Question

ABC is a right-angled triangle with AC = 65 cm and $\angle B={90}^{\circ }$ If r = 7 cm if area of triangle ABC is abc (abc is three digit number) then $(a-c)$ is

• A. 1
• B. 0
• C. 2
• D. 3

Answer 1

The correct option is A 1

$r=7$ cm and $R=AC/2\sin B=32.5$ cm; Again using $r=4R\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}$
$\Rightarrow r=R[\cos A+\cos B+\cos C-1]\Rightarrow \cos A+\cos {90}^{\circ }+\cos ({90}^{\circ }-A)-1=\frac{7}{32.5}$
$\Rightarrow \cos A+\sin A=\frac{39.5}{32.5}=\frac{79}{65}$
$\Rightarrow \sin 2A=\frac{{79}^2-{65}^2}{{65}^2}$
Now $△=2R^2\sin A\sin C=R^2\sin 2A={\left(\frac{65}{2}\right)}^2\times \frac{{79}^2-{65}^2}{{65}^2}=\frac{144\times 14}{4}=72\times 7=504$ sq.units