In a triangle $A B C,$ the sides are of length $17, 25$ and $28$ units. Then the length of the largest altitude is

26.6

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27

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26

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\frac{420}{17}

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Solution

The correct option is D $\frac{420}{17}$
Image
Let $A B = 17, B C = 25, A C = 28$
$s = \frac{17 + 25 + 28}{2} = 35 \Rightarrow Δ = \sqrt{s (s - a) (s - b) (s - c)} = \sqrt{35 \times 18 \times 10 \times 7} = 210$

As the area is constant, the largest altitude is corresponding to the smallest side
$\Rightarrow \frac{1}{2} \times C F \times A B = 210 \Rightarrow C F = \frac{420}{17}$

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In a triangle ABC, the sides are of length 17, 25 and 28 units. Then the length of the largest altitude is

The correct option is D 42017Image Let AB=17, BC=25, AC=28 s=17+25+282=35⇒Δ=√s(s−a)(s−b)(s−c)=√35×18×10×7=210 As the area is constant, the largest altitude is corresponding to the smallest side ⇒12×CF×AB=210⇒CF=42017

In a triangle $A B C,$ the sides are of length $17, 25$ and $28$ units. Then the length of the largest altitude is