Question

The side of a triangle are of length $20,21$ and $29$ units. The sum of the lengths of altitude will be

• A. $\frac{1609}{29}$ units
• B. $49$ units
• C. $\frac{1609}{21}$ units
• D. $40$ units

Answer 1

The correct option is A $\frac{1609}{29}$ units

$s=\frac{20+21+29}{2}=35$
$\Delta =\sqrt{35\times 15\times 14\times 6}$
$=\sqrt{5\times 7\times 3\times 5\times 7\times 2\times 2\times 3}$
$=5\times 7\times 3\times 2=210$
$210=\frac{1}{2}{A}_{1}20$ $210=\frac{1}{2}{A}_{2}29$
$A_1=21$ $A_2=\frac{420}{29}$
$210=\frac{1}{2}{A}_3\times 21$ $A_3=20$
$A_1+A_2+A_3=21+20+\frac{420}{29}=41+\frac{420}{29}$
$=\frac{1189+420}{29}=\frac{1609}{29}$