The sides of a triangle are 35 cm, 54 cm and 61 cm. The length of its longest altitude is Options: A . &nbsp 24 √ 5 c m B . &nbsp 48 √ 5 c m C . &nbsp 56 √ 5 c m D . &nbsp 108 c m

Answer: Option A : A Given, a = 35 cm ; b = 54 cm ; c = 61 cm . Semi-perimeter s = a + b + c 2 = 35 + 54 + 61 2 = 75 cm A ( △ ) = √ s ( s − a ) ( s − b ) ( s − c ) = √ 75 ( 75 − 35 ) ( 75 − 54 ) ( 75 − 61 ) = √ 75 × 40 × 21 × 14 A ( △ ) = 420 √ 5 cm 2 . Now, A ( △ ) = 1 2 × base × altitude The altitude corresponding to shortest base is longest. The shortest side = 35 cm. ∴ 420 √ 5 = 1 2 × 35 × altitude altitude = 2 × 420 √ 5 35 = 24 √ 5 cm .

The sides of a triangle are 35 cm, 54 cm and 61 cm. The length of its longest

Question

The sides of a triangle are 35 cm, 54 cm and 61 cm. The length of its longest altitude is

Options:

A .

24 \sqrt{5} c m

B .

48 \sqrt{5} c m

C .

56 \sqrt{5} c m

D .

108 c m

Answer: Option A
:
A

Given, $a = 35 cm; b = 54 cm; c = 61 cm .$
Semi-perimeter
$s = \frac{a + b + c}{2} = \frac{35 + 54 + 61}{2} = 75 cm$
$A (△) = \sqrt{s (s - a) (s - b) (s - c)} = \sqrt{75 (75 - 35) (75 - 54) (75 - 61)} = \sqrt{75 \times 40 \times 21 \times 14}$
$A (△) = 420 \sqrt{5} {cm}^{2} .$
Now, $A (△) = \frac{1}{2} \times base \times altitude$
The altitude corresponding to shortest base is longest.
The shortest side = 35 cm.
$∴ 420 \sqrt{5} = \frac{1}{2} \times 35 \times altitude altitude = \frac{2 \times 420 \sqrt{5}}{35} = 24 \sqrt{5} cm .$

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