
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
216.3k+ 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
JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

Electricity and Magnetism Explained: Key Concepts & Applications

Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

JEE Energetics Important Concepts and Tips for Exam Preparation

JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2026: Application Form Open, Exam Dates, Syllabus, Eligibility & Question Papers

Derivation of Equation of Trajectory Explained for Students

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Angle of Deviation in a Prism

Understanding Collisions: Types and Examples for Students

How to Convert a Galvanometer into an Ammeter or Voltmeter

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

NCERT Solutions for Class 11 Maths Chapter 10 Conic Sections

NCERT Solutions for Class 11 Maths Chapter 9 Straight Lines

NCERT Solutions For Class 11 Maths Chapter 8 Sequences And Series

Ideal and Non-Ideal Solutions Explained for Class 12 Chemistry

Degree of Dissociation: Meaning, Formula, Calculation & Uses

