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If the sides of a triangle are 17,25 and 28, then find the greatest length of the altitude.

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\frac{410}{17}

\frac{420}{17}

\frac{430}{17}

\frac{440}{17}

answer is B.

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Detailed Solution

Given sides or triangle = 17, 25, and 28.
Question Image

It is known that altitude is perpendicular to the side which is shortest.
Here, the area of the triangle

Δ = \frac{1}{2} AD \times BC

AD = \frac{2 Δ}{17}

Semi perimeter of triangle,

s = \frac{17 + 25 + 28}{2}

s = 35

Now by herons formula,

Δ^{2} = s (s - 17) (s - 25) (s - 28)

Δ^{2} = 35 (35 - 17) (35 - 25) (35 - 28)

Δ^{2} = 210^{2}

So,

Δ = 210

Now,

AD = \frac{2 \times 210}{17}

\Rightarrow AD = \frac{420}{17}

Correct option is 2.

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