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Schach [20]
2 years ago
6

A triangle has three sides 35cm 54 cm and 61 cm find its area Also find the smallest of its altitude ​

Mathematics
2 answers:
Nitella [24]2 years ago
6 0

Answer:

Here you go.Hope this help!!

I am Lyosha [343]2 years ago
3 0

Answer:

Area of given triangle is 939.15cm² and smallest altitude is 30.8cm

<h3>Solution:</h3>

We are given three sides of a triangle, Let the sides be :

  • ( a ) = 35 cm

  • ( b ) = 54 cm

  • ( c ) = 61 cm

We can find the area of the triangle with its three sides using Heron's Formula

  • <u>Heron's </u><u>Formula</u>

Heron's formula was founded by hero of Alexandria, for finding the area of triangle in terms of the length of its sides. Heron's formula can be written as:

\sf{   \pmb { \longrightarrow \:  \sqrt{s(s - a)(s - b)(s - c)} }}

where ( s ) :

\sf \longrightarrow s = \dfrac{a + b + c}{2}

Therefore, for the given triangle first we will calculate ( s )

\begin {aligned}\quad & \quad \longmapsto  \sf  s =  \dfrac{a + b + c}{2}  \\  & \quad \longmapsto  \sf s =  \dfrac{35 + 54 + 61}{2}  \\ & \quad \longmapsto  \sf s =  \dfrac{150}{2}  \\ & \quad \longmapsto  \sf s  = 75cm \end{aligned}

Now, Area of triangle will be:

\begin{aligned}&:\implies \sf\quad \sf \:  A = \sqrt{s(s - a)(s - b)(s - c)} \\ &:\implies \sf\quad \sf \:  A = \sqrt{75(75 - 35)(75 - 54)(75 - 61)}   \\&:\implies \sf\quad \sf \:  A = \sqrt{75 \times 40 \times 21 \times 14}  \\ &:\implies \sf\quad \sf \:  A = \sqrt{5 \times 5 \times 3 \times 3 \times 2 \times 2 \times 7 \times 7 \times 2 \times 2 \times 5}  \\ &:\implies \sf\quad \sf \:  A =5 \times 3 \times 2 \times 7 \times 2 \sqrt{5}  \\ &:\implies \sf\quad \sf \:  A =420 \times 2.23 \\ &:\implies \sf\quad \sf \boxed{ \pmb{ \sf   A =939.15 {cm}^{2} }} \end{aligned}

Also, we have to find the smallest altitude, and the smallest altitude will be on the longest side. So,

\begin{aligned}&:\implies \sf\quad \sf \:  Area =939.15 \\ &:\implies \sf\quad \sf \:   \dfrac{1}{2}  \times b \times h =939.15 \\ &:\implies \sf\quad \sf \:   \dfrac{1}{2} \times 61 \times h  = 939.15 \\&:\implies \sf\quad \sf \:  h =939.15 \times  \dfrac{2}{61}   \\&:\implies \sf\quad \sf \:  h = \dfrac{1818.3}{61}  \\ &:\implies \sf\quad  \boxed{ \pmb{\sf \:  h =30.79 \: (approx)}} \end{aligned}

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none of these has "no solution"

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velikii [3]

Answer:

1. 1343 years

2. 9 hours

3. 39 years

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If N₀ = 100 then N = 85

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㏑(0.70) = -0.05775t

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14 = e^(xk)

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