boxes don't hae area unless you tell us what surface you are describing.
2-2) Are you SURE it doesn't say "CUBE"?

RBGTHGANH
 
Hello, Asad Abbas!

I assume there are typos.
As written, the problems do not have unique answers.


\(\displaystyle \text{Q1. The }sur\!f\!ace\text{ area of a wooden box is 24 cm}^2\text{ and its volume is 96 cm}^3.\)
. . . . \(\displaystyle \text{ What is the height of the box?}\)

\(\displaystyle \text{There is }no\text{ box with a surface area of 24 and a volume of 96.}\)

\(\displaystyle \text{I must assume that it says: The area }o\!f\:the\:base\text{ of a wooden box is 24 cm}^2.\)


\(\displaystyle \text{Since }V \:=\:L\!\cdot\!W\!\cdot\!H,\,\text{ and }\,L\!\cdot\!W = 24,\;V \:=\:96\)

. . \(\displaystyle \text{we have: }\;96 \;=\;24\!\cdot\!H \quad\Rightarrow\quad H \:=\:4\text{ cm.}\)



\(\displaystyle \text{Q2. The volume of a cuboid is 64 cm}^3\text{, while the height of the box is 4 cm.}\)
. . . . \(\displaystyle \text{What is the }sur\!f\!ace\text{ area of the box? \; What is the length of each side of the cuboid?}\)

tkhunny is correct . . . It could be a cube.

But then the last question is silly, isn't it?
If the height is 4 cm, then the length and width are also 4 cm ... of course.

The problem does not state that the dimension are integers.
. . Hence, there are several brizillion solutions.


\(\displaystyle \text{If we assume that the dimensions are integers and }L \ge W\)
. . \(\displaystyle \text{there are still }three\text{ solutions.}\)

. . \(\displaystyle \begin{array}{cc} (L,W,H) & \text{S.A.} \\ \hline (16,1,4) & 168 \\ (8,2,4) & 112 \\ (4,4,4) & 96 \end{array}\)

 
Asad Abbas said:
I know it sounds silly when you read it but I haven't been taught how to do the above 2 questions.Please help me.
WHY haven't you been taught? Lazy teacher? YOU skipped classes?
 
Denis

Based on previous posts, Asad Abbas is trying to get into a school or class, and he is preparing on his own for the entrance exam. Admirable ambition for a 7th grader, but perhaps not wise to cram to get into a class for which he does not seem fully prepared.
 
Actually JeffM I already gave the exam and this is homework.Denis I don't have a lazy teacher and neither do I skip class.Thanks sorobon.
 
That really hurts.

Note to OP: The less information you provide, the more we have to guess. If you would like us to jump to actual and correct conclusions, every time, you must provide sufficient information. We cannot read minds.

What else have you? Let's do some good.
 
Asad Abbas said:
Actually JeffM I already gave the exam and this is homework.Denis I don't have a lazy teacher and neither do I skip class.Thanks sorobon.
OK....you're saying this is now homework; fine:
well then, WHY is your teacher giving you homework that you don't understand?
Has your teacher taught the material, and you forgot or didn't understand?
Plus: are you looking for hints, or for FULL solutions as given by Soroban?
 
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