Question

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

Answer 1

Hint: We were given that ABC is a right-angled triangle with $$AC=65cm$$ , $$\mathrm{\angle }B=90{}^{\circ }$$ 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 $${\displaystyle \frac{a}{\sin A}}={\displaystyle \frac{b}{\sin B}}={\displaystyle \frac{c}{\sin C}}=2R$$. 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 $$r=R[\cos A+\cos B+\cos C-1]$$. Now we can find the value of angle A. We know that if A, B, and C are angles of a triangle and $$\mathrm{\Delta }$$ is the area of the triangle, then $$\mathrm{\Delta }=2R^2\sin A\mathrm{sinC}$$. Now we can find the value of $$\mathrm{\Delta }$$. 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 $$\mathrm{\Delta }$$. Now let us find the value of $$(a-c)$$.

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 $$AC=65cm$$ , $$\mathrm{\angle }B=90{}^{\circ }$$ 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 $${\displaystyle \frac{a}{\sin A}}={\displaystyle \frac{b}{\sin B}}={\displaystyle \frac{c}{\sin C}}=2R$$.
Now we will apply sine rule for the triangle ABC, then we get
$${\displaystyle \frac{a}{\sin A}}={\displaystyle \frac{65}{\sin 90}}={\displaystyle \frac{c}{\sin C}}=2R.....(1)$$
From equation (1), we can write
$$\begin{array}{rl}& \Rightarrow {\displaystyle \frac{65}{\sin 90}}=2R\\ & \Rightarrow 2R=65\\ & \Rightarrow R={\displaystyle \frac{65}{2}}....(2)\end{array}$$
We know that the angles of a triangle is equal to $$180{}^{\circ }$$.
So, we can write
$$\begin{array}{rl}& \Rightarrow \mathrm{\angle }A+\mathrm{\angle }B+\mathrm{\angle }C=180{}^{\circ }\\ & \Rightarrow \mathrm{\angle }A+90{}^{\circ }+\mathrm{\angle }C=180{}^{\circ }\\ & \Rightarrow \mathrm{\angle }A+\mathrm{\angle }C=90{}^{\circ }\\ & \Rightarrow \mathrm{\angle }A=90{}^{\circ }-\mathrm{\angle }C.....(3)\end{array}$$
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 $$r=R[\cos A+\cos B+\cos C-1]$$.
Now we will apply this formula for the triangle ABC, then we get
$$\Rightarrow r=R[\cos A+\cos B+\cos C-1].....(4)$$
Now let us substitute equation (1), equation (2) and equation (3) in equation (4), then we get
$$\Rightarrow r=\left({\displaystyle \frac{65}{2}}\right)[\cos A+\cos 90+\cos (90-A)-1]$$
We know that $$\cos (90-A)=\sin A$$.
So, now we will write
$$\Rightarrow r=\left({\displaystyle \frac{65}{2}}\right)[\cos A+\mathrm{sinA}-1]$$
We know that $$r=7cm$$. So, we will write
$$\begin{array}{rl}& \Rightarrow {\displaystyle \frac{14}{65}}=\cos A+\mathrm{sinA}-1\\ & \Rightarrow \cos A+\sin A=1+{\displaystyle \frac{14}{65}}\\ & \Rightarrow \cos A+\sin A={\displaystyle \frac{79}{65}}.......(5)\end{array}$$
Now we will square equation (5) on both sides, then we get
$$\Rightarrow {(\cos A+\mathrm{sinA})}^2={\left({\displaystyle \frac{79}{65}}\right)}^2$$
We know that $${(a+b)}^2=a^2+b^2+2ab$$
$$\Rightarrow {\cos }^{2}A+{\sin }^{2}A+2\mathrm{sinAcosA}={\left({\displaystyle \frac{79}{65}}\right)}^2$$
We know that $$\sin 2A=2\sin A\cos A$$.
$$\Rightarrow {\cos }^{2}A+{\sin }^{2}A+\sin 2A={\left({\displaystyle \frac{79}{65}}\right)}^2$$
We know that $${\cos }^{2}A+{\sin }^{2}A=1$$.
$$\Rightarrow 1+\sin 2A={\left({\displaystyle \frac{79}{65}}\right)}^2$$
$$\begin{array}{rl}& \Rightarrow \sin 2A={\left({\displaystyle \frac{79}{65}}\right)}^2-1\\ & \Rightarrow \sin 2A={\displaystyle \frac{{\left(79\right)}^2-{\left(65\right)}^2}{{\left(65\right)}^2}}.....(6)\end{array}$$
We know that if A, B and C are angles of a triangle and $$\mathrm{\Delta }$$ is the area of the triangle, then
$$\mathrm{\Delta }=2R^2\sin A\mathrm{sinC}$$.
So, let us assume the area of the triangle is equal to $$\mathrm{\Delta }$$.
$$\Rightarrow \mathrm{\Delta }=2R^2\sin A\mathrm{sinC}.....(7)$$
So, let us substitute equation (2), equation (3) in equation (7), then we get
$$\begin{array}{rl}& \Rightarrow \mathrm{\Delta }=2{\left({\displaystyle \frac{65}{2}}\right)}^2\sin A\sin (90-A)\\ & \Rightarrow \mathrm{\Delta }=2{\left({\displaystyle \frac{65}{2}}\right)}^2\sin A\cos A\\ & \Rightarrow \mathrm{\Delta }={\left({\displaystyle \frac{65}{2}}\right)}^2(2\sin A\cos A)\\ & \Rightarrow \mathrm{\Delta }={\left({\displaystyle \frac{65}{2}}\right)}^2(\sin 2A).....(8)\end{array}$$
Now let us substitute equation (6) in equation (8), then we get
$$\Rightarrow \mathrm{\Delta }={\left({\displaystyle \frac{65}{2}}\right)}^2\left({\displaystyle \frac{{\left(79\right)}^2-{\left(65\right)}^2}{{\left(65\right)}^2}}\right)$$
$$\Rightarrow \mathrm{\Delta }={\left({\displaystyle \frac{1}{2}}\right)}^2({\left(79\right)}^2-{\left(65\right)}^2)$$
We know that $$a^2-b^2=(a+b)(a-b)$$.
$$\begin{array}{rl}& \Rightarrow \mathrm{\Delta }={\left({\displaystyle \frac{1}{2}}\right)}^2(79+65)(79-65)\\ & \Rightarrow \mathrm{\Delta }={\displaystyle \frac{\left(144\right)\left(14\right)}{4}}\\ & \Rightarrow \mathrm{\Delta }=504.....(9)\end{array}$$
So, it is clear that the value of $$\mathrm{\Delta }$$ is equal to 504.
From the question, it is clear that abc is equal to 504.
$$\begin{array}{rl}& a=5....(10)\\ & b=0.....(11)\\ & c=4.....(12)\end{array}$$
Then from equation (10) and equation (11), we will get
$$\begin{array}{rl}& \Rightarrow a-c=5-4\\ & \Rightarrow a-c=1.....(10)\end{array}$$
So, the value of $$(a-c)$$ 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 $${\displaystyle \frac{a}{\sin A}}={\displaystyle \frac{b}{\sin B}}={\displaystyle \frac{c}{\sin C}}=2R$$. 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.