Question

If in a triangle ABC, right angle at B, $s - a = 3$ and $s - c = 2$, then
A. $a = 2;c = 3$
B. $a = 3;c = 4$
C. $a = 4;c = 3$
D. $a = 6;c = 8$

Answer 1

Hint: First of all this is a very simple and a very easy problem. This problem deals with a right angled angle in which given a certain amount of information regarding its semi-perimeter. Semi perimeter is the half of the sum of lengths of all the sides of the triangle. In order to solve this problem we need to be familiar with the Pythagoras theorem which is given by:
$ \Rightarrow {a^2} + {c^2} = {b^2}$
Where b is the length of hypotenuse, and a, c are the other sides of the right angled triangle.
The Pythagoras theorem is only applicable to right angles triangles.

Complete answer:
Given that there is a right angled triangle, right angled at B.
Visualizing the right angled triangle as given below:

Let the hypotenuse of the right angled triangle = b
The base of the right angled triangle, which is the length of the side BC = a
The height of the right angled triangle, which is the length of the side AB = c
Here given that $s - a = 3$ and $s - c = 2$.
From the above two expressions, equating the value of $s$, as given below:
From $s - a = 3$, the expression below is obtained.
$ \Rightarrow s = 3 + a$ and also
From $s - c = 2$, the expression below is obtained.
$ \Rightarrow s = 2 + c$
Here equating the expressions obtained for $s$, as given below:
$ \Rightarrow 3 + a = 2 + c$
$\therefore c = a + 1$
Hence obtained the expression for $c$is obtained, which will be used later.
Here $s$ is the semi-perimeter of the given triangle, which is given by:
$ \Rightarrow s = \dfrac{{a + b + c}}{2}$
Now expanding the expression $s - a = 3$, as given below:
$ \Rightarrow s - a = \dfrac{{a + b + c}}{2} - a$
$ \Rightarrow \dfrac{{b + c - a}}{2} = 3$
$\therefore b + c - a = 6$
Now expanding the expression $s - c = 2$, as given below:
$ \Rightarrow s - c = \dfrac{{a + b + c}}{2} - c$
$ \Rightarrow \dfrac{{a + b - c}}{2} = 2$
$\therefore a + b - c = 4$
Solving the both equations obtained above, to get the value of b, as given below:
$ \Rightarrow b + c - a = 6$
$ \Rightarrow a + b - c = 4$
$ \Rightarrow 2b = 10$
$\therefore b = 5$
Here using the Pythagoras theorem for the given triangle ABC, which is given by:
$ \Rightarrow {a^2} + {c^2} = {b^2}$
Here we obtained that $c = a + 1$ and $b = 5$, now substituting these in the Pythagoras theorem as given below:
$ \Rightarrow {a^2} + {(a + 1)^2} = {5^2}$
$ \Rightarrow {a^2} + {a^2} + 2a + 1 = 25$
By expanding the ${(a + 1)^2}$ term and simplifying as given below:
$ \Rightarrow 2{a^2} + 2a - 24 = 0$
Dividing the above equation by 2, as given below:
$ \Rightarrow {a^2} + a - 12 = 0$
Solving the above quadratic equation as given below:
$ \Rightarrow a = \dfrac{{ - 1 \pm \sqrt {1 - 4(1)( - 12)} }}{{2(1)}}$
$ \Rightarrow a = \dfrac{{ - 1 \pm \sqrt {49} }}{2}$
The square root of 49 is given by $\sqrt {49} = 7$
$ \Rightarrow a = \dfrac{{ - 1 \pm 7}}{2}$
$\therefore a = \dfrac{6}{2};a = \dfrac{{ - 8}}{2}$
$\because a$ cannot be negative, as it is the length of a side of the triangle.
$\therefore a = 3$ and hence $c$ is given by:
$ \Rightarrow c = a + 1$
$ \Rightarrow c = 3 + 1 = 4$
$\therefore c = 4$
Thus $a = 3;c = 4$

Final Answer: $a = 3;c = 4$

Note:
Here this problem can be solved in another way also, which is described here. In the problem after finding the value of the hypotenuse of the triangle which is b= 5, hence as we know that this is one of the pythagorean triplet which is of (3,4 and 5). Hence the other two sides have to be 3 and 4. So here $a + c = 7$. So we already found that $b + c - a = 6$ from $s - a = 3$, here we substitute the value of b and we get the equation as $c - a = 1$. So now we have two equations and two variables, which are $a + c = 7$ and $c - a = 1$, hence solve these equations to get the values of a and c.