
A pencil box in the shape of a cuboid is $9cm$longer than it is wide and $1cm$ higher than it is wide. The longest pencil which can be fitted into the case is $13cm$ long. Find the dimensions of the pencil case.
Answer
578.7k+ views
Hint:
For this question , first suppose the length of breadth in the form of $x$, then apply the condition of the question to get the values of length and height of the box. At last to find the value of $x$ by using the formula of diagonal of cuboid and then find the value of length, breadth and height.
Complete step by step solution:
To get the dimensions of the pencil case , let us suppose the breadth of the pencil case be $x\,cm$.
Then from the condition of question length will be $(x+9)cm$ and height will be $(x+1)cm$.
We know that the formula of diagonal of a cuboid such that
Diagonal of cuboid=the longest pencil fit over the box
=$\sqrt{{{l}^{2}}+{{b}^{2}}+{{h}^{2}}}$
l for length, b for breadth, h for height.
Now on putting all values of length, breadth and height, we get
$13=\sqrt{{{x}^{2}}+{{\left( x+1 \right)}^{2}}+{{\left( x+9 \right)}^{2}}}$
On squaring both sides, we get
$\Rightarrow {{13}^{2}}={{x}^{2}}+{{(x+1)}^{2}}+{{(x+9)}^{2}}$
On expanding the above term by using identity, we get
$\Rightarrow 169={{x}^{2}}+{{x}^{2}}+2x+1+{{x}^{2}}+18x+81$
On calculating we get
$\Rightarrow 3{{x}^{2}}+20x-261=0$
Now we have a quadratic equation and by solving it by split the middle term method, we get
$\Rightarrow 3{{x}^{2}}+29x-9x-261=0$
On taking common terms out,
$\Rightarrow x(x+29)-3(x+29)=0$
On taking $x+29$as common,
$\Rightarrow (x+29)(x-3)=0$
Now on equating the above terms equal to zero, we get
$
\Rightarrow (x+29)=0\Rightarrow x=-29 \\
\Rightarrow (x-3)=0\Rightarrow x=3 \\
$
We know that dimensions of a cuboid are never negative, so that we have to take $x=3$ as length of breadth.
Therefore, length = $x+9=3+9=12cm$
Height =$x+1=3+1=4cm$
Note:
In such a type of problem, sometimes students get confused between the formulas of cube and cuboid, so that they should be well aware about the difference , otherwise they make mistakes. In case of mensuration questions we would suggest that write all the formulas in a page and learn them. It'll also help you to revise formulas in exam time.
For this question , first suppose the length of breadth in the form of $x$, then apply the condition of the question to get the values of length and height of the box. At last to find the value of $x$ by using the formula of diagonal of cuboid and then find the value of length, breadth and height.
Complete step by step solution:
To get the dimensions of the pencil case , let us suppose the breadth of the pencil case be $x\,cm$.
Then from the condition of question length will be $(x+9)cm$ and height will be $(x+1)cm$.
We know that the formula of diagonal of a cuboid such that
Diagonal of cuboid=the longest pencil fit over the box
=$\sqrt{{{l}^{2}}+{{b}^{2}}+{{h}^{2}}}$
l for length, b for breadth, h for height.
Now on putting all values of length, breadth and height, we get
$13=\sqrt{{{x}^{2}}+{{\left( x+1 \right)}^{2}}+{{\left( x+9 \right)}^{2}}}$
On squaring both sides, we get
$\Rightarrow {{13}^{2}}={{x}^{2}}+{{(x+1)}^{2}}+{{(x+9)}^{2}}$
On expanding the above term by using identity, we get
$\Rightarrow 169={{x}^{2}}+{{x}^{2}}+2x+1+{{x}^{2}}+18x+81$
On calculating we get
$\Rightarrow 3{{x}^{2}}+20x-261=0$
Now we have a quadratic equation and by solving it by split the middle term method, we get
$\Rightarrow 3{{x}^{2}}+29x-9x-261=0$
On taking common terms out,
$\Rightarrow x(x+29)-3(x+29)=0$
On taking $x+29$as common,
$\Rightarrow (x+29)(x-3)=0$
Now on equating the above terms equal to zero, we get
$
\Rightarrow (x+29)=0\Rightarrow x=-29 \\
\Rightarrow (x-3)=0\Rightarrow x=3 \\
$
We know that dimensions of a cuboid are never negative, so that we have to take $x=3$ as length of breadth.
Therefore, length = $x+9=3+9=12cm$
Height =$x+1=3+1=4cm$
Note:
In such a type of problem, sometimes students get confused between the formulas of cube and cuboid, so that they should be well aware about the difference , otherwise they make mistakes. In case of mensuration questions we would suggest that write all the formulas in a page and learn them. It'll also help you to revise formulas in exam time.
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