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

Answer: (B)

$(s-a)=3;(s-c)=2$

$3+a=2+c$

$a+1=c---eqn.1$

$s=\frac{a+b+c}{2}s-a=\frac{b+c-a}{2}2\left(3\right)=b+1\Rightarrow 6=b+1\Rightarrow b=5s-c=\frac{b+a-c}{2}\Rightarrow 4=b-1\Rightarrow b=5$

$a^2+c^2=b^2$ (Pythagoras theorem)

$\Rightarrow a^2+(a+1{)}^2=25$

$\Rightarrow a^2+a^2+2a+1=25$

$\Rightarrow 2a^2+2a-24=0\Rightarrow a=3$

$\Rightarrow c=a+1=4$