Question

If area of a right triangle is $180c{m}^2$ and 9 cm is one of the sides containing the right angle, then the length of altitude on the hypotenuse of the triangle is equal to

• A. $\frac{360}{41}$ cm
  
• B. $\frac{180}{\sqrt{21}}$ cm
  
• C. 15 cm
• D. 29 cm

Answer 1

The correct option is A $\frac{360}{41}$ cm


Let ABC be the right angle triangle with BC = 9 cm and area of $\Delta ABC=180c{m}^2$.

Area of ABC = $\frac{1}{2}\times AB\times BC=180c{m}^2$
$\frac{1}{2}\times AB\times 9=180$
$AB=40cm$
Now, in ΔABC, by Pythagoras Theorem,
$A{C}^2=A{B}^2+B{C}^2$
$A{C}^2=(40{)}^2+(9{)}^2$
$A{C}^2=1600+81$
$A{C}^2=1681$
$AC=\sqrt{1681}=41cm$ .........(i)
Area of $\Delta ABC=\frac{1}{2}\times Base\times Height$
$\frac{1}{2}\times AC\times BD=180$
$\frac{1}{2}\times 41\times BD=180$ [from(i)]
$BD=\frac{360}{41}cm$
Thus, the length of altitude on the hypotenuse of the triangle is equal to $\frac{360}{41}cm$ .
Hence, the correct answer is option (a).