
If in a triangle $ABC$ right angled at $B$, $s-a=3,\,s-c=2$, then the values of $a$ and $c$ are respectively.
A. $2,3$
B. $3,4$
C. $4,3$
D. $6,8$
Answer
162.3k+ views
Hint: To solve this question, we will use the formula of semi perimeter of the triangle. We will substitute the value of $s$ in given equations $s-a=3,\,s-c=2$ and simplify it. After simplifying we will get two equations from which we will get the value of one of the variables. Then using substitution method and simplifying we will get the value of $a$ and $c$
Formula Used: Semi perimeter $s=\dfrac{a+b+c}{2}$.
Complete step by step solution: We are given a triangle $ABC$ right angled at $B$ with $s-a=3,\,s-c=2$ and we have to determine the value of $a$ and $c$.
We will first take the equation $s-a=3$ and substitute the value of $s$.
$\begin{align}
& \dfrac{a+b+c}{2}-a=3 \\
& a+b+c-2a=6 \\
& b+c-a=6.....(i)
\end{align}$
Now we will take the equation $s-c=2$ and substitute the value of $s$.
$\begin{align}
& s-c=2 \\
& \dfrac{a+b+c}{2}-c=2 \\
& a+b-c=4...(ii)
\end{align}$
Adding equations (i) and (ii).
$\begin{align}
& b+c-a+a+b-c=6+4 \\
& 2b=10 \\
& b=5
\end{align}$
As a given triangle$ABC$ right angled at $B$, we can use the Pythagoras theorem. So,
$\begin{align}
& {{5}^{2}}={{a}^{2}}+{{c}^{2}} \\
& 25={{a}^{2}}+{{c}^{2}}....(iii) \\
\end{align}$
Using equation (ii),
$\begin{align}
& a-c+5=4 \\
& a-c=-1.....(iv) \\
\end{align}$
Squaring both sides,
$\begin{align}
& {{(a-c)}^{2}}=1 \\
& {{a}^{2}}+{{c}^{2}}-2ac=1 \\
\end{align}$
We will now substitute the equation (iii) in the above equation.
$\begin{align}
& 25-2ac=1 \\
& 24=2ac \\
& ac=12 \\
& a=\dfrac{12}{c}
\end{align}$
We will now substitute the value $a=\dfrac{12}{c}$ in equation (iv).
$\begin{align}
& \dfrac{12}{c}-c=-1 \\
& 12-{{c}^{2}}=-c \\
& {{c}^{2}}-c-12=0 \\
& (c-4)(c+3)=0 \\
& c=-3,4
\end{align}$
We will consider the value of $c$ as $c=4$ because the value of the length cannot be negative.
Now we will substitute the value of $c$ in equation (iv) and derive the value of $a$.
$\begin{align}
& a-c=-1 \\
& a-4=-1 \\
& a=3
\end{align}$
The value of $a$ and $c$ are $a=3$ and $c=4$ when the triangle $ABC$ is right angled at $B$ and $s-a=3,\,s-c=2$. Hence the correct option is (B).
Note: The Pythagoras theorem is only applicable for a right angled triangle and it depicts the relationship between the three sides of the triangle.
Formula Used: Semi perimeter $s=\dfrac{a+b+c}{2}$.
Complete step by step solution: We are given a triangle $ABC$ right angled at $B$ with $s-a=3,\,s-c=2$ and we have to determine the value of $a$ and $c$.
We will first take the equation $s-a=3$ and substitute the value of $s$.
$\begin{align}
& \dfrac{a+b+c}{2}-a=3 \\
& a+b+c-2a=6 \\
& b+c-a=6.....(i)
\end{align}$
Now we will take the equation $s-c=2$ and substitute the value of $s$.
$\begin{align}
& s-c=2 \\
& \dfrac{a+b+c}{2}-c=2 \\
& a+b-c=4...(ii)
\end{align}$
Adding equations (i) and (ii).
$\begin{align}
& b+c-a+a+b-c=6+4 \\
& 2b=10 \\
& b=5
\end{align}$
As a given triangle$ABC$ right angled at $B$, we can use the Pythagoras theorem. So,
$\begin{align}
& {{5}^{2}}={{a}^{2}}+{{c}^{2}} \\
& 25={{a}^{2}}+{{c}^{2}}....(iii) \\
\end{align}$
Using equation (ii),
$\begin{align}
& a-c+5=4 \\
& a-c=-1.....(iv) \\
\end{align}$
Squaring both sides,
$\begin{align}
& {{(a-c)}^{2}}=1 \\
& {{a}^{2}}+{{c}^{2}}-2ac=1 \\
\end{align}$
We will now substitute the equation (iii) in the above equation.
$\begin{align}
& 25-2ac=1 \\
& 24=2ac \\
& ac=12 \\
& a=\dfrac{12}{c}
\end{align}$
We will now substitute the value $a=\dfrac{12}{c}$ in equation (iv).
$\begin{align}
& \dfrac{12}{c}-c=-1 \\
& 12-{{c}^{2}}=-c \\
& {{c}^{2}}-c-12=0 \\
& (c-4)(c+3)=0 \\
& c=-3,4
\end{align}$
We will consider the value of $c$ as $c=4$ because the value of the length cannot be negative.
Now we will substitute the value of $c$ in equation (iv) and derive the value of $a$.
$\begin{align}
& a-c=-1 \\
& a-4=-1 \\
& a=3
\end{align}$
The value of $a$ and $c$ are $a=3$ and $c=4$ when the triangle $ABC$ is right angled at $B$ and $s-a=3,\,s-c=2$. Hence the correct option is (B).
Note: The Pythagoras theorem is only applicable for a right angled triangle and it depicts the relationship between the three sides of the triangle.
Recently Updated Pages
If there are 25 railway stations on a railway line class 11 maths JEE_Main

Minimum area of the circle which touches the parabolas class 11 maths JEE_Main

Which of the following is the empty set A x x is a class 11 maths JEE_Main

The number of ways of selecting two squares on chessboard class 11 maths JEE_Main

Find the points common to the hyperbola 25x2 9y2 2-class-11-maths-JEE_Main

A box contains 6 balls which may be all of different class 11 maths JEE_Main

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

JoSAA JEE Main & Advanced 2025 Counselling: Registration Dates, Documents, Fees, Seat Allotment & Cut‑offs

NIT Cutoff Percentile for 2025

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

NCERT Solutions for Class 11 Maths Chapter 4 Complex Numbers and Quadratic Equations

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

Degree of Dissociation and Its Formula With Solved Example for JEE

Free Radical Substitution Mechanism of Alkanes for JEE Main 2025
