# A 34cm long wire is bent into a rectangle. The length of its diagonal is 13cm. What are the lengths of the sides of the rectangle ?

Last updated date: 27th Mar 2023

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Answer

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Hint: The wire is being bent to form a new shape, so the sum of the lengths that make up the new shape will be equal to the total length of the wire. Therefore, the perimeter of the rectangle will be equal to the length of the wire provided to us.

The total length of the wire that is provided to us = 34cm.

Now, just imagine. If we take a wire and bend it into some shape, then what would the length of the wire ultimately become for the shape ?

It’ll become its perimeter. This is because the perimeter is essentially a measure of the sum of the sides of the polygon, whose perimeter is being measured. Therefore, whatever shape the wire has been bent into, the sum of its sides, if it’s a polygon, will definitely be equal to the length of the wire, because after all the sides are made by bending the wire itself.

Therefore, we know that the perimeter of the rectangle = the total length of the wire.

Let’s assume that the rectangle’s length is $l$ cm, and its breadth is $b$ cm.

The perimeter of a rectangle $P=2(l+b)$.

Therefore, since the total length of the wire = the perimeter of the rectangle, we can say that $P=34cm$, which is the total length of the wire given to us.

$\begin{align}

& \Rightarrow 2(l+b)cm=34cm \\

& \Rightarrow 2(l+b)=34 \\

& \Rightarrow (l+b)=17 \\

\end{align}$

Let’s call the equation $l+b=17$ equation (1).

Now, the next information that has been given to us concerns the diagonal of the rectangle we formed by bending the wire.

We know that, the length of the diagonal of a rectangle having length $l$ cm and breadth $b$ cm = $\sqrt{{{l}^{2}}+{{b}^{2}}}cm$, and in this case, we are told that the length of the diagonal of the rectangle = 13cm.

Thus, we can say that, $\begin{align}

& \sqrt{{{l}^{2}}+{{b}^{2}}}cm=13cm \\

& \Rightarrow \sqrt{{{l}^{2}}+{{b}^{2}}}=13 \\

\end{align}$

Squaring both sides, we get :

$\Rightarrow ({{l}^{2}}+{{b}^{2}})=169$…………….. (2)

Now, we have two variables to find out, and we also have two equations concerning those variables.

Let’s square both sides of equation (1) and see what we get.

Doing so, we’ll have :

$\begin{align}

& {{(l+b)}^{2}}=289 \\

& \Rightarrow {{l}^{2}}+{{b}^{2}}+2lb=289 \\

\end{align}$

Here, we can easily substitute the value of $({{l}^{2}}+{{b}^{2}})$ from equation (2).

Doing so, we’ll get :

$\begin{align}

& {{l}^{2}}+{{b}^{2}}+2lb=289 \\

& \Rightarrow 169+2lb=289 \\

& \Rightarrow 2lb=120 \\

& \Rightarrow lb=60 \\

& \Rightarrow l=\dfrac{60}{b} \\

\end{align}$

Now, we have expressed $l$ in terms of $b$, and can substitute this value back into equation (1), and obtain a quadratic in $b$.

Substituting for $l$ in equation (1), we get :

$\begin{align}

& l+b=17 \\

& \Rightarrow \dfrac{60}{b}+b=17 \\

& \Rightarrow \dfrac{60+{{b}^{2}}}{b}=17 \\

& \Rightarrow 60+{{b}^{2}}=17b \\

& \Rightarrow {{b}^{2}}-17b+60=0 \\

& \Rightarrow {{b}^{2}}-12b-5b+60=0 \\

& \Rightarrow b(b-12)-5(b-12)=0 \\

& \Rightarrow (b-5)(b-12)=0 \\

& \Rightarrow b=5,12 \\

\end{align}$

Now that we have two possible values of $b$, let’s substitute for $b$ in the expression where we wrote $l$ in terms of $b$, and find out the corresponding values of $l$.

Doing so, we get :

$\begin{align}

& b=5,12 \\

& l=\dfrac{60}{b} \\

& \Rightarrow l=\dfrac{60}{5},\dfrac{60}{12} \\

& \Rightarrow l=12,5 \\

\end{align}$

Therefore, we have two values of $l$ corresponding to each $b$ as well.

If you notice the values of the dimensions are essentially the same, one is 12cm and the other is 5cm. It’s on us to choose which is $l$ and which is $b$.

Now, we know that the longer side in a rectangle is called the length, and hence, we’ll consider the length of this rectangle to be 12cm. The shorter side is called the breadth, and hence, the breadth of this rectangle = 5cm.

Hence, the lengths of the sides of the rectangle are 12cm and 5cm.

Note: Be very careful while squaring both sides of equation (1). Note that, while squaring the LHS of equation (1), we have used the formula that says that ${{(a+b)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$, and you should be very careful while squaring, because often students miss the 2ab term, and end up getting wrong results. Thus, the result is ${{(a+b)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$, not $$${{(a+b)}^{2}}={{a}^{2}}+{{b}^{2}}$, so squaring (1) on both sides, will not give you equation (2), in any case.

The total length of the wire that is provided to us = 34cm.

Now, just imagine. If we take a wire and bend it into some shape, then what would the length of the wire ultimately become for the shape ?

It’ll become its perimeter. This is because the perimeter is essentially a measure of the sum of the sides of the polygon, whose perimeter is being measured. Therefore, whatever shape the wire has been bent into, the sum of its sides, if it’s a polygon, will definitely be equal to the length of the wire, because after all the sides are made by bending the wire itself.

Therefore, we know that the perimeter of the rectangle = the total length of the wire.

Let’s assume that the rectangle’s length is $l$ cm, and its breadth is $b$ cm.

The perimeter of a rectangle $P=2(l+b)$.

Therefore, since the total length of the wire = the perimeter of the rectangle, we can say that $P=34cm$, which is the total length of the wire given to us.

$\begin{align}

& \Rightarrow 2(l+b)cm=34cm \\

& \Rightarrow 2(l+b)=34 \\

& \Rightarrow (l+b)=17 \\

\end{align}$

Let’s call the equation $l+b=17$ equation (1).

Now, the next information that has been given to us concerns the diagonal of the rectangle we formed by bending the wire.

We know that, the length of the diagonal of a rectangle having length $l$ cm and breadth $b$ cm = $\sqrt{{{l}^{2}}+{{b}^{2}}}cm$, and in this case, we are told that the length of the diagonal of the rectangle = 13cm.

Thus, we can say that, $\begin{align}

& \sqrt{{{l}^{2}}+{{b}^{2}}}cm=13cm \\

& \Rightarrow \sqrt{{{l}^{2}}+{{b}^{2}}}=13 \\

\end{align}$

Squaring both sides, we get :

$\Rightarrow ({{l}^{2}}+{{b}^{2}})=169$…………….. (2)

Now, we have two variables to find out, and we also have two equations concerning those variables.

Let’s square both sides of equation (1) and see what we get.

Doing so, we’ll have :

$\begin{align}

& {{(l+b)}^{2}}=289 \\

& \Rightarrow {{l}^{2}}+{{b}^{2}}+2lb=289 \\

\end{align}$

Here, we can easily substitute the value of $({{l}^{2}}+{{b}^{2}})$ from equation (2).

Doing so, we’ll get :

$\begin{align}

& {{l}^{2}}+{{b}^{2}}+2lb=289 \\

& \Rightarrow 169+2lb=289 \\

& \Rightarrow 2lb=120 \\

& \Rightarrow lb=60 \\

& \Rightarrow l=\dfrac{60}{b} \\

\end{align}$

Now, we have expressed $l$ in terms of $b$, and can substitute this value back into equation (1), and obtain a quadratic in $b$.

Substituting for $l$ in equation (1), we get :

$\begin{align}

& l+b=17 \\

& \Rightarrow \dfrac{60}{b}+b=17 \\

& \Rightarrow \dfrac{60+{{b}^{2}}}{b}=17 \\

& \Rightarrow 60+{{b}^{2}}=17b \\

& \Rightarrow {{b}^{2}}-17b+60=0 \\

& \Rightarrow {{b}^{2}}-12b-5b+60=0 \\

& \Rightarrow b(b-12)-5(b-12)=0 \\

& \Rightarrow (b-5)(b-12)=0 \\

& \Rightarrow b=5,12 \\

\end{align}$

Now that we have two possible values of $b$, let’s substitute for $b$ in the expression where we wrote $l$ in terms of $b$, and find out the corresponding values of $l$.

Doing so, we get :

$\begin{align}

& b=5,12 \\

& l=\dfrac{60}{b} \\

& \Rightarrow l=\dfrac{60}{5},\dfrac{60}{12} \\

& \Rightarrow l=12,5 \\

\end{align}$

Therefore, we have two values of $l$ corresponding to each $b$ as well.

If you notice the values of the dimensions are essentially the same, one is 12cm and the other is 5cm. It’s on us to choose which is $l$ and which is $b$.

Now, we know that the longer side in a rectangle is called the length, and hence, we’ll consider the length of this rectangle to be 12cm. The shorter side is called the breadth, and hence, the breadth of this rectangle = 5cm.

Hence, the lengths of the sides of the rectangle are 12cm and 5cm.

Note: Be very careful while squaring both sides of equation (1). Note that, while squaring the LHS of equation (1), we have used the formula that says that ${{(a+b)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$, and you should be very careful while squaring, because often students miss the 2ab term, and end up getting wrong results. Thus, the result is ${{(a+b)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$, not $$${{(a+b)}^{2}}={{a}^{2}}+{{b}^{2}}$, so squaring (1) on both sides, will not give you equation (2), in any case.

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