Question

The sides of a triangle are 50 cm, 78 cm and 112 cm. The smallest altitude is

• A. 20 cm
  
• B. 30 cm
  
• C. 40 cm
  
• D. 50 cm

Answer 1

The correct option is B

30 cm

The sides of a triangle are 50 cm, 78 cm, and 16 cm. The altitude to the largest side is 112 cm
$\therefore s=\frac{a+b+c}{2}=\frac{50+78+112}{2}$
= $\frac{240}{2}=120$
$\therefore Area=\sqrt{s(s-a)(s-b)(s-c)}$
=$\sqrt{120(120-50)(120-78)(120-112)}$
= $\sqrt{120\times 70\times 42\times 8}$
= $\sqrt{10\times 3\times 2\times 10\times 7\times 7\times 3\times 2\times 2\times \times 2\times 2}$
= $10\times 7\times 3\times 2\times 2\times 2=1680c{m}^2$
$\therefore$ Shortest altitude = $\frac{Area\times 2}{longestbase}$

= $\frac{1680\times 2}{112}=30cm$