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Question

Answers

(a) 5, 5

(b) 6, 4

(c) 7, 3

(d) 8, 2

Answer
Verified

Let us assume that the length of the diagonals is x and y.

In the below diagram, we have shown a rhombus ABCD in which the diagonals intersect at point E at right angles.

In the above diagram, let us represent AC as x and BD as y.

Now, sum of the diagonals is given as 10 cm so adding the lengths of the diagonals that we have assumed above and equating it to 10 we get,

$x+y=10$ …………… Eq. (1)

The two diagonals divide the rhombus into four right angled triangles.

In $\Delta AED$, we are going to use Pythagoras theorem:

${{\left( AD \right)}^{2}}={{\left( AE \right)}^{2}}+{{\left( ED \right)}^{2}}$

Now, AE is half of AC and DE is half of DB. Length of AC and DB is x and y respectively.

$\begin{align}

& {{\left( AD \right)}^{2}}={{\left( \dfrac{x}{2} \right)}^{2}}+{{\left( \dfrac{y}{2} \right)}^{2}} \\

& \Rightarrow {{\left( AD \right)}^{2}}=\dfrac{{{x}^{2}}+{{y}^{2}}}{4} \\

\end{align}$

Taking square root on both the sides we get,

$AD=\dfrac{\sqrt{{{x}^{2}}+{{y}^{2}}}}{2}$

AD is one of the sides of the rhombus and we know that in rhombus all the sides are equal so the remaining 3 sides of the rhombus are of same length as AD.

Perimeter of the rhombus is equal to the sum of all the sides so the perimeter of the rhombus is equal to multiplication of length of AD by 4.

Perimeter of the rhombus is equal to:

$\begin{align}

& 4\left( \dfrac{\sqrt{{{x}^{2}}+{{y}^{2}}}}{2} \right) \\

& =2\sqrt{{{x}^{2}}+{{y}^{2}}} \\

\end{align}$

We have given the perimeter of rhombus as $4\sqrt{13}cm$ so equating the above perimeter to $4\sqrt{13}cm$ we get,

$2\sqrt{{{x}^{2}}+{{y}^{2}}}=4\sqrt{13}$

Squaring on both the sides of the above equation we get,

$4\left( {{x}^{2}}+{{y}^{2}} \right)=16\left( 13 \right)$

Dividing 4 on both the sides we get,

$\begin{align}

& \left( {{x}^{2}}+{{y}^{2}} \right)=\dfrac{16\left( 13 \right)}{4} \\

& \Rightarrow \left( {{x}^{2}}+{{y}^{2}} \right)=4\left( 13 \right) \\

& \Rightarrow \left( {{x}^{2}}+{{y}^{2}} \right)=52 \\

\end{align}$

Now, from eq. (1) $x+y=10$ we can write $x=10-y$ and then substituting this value of x in the above equation we get,

\[\begin{align}

& \left( {{x}^{2}}+{{y}^{2}} \right)=52 \\

& \Rightarrow {{\left( 10-y \right)}^{2}}+{{y}^{2}}=52 \\

& \Rightarrow 100+{{y}^{2}}-20y+{{y}^{2}}=52 \\

& \Rightarrow 2{{y}^{2}}-20y+48=0 \\

\end{align}\]

Dividing 2 on both the sides of the equation we get,

${{y}^{2}}-10y+24=0$………… Eq. (2)

Using factorization method we can find the factors of y as follows:

$\begin{align}

& {{y}^{2}}-6y-4y+24=0 \\

& \Rightarrow y\left( y-6 \right)-4\left( y-6 \right)=0 \\

\end{align}$

Taking y – 6 as common from the above equation we get,

$\left( y-6 \right)\left( y-4 \right)=0$

Equating $\left( y-6 \right)\And \left( y-4 \right)$ separately to 0 we get,

$\begin{align}

& y-6=0 \\

& \Rightarrow y=6 \\

& y-4=0 \\

& \Rightarrow y=4 \\

\end{align}$

From the above, we have got two values of y i.e. 4 and 6.

Now, substituting the value of y as 4 in eq. (1) we get,

$\begin{align}

& x+y=10 \\

& \Rightarrow x+4=10 \\

& \Rightarrow x=6 \\

\end{align}$

Substituting the value of y as 6 in eq. (1) we get,

$\begin{align}

& x+6=10 \\

& \Rightarrow x=4 \\

\end{align}$

Hence, we have got two pairs of x and y values (x, y) i.e. (4, 6) and (6, 4).

${{y}^{2}}-10y+24=0$

We know the method to find the factors of y as first of all we write the prime factorization of 24 then we add and subtract the factors in such a way that it is equal to 10 so there are two ways to write 10 one is $\left( 12-2 \right)$ and other is $\left( 6+4 \right)$ as you can see that if we multiply 12 and 2 we will get 24. Similarly, if we multiply 6 and 4 we get 24 so the expressions are fulfilling the conditions of the factorization method but if you write 10 as $\left( 12-2 \right)$ then the expression will look as:

$\begin{align}

& {{y}^{2}}-\left( 12-2 \right)y+24=0 \\

& \Rightarrow {{y}^{2}}-12y+2y+24=0 \\

\end{align}$

Now, when you take y as common from the first two expressions and 2 from the last two expressions then we get,

$y\left( y-12 \right)+2\left( y+12 \right)=0$

Here, you can see we are not getting any common factor so this is the mistake that could happen so make sure you have written the right set of factors of 24.

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