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**Hint:**We were given that ABC is a right-angled triangle with \[AC=65cm\] , \[\angle B=90{}^\circ \] and \[r=7cm\]. We know that if A, B, and C are angles of a triangle and a, b and c are the lengths of the sides of a triangle and R is the circumradius then \[\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}=2R\]. Now by using the sine rule, we can find the value of R. We know that the sum of angles of a triangle is equal to \[180{}^\circ \]. Now we will find the relation between A and B. We know that if r is inradius of a triangle, A, B and C are angles of a triangle and R is circumradius of a circle then \[r=R\left[ \cos A+\cos B+\cos C-1 \right]\]. Now we can find the value of angle A. We know that if A, B, and C are angles of a triangle and \[\Delta \] is the area of the triangle, then \[\Delta =2{{R}^{2}}\sin A\operatorname{sinC}\]. Now we can find the value of \[\Delta \]. From the question, we were given that the value of the triangle is equal to abc. So, let us find the values of a, b and c by comparing abc with \[\Delta \]. Now let us find the value of \[\left( a-c \right)\].

**Complete step-by-step solution**Let us represent the triangle from the given details of the question. From the question, we were given that ABC is a right-angled triangle with \[AC=65cm\] , \[\angle B=90{}^\circ \] and \[r=7cm\].

We know that if A, B, and C are angles of a triangle and a, b and c are the lengths of the sides of a triangle and R is the circumradius then \[\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}=2R\].

Now we will apply sine rule for the triangle ABC, then we get

\[\dfrac{a}{\sin A}=\dfrac{65}{\sin 90}=\dfrac{c}{\sin C}=2R.....(1)\]

From equation (1), we can write

\[\begin{align}

& \Rightarrow \dfrac{65}{\sin 90}=2R \\

& \Rightarrow 2R=65 \\

& \Rightarrow R=\dfrac{65}{2}....(2) \\

\end{align}\]

We know that the angles of a triangle is equal to \[180{}^\circ \].

So, we can write

\[\begin{align}

& \Rightarrow \angle A+\angle B+\angle C=180{}^\circ \\

& \Rightarrow \angle A+90{}^\circ +\angle C=180{}^\circ \\

& \Rightarrow \angle A+\angle C=90{}^\circ \\

& \Rightarrow \angle A=90{}^\circ -\angle C.....(3) \\

\end{align}\]

We know that if r is inradius of a triangle, A, B and C are angles of a triangle and R is circumradius of a circle then \[r=R\left[ \cos A+\cos B+\cos C-1 \right]\].

Now we will apply this formula for the triangle ABC, then we get

\[\Rightarrow r=R\left[ \cos A+\cos B+\cos C-1 \right].....(4)\]

Now let us substitute equation (1), equation (2) and equation (3) in equation (4), then we get

\[\Rightarrow r=\left( \dfrac{65}{2} \right)\left[ \cos A+\cos 90+\cos \left( 90-A \right)-1 \right]\]

We know that \[\cos \left( 90-A \right)=\sin A\].

So, now we will write

\[\Rightarrow r=\left( \dfrac{65}{2} \right)\left[ \cos A+\operatorname{sinA}-1 \right]\]

We know that \[r=7cm\]. So, we will write

\[\begin{align}

& \Rightarrow \dfrac{14}{65}=\cos A+\operatorname{sinA}-1 \\

& \Rightarrow \cos A+\sin A=1+\dfrac{14}{65} \\

& \Rightarrow \cos A+\sin A=\dfrac{79}{65}.......(5) \\

\end{align}\]

Now we will square equation (5) on both sides, then we get

\[\Rightarrow {{\left( \cos A+\operatorname{sinA} \right)}^{2}}={{\left( \dfrac{79}{65} \right)}^{2}}\]

We know that \[{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab\]

\[\Rightarrow {{\cos }^{2}}A+{{\sin }^{2}}A+2\operatorname{sinAcosA}={{\left( \dfrac{79}{65} \right)}^{2}}\]

We know that \[\sin 2A=2\sin A\cos A\].

\[\Rightarrow {{\cos }^{2}}A+{{\sin }^{2}}A+\sin 2A={{\left( \dfrac{79}{65} \right)}^{2}}\]

We know that \[{{\cos }^{2}}A+{{\sin }^{2}}A=1\].

\[\Rightarrow 1+\sin 2A={{\left( \dfrac{79}{65} \right)}^{2}}\]

\[\begin{align}

& \Rightarrow \sin 2A={{\left( \dfrac{79}{65} \right)}^{2}}-1 \\

& \Rightarrow \sin 2A=\dfrac{{{\left( 79 \right)}^{2}}-{{\left( 65 \right)}^{2}}}{{{\left( 65 \right)}^{2}}}.....(6) \\

\end{align}\]

We know that if A, B and C are angles of a triangle and \[\Delta \] is the area of the triangle, then

\[\Delta =2{{R}^{2}}\sin A\operatorname{sinC}\].

So, let us assume the area of the triangle is equal to \[\Delta \].

\[\Rightarrow \Delta =2{{R}^{2}}\sin A\operatorname{sinC}.....(7)\]

So, let us substitute equation (2), equation (3) in equation (7), then we get

\[\begin{align}

& \Rightarrow \Delta =2{{\left( \dfrac{65}{2} \right)}^{2}}\sin A\sin \left( 90-A \right) \\

& \Rightarrow \Delta =2{{\left( \dfrac{65}{2} \right)}^{2}}\sin A\cos A \\

& \Rightarrow \Delta ={{\left( \dfrac{65}{2} \right)}^{2}}\left( 2\sin A\cos A \right) \\

& \Rightarrow \Delta ={{\left( \dfrac{65}{2} \right)}^{2}}\left( \sin 2A \right).....(8) \\

\end{align}\]

Now let us substitute equation (6) in equation (8), then we get

\[\Rightarrow \Delta ={{\left( \dfrac{65}{2} \right)}^{2}}\left( \dfrac{{{\left( 79 \right)}^{2}}-{{\left( 65 \right)}^{2}}}{{{\left( 65 \right)}^{2}}} \right)\]

\[\Rightarrow \Delta ={{\left( \dfrac{1}{2} \right)}^{2}}\left( {{\left( 79 \right)}^{2}}-{{\left( 65 \right)}^{2}} \right)\]

We know that \[{{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)\].

\[\begin{align}

& \Rightarrow \Delta ={{\left( \dfrac{1}{2} \right)}^{2}}\left( 79+65 \right)\left( 79-65 \right) \\

& \Rightarrow \Delta =\dfrac{\left( 144 \right)\left( 14 \right)}{4} \\

& \Rightarrow \Delta =504.....(9) \\

\end{align}\]

So, it is clear that the value of \[\Delta \] is equal to 504.

From the question, it is clear that abc is equal to 504.

\[\begin{align}

& a=5....(10) \\

& b=0.....(11) \\

& c=4.....(12) \\

\end{align}\]

Then from equation (10) and equation (11), we will get

\[\begin{align}

& \Rightarrow a-c=5-4 \\

& \Rightarrow a-c=1.....(10) \\

\end{align}\]

So, the value of \[\left( a-c \right)\] is equal to 1.

**Hence, option A is correct.**

**Note:**Students should know that if A, B, and C are angles of a triangle and a, b and c are the lengths of the sides of a triangle and R is the circumradius then \[\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}=2R\]. Students should also avoid calculation mistakes while solving this problem. If a small mistake is made, then we cannot get the correct answer to this problem.

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