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# ABC is a right – angled triangle with $AC=65cm$ and $\angle B=90{}^\circ$. If $r=7cm$and the area of triangle is equal to abc then $\left( a-c \right)$ is \begin{align} & (A)\text{ 1} \\ & (B)\text{ 0} \\ & (C)\text{ 2} \\ & (D)\text{ 3} \\ \end{align}

Last updated date: 12th Sep 2024
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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.